Nicola Scafetta: Jupiter’s orbital eccentricity may drive ~60yr and millennial climate cycles.

Posted: September 24, 2020 by tallbloke in solar system dynamics
Figure 3. The 60‐year eccentricity function (blue) of Jupiter (see Figure 2) against: (a) the HadCRUT global surface temperature record (Morice et al., 2012) detrended of its quadratic polynomial fit f(t) ¼ a(t − 1850)2 + b (cf. Scafetta, 2010, 2016) (correlation coefficient r^2 = 0:5, p < 0.01); (b) the 5‐year running average of the Indian summer monsoon rainfall from 1813 to 1998 (Agnihotri & Dutta, 2003) (correlation coefficient r^2 = 0:5, p < 0.01)


 Plain Language Summary 

The physical origin of the modulation of the cloud system and of many of the Earth’s climate oscillations from the decadal to the millennial timescales is still unclear, despite its importance in climate science. One of the most prominent oscillations has a period of about 60 years and is found in a number of geophysical records such as temperature reconstructions, aurora sights, Indian rainfalls, ocean climatic records, and in many others. These oscillations might emerge from the internal variability of the climate system, but increasing evidence also points toward a solar or astronomical origin.

Herein we speculate whether the oscillations of the orbits of the planetary system could modulate the interplanetary dust flux falling on the Earth, then modifying the cloud coverage. We find that the orbital eccentricity of Jupiter presents a strong 60‐year oscillation that is well correlated with several climatic records and with the 60‐year oscillation found in long meteorite fall records since the 7th century. Since meteorite falls are the most macroscopic aspect of infalling space dust, we conclude that the interplanetary dust should modulate the formation of the clouds and, thus, drive climate changes.

Scafetta, N., Milani, F., & Bianchini, A. (2020).
A 60‐year cycle in the Meteorite fall frequency suggests a possible
interplanetary dust forcing of the Earth’s climate driven by planetary
oscillations.

Geophysical Research Letters, 47, e2020GL089954.
https://agupubs.onlinelibrary.wiley.com/doi/10.1029/2020GL089954

Personal study copy here

Comments
  1. pochas94 says:

    Kudos to Scafetta for pushing the science forward undaunted. I nominate cosmic rays for additional consideration. They dry out the radiating zone, lower the emitting height, and cool both the emitting altitude and the surface.

  2. tallbloke says:

    Yes, Scafetta leaves the Svensmark theory on the table in the paper, and offers his Jupiter-directed interplanetary dust as an additional mechanism.

  3. oldbrew says:

    Re. auroras:

    A shared frequency set between the historical mid-latitude aurora records and the global surface temperature (2012)
    Nicola Scafetta

    The existence of a natural 60-year modulation of the global surface temperature induced by astronomical mechanisms, by alone, would imply that at least 60-70% of the warming observed since 1970 has been naturally induced. Moreover, the climate may stay approximately stable during the next decades because the 60-year cycle has entered in its cooling phase.

    Paper here.

  4. tallbloke says:

    “Thousands of tons of cosmic dust are estimated to reach the Earth’s surface every year,[3] with most grains having a mass between 10−16 kg (0.1 pg) and 10−4 kg (100 mg).[3] ”

    https://en.wikipedia.org/wiki/Cosmic_dust

  5. Recent generations of the world’s scientists seem to have absorbed an exaggerated dependence upon models founded upon suggestive graphs like the above. The first point to be made, always, is how good is the r-square (correllation coefficient, to we older scientists), which tells one directly how much of the variation in the supposed-dependent variable is due to variations in the supposed causal variable. The above graph fails — in this older-generation scientist’s view — by not giving that r-square value (“r 1/4 0:5, p<0.01" is non-informative, even meaningless to the directly-informative — and simple(!) — r-square value). The observed variation — the noise — in the temperature record is suspicious, and suggests the true r-square value of the comparison is small, and hence untrustworthy.

    Second, the temperature record is "detrended of its quadratic polynomial fit", which obviously (looking at the graph) means the global warming signal has been removed. So, in Scarfatta's model, there is no global warming, and his presumed cause of global warming, cosmic dust seeding of clouds, is disallowed.

    The supposed 60-year temperature cycle has been famously explained as multidecadal ocean temperature oscillations, but that is with the global warming signal included. And with that inclusion, is the only real motivation for even believing in the global temperature records of HADCRUT and others over the last century. But there is plenty of evidence of incompetent and/or criminal handling of the temperature records, by politically compromised and debased climate scientists. So in reality, the multidecadal ocean oscillations theory is itself suspect, and with it that 60-year supposed cycle in the global mean surface temperature.

    Sorry, but Scarfatta is playing with pretty stones on the beach, while ignorant of the great ocean of truth nearby. It is all a great waste of peoples' time and energy…par for the course today. In reality, there is no global warming, and thus no 60 year cycle in that global warming, and no control of the global mean temperature either by ocean oscillations or by cosmic dust variations.

    And that of course is what I, starting with my Venus/Earth temperatures comparison, brought out 10 years ago: There is no valid global climate science, and no competent climate scientists.

  6. oldbrew says:

    Just came across an article about Nobel prize winner CTR Wilson…includes a dust-related experiment.

    The cloud chamber and CTR Wilson’s legacy to atmospheric science

    Introduction and early life

    2011 is the centenary year of the short
    paper (Wilson, 1911) first describing the
    cloud chamber, the device for visualising
    high-energy charged particles which earned
    the Scottish physicist Charles Thomas Rees
    (‘CTR’) Wilson the 1927 Nobel Prize for physics.
    His many achievements in atmospheric
    science, some of which have current relevance,
    are briefly reviewed here.

    CTR Wilson’s lifetime of scientific research work
    was principally in atmospheric electricity at
    the Cavendish Laboratory, Cambridge; he
    was Reader in Electrical Meteorology from
    1918 and Jacksonian Professor from 1925 to
    1935. However, he is immortalised in physics
    for his invention of the cloud chamber,
    because of its great significance as an early
    visualisation tool for particles such as cosmic rays
    (Galison, 1997).
    . . .
    In summary, CTR Wilson’s visualisation
    techniques for particle physics concerned
    microscopic cloud processes, whereas his
    synthesis of atmospheric electricity unravelled
    invisible atmospheric properties on a
    global scale. Half a century after his death,
    it is a tribute to his painstaking reasoning
    and wonderful experimental ingenuity that
    both his principal scientific achievements
    still influence physics education and atmospheric
    electricity research.

    Click to access Harrison2011_CTRWilson_Weather.pdf

    Only a few pages but worth a read IMO.

  7. tallbloke says:

    If Harry had taken the time to read the linked full paper, he’d have found that …..

    Actually, I can’t be bothered to spoon feed info to people who know it all without reading it.

    Suffice to say the 60 year cycle with detrended T data is supplemented by the millennial cycle which accounts for the centennial ups and downs apparent in longer term T reconstructions..

  8. Paul Vaughan says:

    The following commentary has NOTHING to do with planets, orbits, or temperatures.
    It’s just the conventional paradigm.

    r^2 alone does NOT determine what statisticians call “statistical significance”.
    A low r^2 can be highly statistically significant.
    A high r^2 can fail to be statistically significant.

  9. Paul Vaughan says:

    Irretrievable: 2 comments vanished.

    Political commentary on the “60 year linear component” of climate variables — motivated not by climate so far as I can tell but maybe fear that western financial sanctions won’t have a valid military guarantor by 2030 or 2036 — ignores the more general context of nonlinear slip cycles.

    Alert readers may have noticed.

    “The number filter” readily accepts political rants (even if they are wild). Johnson’s, Macron’s, & Biden’s people will be happy to note: At worst IT’s a very strong participation turn-off and IT’s at least a cause of major “behavior modification” (ware IT may be inverted totalitarianism).

    I have some “mysterious” (only “thanks” to the west turn 2030 “backup energy” plan, but otherwise acutely precise) calculations I would share, but for sure they won’t pass the filter.

    Harry would no doubt be left with unanswered questions about the block of numbers and IT’s creative decode.

  10. tallbloke says:

    Paul, was this a wordpress issue? No comments in spam to be found, which is where they end up when the ‘vanish’ on submission.

  11. Paul Vaughan says:

    One was on the recent XR thread and one on the 2400 year thread. I must have an out-of-date e-mail address for you.

  12. ren says:

    Cosmic dust can be held in the upper stratosphere for a long time by jet currents. It should be concentrated in the equatorial belt, which may limit the heating of the tropical ocean.

  13. oldbrew says:

    Scafetta’s paper says:
    The periodogram shows that the eccentricity function is characterized by a prominent 922‐year oscillation. Jupiter’s eccentricity also shows another prominent 60‐year oscillation (apparently made of two cycles at about 57.1 and 60.9 years), which is slightly modulated by the 20‐year conjunction cycle with Saturn.

    The 922-year oscillation looks a lot like the lunar precession period at 1/7th of 6441 tropical years, or 104 apsidal cycles, as described at the Talkshop here:

    For every 104 apsidal cycles, all numbers except SM slip by -1 from being multiples of 104, i.e. a precession. So after 7*104 LAC all the other totals except SM are ‘reduced’ by 7 each.

    https://tallbloke.wordpress.com/2017/10/15/lunar-precession-update/
    [SM = synodic month]

    6441/7 = 920 + 1/7 tropical years.

    To explain it another way, the formula of 55 full moon cycles = 7 apsidal cycles = 62 tropical years (55+7=62) is almost true. But the more accurate formula requires multiplying by 104:
    55*104, -7 FMC = 7*104 apsidal = 62*104, -7 TY

    Hence 1/7th of that is one lunar precession period, or whatever you want to call it.
    7 apsidal cycles = 766 synodic months.
    – – –
    Re the 60.9 years, a suggestion would be that this is really the time of 360 degrees of rotation of the Jupiter-Saturn conjunction (~61.05 tropical years), another precession period.

  14. tallbloke says:

    OB. Good spot. Considering the relative masses of Jupiter and our moon, it seems likely that it’s Jupiter’s 922yr eccentricity cycle which has shaped our moon’s precession cycle. The interesting question is whether it actually our moon which is channelling the distribution of cosmic dust in the upper atmosphere, as well as Jupiter channeling the particulate masses in our direction from interplanetary space.

  15. oldbrew says:

    Eccentricity and apsides…

    ECCENTRICITY EXCITATION AND APSIDAL RESONANCE CAPTURE IN THE
    PLANETARY SYSTEM u ANDROMEDAE [2002]

    This paper lays the groundwork for understanding the
    origin of the large eccentricities and the apsidal alignment
    exhibited by the orbits of planets C and D in u And. Our
    main result is that the eccentricity of planet C and the
    locking of orbital apsides are both consequences of the slow
    growth of the eccentricity of planet D. The latter eccentricity,
    in turn, was driven by an external agent—plausibly a
    primordial circumstellar disk lying exterior to the orbit of
    planet D—that acted over timescales exceeding 104 yr. We
    play our scenario out and explain the mechanics of apsidal
    resonance capture in [section] 2.

    Paper: ‘https://iopscience.iop.org/article/10.1086/341617/pdf’

  16. oldbrew says:

    From another Scafetta paper:
    Multiscale Analysis of the Instantaneous Eccentricity Oscillations of the Planets of the Solar System from 13 000 BC to 17 000 AD

    The eccentricity function of the orbit of Jupiter presents large oscillations with periods of about 60 and 900- 960 years, mostly due to the interaction with Saturn. These oscillations also correspond to oscillations found in several geophysical records. The eccentricity functions of Uranus and Neptune are characterized by a large 4300-year oscillation. The eccentricity function of Pluto is characterized by a large nearly 20000-year modulation.

    https://link.springer.com/article/10.1134/S1063773719110094

    The U-N eccentricity of 4300 years is very similar to:
    (U-N * N) / (U-N – N) = 4270.11~ years

    No. of N is 1 greater than no. of U-N in this period, so the planetary orientation relative to the Sun repeats, being a multiple of 360 degrees of movement.

  17. oldbrew says:

    4270 * 13/12 = ~4626

  18. Paul Vaughan says:

    4270 is the aliquot sum for B (baby monster 4370). Way too much is piling up to share — supremely rich vein.

  19. Paul Vaughan says:

    =
    φ(n)
    Euler Totient
    1584
    λ(n)
    Carmichael Lambda
    396

    There are 1,584 positive integers (less than 4,370) that are coprime with 4,370.
    =
    https://metanumbers.com/4370

    You’ll find those numbers buried in 104-yielding levels (once the moderator frees the most recent filter victim on 2400).

    Fits in what they call simple sporadic 5-group or Mathieu group M11 with order 7920.

    It’s a monstrous insight avalanche. The filter is a big problem. It hates round-off bars and math.

    Weather by dark deception or dark ignorance (dark either way) the so-called “experts” savagely misdirected US.

    Beyond B there are 2 ways to construct 1/(U-N) with M. When I start to post the fifth-roots stuff the filter will always be full.

    More mundane but should be noted: Scafetta should be looking at 66 not 61. You can’t differentiate between them with 2 waves, but it’s clear when you go back further.

    “EU’V-E got a monde stir in Eur. Paris Sol.
    The walls are closing in again: O[r]well.”
    — Queens of the Stone Age

  20. Paul Vaughan says:

    UN Seek Cure IT Count Sol.integrity

    General Ramanujan points to sharp spikes in discrete-continuous relations.

    For example, consider level 54 “almost-integer” (math lingo — what they call it) spike on 5th-root scale:

    R(p) = ⌊(e^π*54^(1/5))^(1/p)⌉^p – e^π*54^(1/5)

    Pervasive round-off brackets of discrete-continuous relations needn’t freeze reader awareness in blank-stares of ignorance.

    Link to easy answers for an example. If you can round off a number, you’re in.

    R(3) = -71.0083263199169 = ⌊(e^π*54^(1/5))^(1/3)⌉^3 – e^π*54^(1/5) = 10^3 – e^π*54^(1/5)

    Note 2 embedded links in the expression.

    Bracket expressions conservatively for the calculator — otherwise the automation is prone to silently making incorrect organizational changes (downscale fractal of misled big tech, no. doubt).

    At the end of the first link change the example 3 to 5.
    Round off the result to 4. That gives new input 4^5 for the beginning of the second link.

    Repeat with 10 at the right end of the first link.
    The result rounds off too 2 so 2^10 goes.in to the far left of the second link. Find:

    R(5) = R(10) = -47.0083263199169

    59 = ( 71 + 47 ) / 2
    Recall M: 47*59*71.

    Check R(6) / 2 for a clue as to how we nearly won over U-N (wins even tie won).

    DCoy daze review:
    171.406964273337 = 50 / ( 1/(2*(13*11*7*5*3)^(1/4)) + 2/(2*(13*11*7*5*3)^(1/4)) + 1/(2*(7*5*3)^(1/4)) )
    47.0085558422417 = ( 2 / ( 1/(2*(13*11*7*5*3)^(1/4)) + 2/(2*(13*11*7*5*3)^(1/4)) + 1/(2*(7*5*3)^(1/4)) ) ) ^ 2

    TB’11 also remember from way back in the day: (φ√5)^4 = 25*φ^4.
    171.352549156242 = 25*1.61803398874989^4

    Monde Stir’11 each let US anyon?
    No. Eur. CRude type 0! symbols O[r]We’11 build a C[ENSO]Rship mess tory, in.deed.

    Binet and Lucas AImost redirect what curry US IT to 196883196560 = 323 as fall lows:
    25*√(φ^8+Φ^8) = 25*√47
    171.391365010026 = 25*(1.61803398874989^8+0.618033988749895^8)^(1/2)

    EUCRUS: Stand buy northern D-fence 4 clarification.

    Note 646 = 2 * 323

    Half before the US election — and 1/2 after. My way of proving I’m know partisan under threat of financial terror directed by well-off e/11 IT WHO’s winter CR(U-N)chess west turn common folks numb brrrs right down to street-level aware noose weather left or right.

    OB’11 sea fib’n’luc.in top line …sequel too:

    171.391380036748
    = 5^2*(ΦΦ/(1/34/2-1/76/2))^(1/2) = 5^2*(ΦΦ*152*68/(152-68))^(1/2)
    = 5^2*(ΦΦ/(1/646+1/152))^(1/2) = 5^2*(ΦΦ*646*152/(646+152))^(1/2)

    152: wise IT AImost X[R(p)]act weather we sea 2400 or knot?

    171.391365010026 = 5^2*(ROUND(4^2*(59)^(1/2),0)-ROUND(3^2*(71)^(1/2),0))^(1/2)
    = 5^2*(123-76)^(1/2) = 5^2*(47)^(1/2)

    Shh! OK?.in’ 47 Tops B

    Aliquot sum (sum of proper divisors) for 2*5*19*23 = 4370:
    4270 = 1+2+5+10+19+23+38+46+95+115+190+230+437+874+2185

    “Velvet Ears” 123 Flash mnemonIC Back.in Hazard Count: Tie Wan M$y$n D-Po11ace

    171.406545820013 = 5^2*(EXP(54^(1/5)*PI())-ROUND(EXP(54^(1/5)*PI())^(1/5),0)^5)^(1/2)
    164.791478352793 = 1 / ( 1 / 4270 + 1/25/(EXP(54^(1/5)*PI())-ROUND(EXP(54^(1/5)*PI())^(1/5),0)^5)^(1/2) )
    84.016966353458 = 1 / ( 1 / 4270 + 2/25/(EXP(54^(1/5)*PI())-ROUND(EXP(54^(1/5)*PI())^(1/5),0)^5)^(1/2) )
    55.6463431474333 = 1 / (2 / 4270 + 3/25/(EXP(54^(1/5)*PI())-ROUND(EXP(54^(1/5)*PI())^(1/5),0)^5)^(1/2) )

  21. Paul Vaughan says:

    Study with care: easy to derive system at IC ally (it’s only a few trivial steps).

    You’ll recognize not only 2^(1/J/2) & φ^4*2^(1+1/J/2) but exactly why they’re off and by exactly how much. IT’s Ramanujan’s world of AImost.in toujours.

    836.531021854751
    835.546575435631
    0.117820651543 = % “error” (not actu[s]ally un error)

    61.0241156298752
    61.0464822565173
    -0.036638682223 = % “error”

    “Why fall O? 2 “higher” grrOun-D? 11Ost as USware ayaM?” — C-elective Sol

  22. Paul Vaughan says:

    By part tie sun IC top knew trail’ski left M is tory…

    836.531021854751 = φ^4*2^(1+11.8626151546089/2)
    835.546575435627 = 1/(5/29.4474984673838-2/11.8626151546089)

    61.0241156298752 = 2^(11.8626151546089/2)
    61.0464822565173 = 1/(1/11.8626151546089-2/29.4474984673838)

    710003.731008934 = (836.531021854751)*(835.546575435627) / (836.531021854751 – 835.546575435627)
    418.019254422807 = (836.531021854751)*(835.546575435627) / (836.531021854751 + 835.546575435627)
    836.038508845615 = (836.531021854751)*(835.546575435627)/((836.531021854751+835.546575435627)/2)

    71.0003731008934 = 710003.731008934 / 10000
    209.009627211404 = 418.019254422807 / 2

    …distinct from 208.

  23. Paul Vaughan says:

    The Simplest Conventional View

    836.492470214859 = 2/(1/418-1/710000)
    835.508109310293 = 2/(1/418+1/710000)
    61.0213033256461 = Φ^4/(1/418-1/710000)
    11.8624821779712 = 2*LOG(Φ^4/(1/418-1/710000),2)
    29.4471611398395 = 5/((1/418+1/710000)/2+1/LOG(Φ^4/(1/418-1/710000),2))
    61.0458600788091 = 1/(1/2/5/LOG(Φ^4/(1/418-1/710000),2)-(1/418+1/710000)/5)
    19.8648166947302 = 1/(3/2/5/LOG(Φ^4/(1/418-1/710000),2)-(1/418+1/710000)/5/2)
    8.45605035913211 = 1/(7/2/5/LOG(Φ^4/(1/418-1/710000),2)+(1/418+1/710000)/5/2)

    -0.001125389948 = % error average across j, s, beat, & axial with stable bias — sign & magnitude both stable meaning correction’s simple

    Out of curiosity compare the biased 1/(J-S) with:
    19.8643454852672 = 4 / ( 1/(2*(13*11*7*5*3)^(1/4)) + 1/(2*(7*5*3)^(1/4)) )
    0.002372136869 = % error

    Remember that 836 is the smallest untouchable weird number (not making this shh!IT up — it’s number theory lingo).

  24. Paul Vaughan says:

    Hi! K.in.on Boris Bluff

    Too daze note U-N.Doors.cores perfect non.11.in.ear.IT.
    Hears the backs tory: dec.aids a go sum BRIteechairs D-sided school-D B e/11.in.ear.

    28 is perfect.
    proper divisors of 298: 1, 2, 149
    s(298) = 1 + 2 + 149 = 152

    25 + 298 = 323 ——————– / —————————–

    See.in’re:cure.sieve.a11yET?

    From where again did Ramanujan claim many of his sharpest insights came?

    Other perfect numbers include 2, 5, 52, 88, & 96.
    UNassum.in2400 review spill28IT /

    Does every 1 remember whale engine.in’s voice on the ducts of has heard? This ain’t no.ware born Gen.R’a11[y]each let US.

  25. Paul Vaughan says:

    Freedom is Not a Dr.ill

    Ballparks can B Bo ring but knot fin weighin’ best tune where EU can see the playbook writin’ on “the green monster” weather left and right:

    28 = (84)*(42) / (84 + 42) = (1^2+2^2+3^2+…+22^2+23^2+24^2)^(1/2) – 42 = 70 – 42

    58 = 73+(-15)^(1)
    298 = 73+(-15)^(2)

    R(5) = 224.991673680083
    836 / 44 ~= -R(6) / R(2) = s(77)

    Artist IC types can’t be reached with just logic.

    146 = 298 – s(298) = 298 – 152
    s(48) = s(146) = 152 / 2 = 76

    Jovian Giant C-elective:
    Make sure you understand 298 as the (red hot) “others hide” of 25 (chili peppers).
    IT AIn’t 11.in.ear: climb IT SAM pulls cont.in.EU.US exponential with discrete s.wit.ch.

    General Summary

    M B Lee.ch.in jsun: a11 won and the same.

  26. Paul Vaughan says:

    “Spam” in the C[ENSO]Rship

    Can mod phi niche fish file tour?

    104 ~= ⌊(e^√s(298+25)π)^(1/2)⌉^2 – e^√s(298+25
    104 ~= ⌊(e^√s(196883-196560)π)^(1/2)⌉^2 – e^√s(196883-196560

    mod 13

  27. Paul Vaughan says:

    Note the Tall spike above.

    s(323) = 37
    -103.999977946281 = ⌊(e^√37π)^(1/2)⌉^2 – e^√37π

    Scenic aside:
    s(67) = 1
    s(2*67) = 70 ——– Leech
    s(3*67) = 71 ——– M
    s(4*67) = 208
    s(5*67) = 73 ——– lowest prime congruent to 1 mod 24
    s(8*67) = 11*44 = 22^2

    The order of presentation is cryptic, but the ingredients are well-served.

    Review.

  28. Paul Vaughan says:

    Luck key mods comparatively fish pike threw D-baitless lines of clean discretion.

    s(3*43) = 47
    -743.999775171279 = ⌊(e^√43π)^(1/3)⌉^3 – e^√43π
    s(3*67) = 71
    -743.999816894531 = ⌊(e^√67π)^(1/3)⌉^3 – e^√67π

    mod 8

    59 = s((59 mod 24)*s(3*43)) = s(s(20)*s(3*(67-24))/2) = (47+71)/2

  29. Paul Vaughan says:

    IT’s possible no. thing get$11earned bye the right UN tell after USelection.

    s(3*19) = 23 = 71 mod 24 = 47 mod 24
    = 196883 – ( ⌊(e^√19*(2π))^(1/2)⌉^2 – ⌊(e^√19*(2π))⌉ ) / 2
    -743.777680155239 = ⌊(e^√19π)^(1/3)⌉^3 – e^√19π

    What’s left tune out ice?

    s(6) = 6 = 3#

    “There are only four all-harshad numbers: 1, 2, 4, and 6 (The number 12 is a harshad number in all bases except octal).”

  30. Paul Vaughan says:

    Link to List of Aliquot Sums

    11.in.ear think kings ware IT just AIn’t .

    2432 / 836 = s(152/2) / s(76/2)
    84 = ( 836 + s(836) ) / s(34)

    s(30) = 42
    s(5#) = 42 where # indicates primorial
    5# = 5*3*2 product of all primes lower than or equal to 5

    18.6 = 744 / s(44)
    Suggestions-43 distilled moon lightin’ the table.

    Maybe only God Nos.: how many ways can Conway & Norton B written?
    298 = s(104/4)^2 + s(5#)
    = 146 + 2*s(146) = 146 + s(298) = 146 + 104 + s(47^2)

    Monster US aliquot sum phine PR O-ducts:

    φ = 2*cos(s(71^2)/s(59^2)/s(47^2)*8*π)
    2.61803398874989 = (2*cos(s(71^2)/s(59^2)/s(47^2)*s(2*5)*π))^2

    “Freedom I hold dear:
    The autumn moon lights my weigh
    — 11ed 22plan “R(amble)ln”

    Perfect construct$yen a head: list of aliquot sums s(n)

    s(28) = 28

    IT AIn’t just No. 1. WHO nos.?

    s(5*31) = 37 = s(196883-196560)
    s(7*29) = 37
    s(13*23) = 37
    s(17*19) = 37 = s(298+25)

  31. Paul Vaughan says:

    Vague O-Port Tune IT AIn’t

    Typo: “Other perfect numbers include 2, 5, 52, 88, & 96.” — correct shh!UNtouchABLE

    For more insight: Compare perfect with Ore.

    Once you memorize some of the key aliquot sums a whole (nonlinear) framework crystallizes. Remember what Conway and Norton said.

  32. Paul Vaughan says:

    Primorials (including 210) are no mystery at this stage, but there remains opportunity to clarify that 208 & 209 are both distinct and compatible. Above I outlined the most naive conventional model.

    On the 2400 thread I posted a very sharp calculation that no doubt puzzled pretty much anyone who looked at it dismissively — if only because they haven’t yet started to think carefully about discrete-continuous relations and aggregation criteria more generally.

    This insight eluded us for too long:

    208 ~ = slip(slip(24.067904774739,19.8650360864628),19.8650360864628)/2

    Once I saw that I understood the natural role of B & M in solar system stability at a whole new level.

    I used to hesitate — sometimes for months compounding delays from previous months — to post slip cycle calculations because the filter hates them. 2 comments never appeared. Technology failures have forced complete change of how, what — and when — to express.

    208
    209
    distinct
    and compatible

    Similarly we’re now past 5256/”836″/2. If you didn’t notice this yet: check this calculation with the 2 pairs of “836” above (pair from most naive conventional model and pair from Seidelmann (1992)).

    I defined what I mean by slip(x,y) on the 2400 thread. This convention along with chopping comments (especially ones with round-off bars) into fragments seems to keep the filter from blocking.

  33. Paul Vaughan says:

    Wikipedia’s “Kepler Trigon” overview isn’t based on Seidelmann (1992), which gives this slip series:
    1 / 19.8650360864628 = +1J-1S
    1 / 61.0464822565173 = +1J-2S
    1 / 835.546575435627 = -2J+5S
    1 / 2669.94916589798 = -29J+72S
    1 / 13660.3670170363 = +85J-211S
    1 / 117417.893491061 = -454J+1127S
    1 / 290290.064137486 = -4171J+10354S

    On the 2400 thread I applied the generalized Bollinger method to derive slip cycles including 66 & 132.

    Slip cycles are key (in long-run central limit) because circulatory structures and materials are aliasing exponentials. Our education system was designed to brainwash people into becoming dumb linear thinkers (puppets on straightforward strings). Fool me once shame on you fool me twice shame on me sort of thing.

    As some of you realize I no longer concern myself with “climate debate” but the context gives opportunity for astute readers to consider that hierarchies like the one listed at the beginning of this comment fall apart by the 2nd level with loose aggregation criteria. It isn’t necessarily just nonlinear drift on a curve; it’s potentially discrete switch-flipping.

    The comparative study of sets of “almost integer” (math lingo) fits alerts us to the existence of a higher organizing principle pulling threads towards a central limit. With this awakening we can realize several strands approaching (but not reaching) limits. A valid unifying principle accommodates such bundles — i.e. different combinations of pieces giving very-nearly the same thing.

  34. Paul Vaughan says:

    GA11actIC Green Monster Challenge 9801.in the C[ENSO]Rship

    “…and IT’s whisper-D that soon if we a11 ca11 the tune” — 11ed 22plan

    4th comm.in simplification & clarification suggests diagnostic comparison
    of Seidelmann’s (1992)
    2432 = 19 * 2^7
    with
    2436 = 29 * 84
    2436 = 58 * 42 ——— General√(Φ-φ)Ramanujan’s Guide to the 9*11*IX*XI = N.in.8.O.won

  35. Paul Vaughan says:

    Simplifying

    1/Φ = φ = ((1+5^(1/2))/2)

    11.8626176385713 = 1/((1/104+1/(1+(1/104+1/298)*φ^2/12))/20+(1/104+1/298)*φ^2)
    29.4474891061275 = 1/(1/104+1/298)/φ^2

    19.8650473122013 = 20/(1/104+1/(1+(1/104+1/298)*φ^2/12))
    8.45614623658195 = 1/((1/104+1/(1+(1/104+1/298)*φ^2/12))/20+2*(1/104+1/298)*φ^2)

    16.9122924731639 = 2/((1/104+1/(1+(1/104+1/298)*φ^2/12))/20+2*(1/104+1/298)*φ^2)

    104 = 1/((1/104+1/(1+(1/104+1/298)*φ^2/12))-1/(1+(1/104+1/298)*φ^2/12))

  36. Paul Vaughan says:

    Clarifying

    208 & 209 are thus distinct.

    61.0466285002156 = 1/((1/104+1/(1+(1/104+1/298)*φ^2/12))/20-(1/104+1/298)*φ^2)
    835.484250291355 = 1/((1/104+1/298)*φ^2-2*((1/104+1/(1+(1/104+1/298)*φ^2/12))/20-(1/104+1/298)*φ^2))

    61.0241681640784 = 2^(1/((1/104+1/(1+(1/104+1/298)*φ^2/12))/20+(1/104+1/298)*φ^2)/2)
    836.531742004323 = φ^4*2^(1+1/((1/104+1/(1+(1/104+1/298)*φ^2/12))/20+(1/104+1/298)*φ^2)/2)

    836.0076680293 = 2/(((1/104+1/298)*φ^2-2*((1/104+1/(1+(1/104+1/298)*φ^2/12))/20-(1/104+1/298)*φ^2))+1/φ^4/2^(1+1/((1/104+1/(1+(1/104+1/298)*φ^2/12))/20+(1/104+1/298)*φ^2)/2))

    209.001917007325 = 1/(((1/104+1/298)*φ^2-2*((1/104+1/(1+(1/104+1/298)*φ^2/12))/20-(1/104+1/298)*φ^2))+1/φ^4/2^(1+1/((1/104+1/(1+(1/104+1/298)*φ^2/12))/20+(1/104+1/298)*φ^2)/2))/2

    Familiar structures “almost” matching this frame afford rich review with developed hindsight.

  37. Paul Vaughan says:

    Level 7 Keys Concisely

    104 ~= slip(slip(24.067904774739,19.8650360864628),19.8650360864628)/4
    24.067904774739 = ⌊(e^√7π)^(1/p)⌉^p – e^√7π for p=2,3,4,6,12
    Twice the reciprocal of that for p = 1.

    This should help organize simple perspective because that (which keeps higher slips locked better) may look confusing.

    This approximation slips a little at higher-level, but being so simple (parsimony’s why I’m illustrating) it aids comparative study:
    24.067917505387 = 24+2*(1/104+1/298)*φ^2
    24th harmonic (divide period by 24) :
    1.00282989605779 = 1+(1/104+1/298)*φ^2/12
    0.993252365610063 = (104)*(1.00282989605779) / (104 + 1.00282989605779)
    20th subharmonic (multiply period by 20) ~= 1/(J-S)
    Level 7 has some really special properties. There are 2 other ways to estimate 24.0679 (outlined previously). Like 104 it’s a master key.

    For the set j, s, js beat, js axial, u, n, un beat, un axial arising from the last calculations above:
    0.000017200983 = average absolute %error

    The form extending j & s to u & n (based on perfect numbers 28 & 496) was outlined here.

    aside: flashback with hindsight
    5256.08395035091 = φ^4*2^(1+1/((1/104+1/(1+(1/104+1/298)*φ^2/12))/20+(1/104+1/298)*φ^2)/2)*2*π

  38. Paul Vaughan says:

    Tie.in de Rop 4 16 Figures D-light

    Review
    lunar draconic & anomalistic
    5.99685290323073 = beat(0.0754402464065708,0.0745030006844627)
    with anomalistic year
    1814.31362251033 = slip(5.99685290323073,1.00002638193018)

    aliquot sequence for B
    s(4370) = 4270
    s(4270) = 4658
    s(4658) = 2794
    s(2794) = 1814
    s(1814) = 910
    s(910) = 1106
    s(1106) = 814
    s(814) = 554
    s(554) = 280
    s(280) = 440
    ————————– see below
    s(440) = 640
    s(640) = 890
    s(890) = 730
    s(730) = 602
    s(602) = 454
    s(454) = 230
    s(230) = 202
    s(202) = 104
    s(104) = 106
    s(106) = 56
    s(56) = 64
    s(64) = 63
    s(63) = 41
    s(41) = 1
    s(1) = 0

    note:
    1106/2 = 553, 1106/2+554 = 1107, 4428 = 4*1107, & 744 = 640+104 demystify powers below the perfect level 28 4th power split of M & Leech.

    Also note 836 sequence overlap with B:
    s(836) = 844
    ————————–
    s(844) = 640
    s(640) = 890
    s(890) = 730
    s(730) = 602
    s(602) = 454
    s(454) = 230
    s(230) = 202
    s(202) = 104
    s(104) = 106
    s(106) = 56
    s(56) = 64
    s(64) = 63
    s(63) = 41
    s(41) = 1
    s(1) = 0

    Just imagine how much more we have to learn about number theory before we even have the basic aggregation criteria foundations needed to even BEGIN sensible climate stability exploration. Today’s politics are hopelessly intractable (ever since lockdowns became the west turn weapon of homeland financial terror). With monstrous leadership failures ALL across the west — weather left or right — George Polya’s advice is prescient and clear:

    Solve a simpler problem.

    flurry of miscellaneous puzzle pieces:

    s(652) = 49+149+298 = 496 = s(496) ——- perfect
    652/4 = 163 points to Ramanujan again.

    s(25) = 6 —– links to review: s(298) = 152
    s(6) = 6 ——- This is how you recognize perfect numbers.

    s(8128) = 8128 ——- perfect
    4428 = 8128 – 3700 = 4370 + 58 ———- 37 & 58 link precisely to 104

    UNtouchABull Branch

    de Rop fits into the aliquot sequence for the baby monster and 836 is a dead-end branch — what it means to be untouchableoff that.

    Arithmetic Dynamics
    =
    Arithmetic dynamics is a field that amalgamates two areas of mathematics, dynamical systems and number theory. […] A fundamental goal is to describe arithmetic properties in terms of underlying geometric structures.
    =

    For an illustrative example, take a look at how simple this recipe is: […] “Mathematics may not be ready for such problems.” […] “is an extraordinarily difficult problem, completely out of reach of present day mathematics.”

  39. Paul Vaughan says:

    The Green Monster

    104.000044599386
    104.000034332275 = ⌊(e^√58π)^(1/p)⌉^p – e^√58π for p=2,4
    0.000009872218 = % error

    24.0679036739334
    24.067904774739 = ⌊(e^√7π)^(1/p)⌉^p – e^√7π for p=2,3,4,6,12
    -0.000004573749 = % error

    “don’t speak” — no. doubt

  40. Paul Vaughan says:

    SIM pull iff √(Φ-φ) U-N express sun:

    y = 24.067904774739 = ⌊(e^√7π)^(1/p)⌉^p – e^√7π for p=2,3,4,6,12
    x = 104.000034332275 = ⌊(e^√58π)^(1/p)⌉^p – e^√58π for p=2,4

    11.8626151191159=1/((6/y+(1/4)/x)/5+(1/x+1/298)*φ^2)
    11.8626151546089=1/J
    -0.000000299201=%error

    29.4474963123504=1/(1/x+1/298)/φ^2
    29.4474984673838=1/S
    -0.000007318222=%error

    19.8650369676306=5/(6/y+(1/4)/x)
    19.8650360864628=1/(J-S)
    0.000004435773=%error

    8.45614555057611=1/((6/y+(1/4)/x)/5+2*(1/x+1/298)*φ^2)
    8.4561457463176=1/(J+S)
    -0.000002314784=%error

    61.0464998394537=1/((6/y+(1/4)/x)/5-(1/x+1/298)*φ^2)
    61.0464822565173=1/(J-2S)
    0.000028802538=%error

  41. Paul Vaughan says:

    Perfect Fit

    19=s(77)
    37=s(323)

    28=(s(77)+s(323))/2=(19+37)/2

    19=28-9=s(77)
    37=28+9=s(323)

    496=298+49+149
    323=298+25=196883-196560

    196585=196883-298=196560+25
    s(298)=152=d(2,1/2,25)
    s(152)=148=s(d(2,1/2,25))

    248=149-49+148=496/2
    149=72+77
    77=152+248-323=248-171—- halve perfect mnEMonICm$y$nUSwanOvert(U-N)
    72=149-152-248+323=(323-152)-(248-149)=171-99 ————- m$y$nUSgreatsKEY
    171=323-152

    208 = 8128 – 7920 ———– M11
    s(8128) = 8128

    95 = 37 + 58 = 70 + 25 = √(73500/15) + 25 = s(5#)+s(28)+25 = 42+28+25
    s(95) = 25
    s(25) = 6
    s(6) = 6

    hitchhiker miss sol any us prime more real goal act IC guides?

    42 = (84)*(28) / (84 – 28) = (84)*(28)/((84+28)/2) — beat = harmonic mean
    = 70 – 28 = (210/2)*(70) / (210/2 + 70) = (7*5*3*2)*(70/2) / (7*5*3*2 – 70/2)

  42. Paul Vaughan says:

    Reviewing History with 2020 Hindsight

    Nature weaves convergent strands in perfect bundles. This line of communication was previously arrested by filter “misbehavior”.

    =
    the sum of its reciprocals forms a series of unit fractions that converges to 1 more rapidly than any other series of unit fractions with the same number of terms.

    It is possible to interpret the Sylvester sequence as the result of a greedy algorithm for Egyptian fractions, that at each step chooses the smallest possible denominator that makes the partial sum of the series be less than one. Alternatively, the terms of the sequence after the first can be viewed as the denominators of the odd greedy expansion of 1/2.

    […] Curtiss (1922) describes an application of the closest approximations to one by k-term sums of unit fractions, in lower-bounding the number of divisors of any perfect number, and Miller (1919) uses the same property to upper bound the size of certain groups.
    =

    Sylvester’s sequence 2, 3, 7, 43, 1807, … can be viewed as generated by an infinite greedy expansion […] Truncating this sequence to k terms and forming the corresponding Egyptian fraction, e.g. (for k = 4)”

    1 = 1/2+1/3+1/7+1/43+1/2/3/7/43

    s(n) = aliquot sum of n
    s(4370) = 4270

    100 = 4370 – 4270 = 4370 – s(4370)

    70 = √(73500/15) — recall 73500 = 2*36750 key weighted harmonic mean from Laskar

    4200 = s(4370) – 70 = 4270 – 70
    4300 = 4370 – 70

    42 = (s(4370) – √(73500/15)) / (4370-s(4370))
    43 = (4370 – √(73500/15)) / (4370-s(4370))

    The first four of these numbers are one less than the corresponding numbers in Sylvester’s sequence, but then the two sequences diverge.

    1806 = 2*3*7*43 = 42*43
    = ( 4370*s(4370) – √(73500/15)*(4370+s(4370)) + 73500/15 ) / (4370-s(4370))^2

    271 = 496 – 15^2 = ⌊√73500⌉
    s(496) = 496

    489426 = 1806 * 271 ———– count.sol luck D-own sat U-N 4 XRview from fan-wise “monster” seats

  43. Paul Vaughan says:

    Rome and knew join general
    ….each let us group monde stir us sly stable parameter√(Φ-φ)shh!UN:

    d(p,r,k) = ⌊(e^π*k^r)^(1/p)⌉^p – e^π*k^r

    For example
    323.013383387384 = d(2,1/3,59) = ⌊(e^π*59^(1/3))^(1/2)⌉^2 – e^π*59^(1/3)
    and
    0.0133833873842377 = d(1,1/3,59) = ⌊(e^π*59^(1/3))^(1/1)⌉^1 – e^π*59^(1/3)
    such that
    323 = d(2,1/3,59) – d(1,1/3,59)

    Incidentally note us:
    323 = (59-12)*59*(59+12) – 196560 = 196883 – 196560 = 298 + 25

    Standby 4 perfect primorial leech tie.in.

  44. Paul Vaughan says:

    Monster US$green.in 1984=4*s(496)

    before rolling far enough away to allow Stuart to score.”

    =
    496 is most notable for being a perfect number, and one of the earliest numbers to be recognized as such. […] Also related to its being a perfect number, 496 is a harmonic divisor number, since the number of proper divisors of 496 divided by the sum of the reciprocals of its divisors, 1, 2, 4, 8, 16, 31, 62, 124, 248 and 496, (the harmonic mean), yields an integer, 5 in this case.
    […]
    E8 has real dimension 496.

    The number 496 is a very important number in superstring theory. In 1984 […] Green and […] Schwarz realized that one of the necessary conditions for a superstring theory to make sense is that the dimension of the gauge group of type I string theory must be 496. The group is therefore SO(32). Their discovery started the first superstring revolution. It was realized in 1985 that the heterotic string can admit another possible gauge group, namely E8 x E8.
    =
    https://en.wikipedia.org/wiki/496_(number)

    =
    There is a unique complex Lie algebra of type E8, corresponding to a complex group of complex dimension 248. The complex Lie group E8 of complex dimension 248 can be considered as a simple real Lie group of real dimension 496. This is simply connected, has maximal compact subgroup the compact form […] of E8, and has an outer automorphism group of order 2 generated by complex conjugation.
    =
    https://en.wikipedia.org/wiki/E8_(mathematics)#Real_and_complex_forms

    248 = s(298)+96 = s(s(298))+(104+96)/2 = (2*72)+104 = (298-2*77)+104
    = (72+77)+49+(104+96)/4 = (149+49)+(149-49)/2 = s(496)/2

    5=77-72=harmonic mean of proper divisors of 496

    Perfect Moonshine

    “ayaM the jigsaw, man” — “more hue moon than WHO man” whyIT22oN?bei

    3 = √(absolute mean deviation from perfect) won over U-N see cure IT count.sol tip:

    s(28)=(s(323)+s(77))/2=(323+77)-s(298)
    ware p=2, r=1/2, & k=25 for U-N dr. read m$y$n US d(p,r,k)

  45. Paul Vaughan says:

    37 FEAT HI.in Left Fields φ niche-D “a11y” caught sum green monster:

    φ = 2*cos(s(71^2)/s(59^2)/s(47^2)*8*π) where s(n) = aliquot sum for n
    φ = 2*cos(s((59+12)^2)/s(59^2)/s((59-12)^2)*8*π)

    196883 = 71*59*47 = (59+12)*59*(59-12)

    Naive weather left or right, IT’s AI “just” a Que. when so dense.
    Don’t Mayan D-proof: “just” go ON with Bei=lief.in 2+2=5 no. when 4*s(496) = 1984 is AB US sov. “green” morse un out field doors vic[tory] Dev.O[s] lie low.

    “[…] a popular target for right-handed hitters […] the Green Monster was not painted green until 1947 […] Yellow numbers are used to represent in-inning scores […] left-field distortion is offset by the odd shape and generous size of right field […] The placement of the ladder is noteworthy given the fact that it is in fair territory […] a high fly that ricocheted first off the ladder, and then the head of outfielder [___]”

    323 = 71*59*47 – 196560
    s(323) = 37

    “The Coke Bottles on the left light tower were a target for power-hitters […]”

    104 ~= ⌊(e^√37π)^(1/2)⌉^2 – e^√37π

  46. Paul Vaughan says:

    Bei. U-N D!Superst[r]ing

    Reverse-eng.in.ear.ring oldschool “luminary” assembly with 2020 hindsight, we find attached a perfect leech:

    490000 = 15*70^2 / asin(φ/2) * 2π

    Lol! What curry US IT didn’t even (“don’t speak” no. doubt) tell us. Tech no. CR at IC govern ants?ware.in monde stir US $ sly “green”.

    Sylvester’s Sequence (“more hue moon than WHO man” = whyITzoombei) goes as far as possible past the perfect leach, account.in 4 minor deviations from superstring perfectshhU-N.

    1 / 164.791265692394 = (2/1806-1/3)J+(2/1806+1)S-2/1806/15/70^2*asin(φ/2)/π+1/3/s(4370)
    -0.000030309658 = % error

    1 / 84.0168557977672 = 2*((2/1806-1/3)J+(2/1806+1)S-2/1806/15/70^2*asin(φ/2)/π)-1/3/s(4370)
    0.000011754317 = % error

    peer√(Φ-φ)ed dub e/11 ln acts pawn ants yell

    1 / 164.791269677405 = (2/1806-1/3)J+(2/1806+1)S-2/1806/1806/271+1/3/s(4370)
    -0.000027891441 = % error

    1 / 84.0168578694536 = 2*((2/1806-1/3)J+(2/1806+1)S-2/1806/1806/271)-1/3/s(4370)
    0.000014220116 = % error

    4270 = s(4370) gives the Bay’s first-step B-estimate for U & N as does 4428 = 4370+58 = 8128-3700 for J & S.

    “on my blue PR(INT) IT’SIM F(U-N) IC; spin that record B — Lady G “just” D-ants

    Anchoring thus contrasts oldschool “green” leach attached to mainstream mindset verse USSylvester’s double exponential greed sequence.

    The 1st pair of U & N estimates above is based on J & S estimates.
    The 2nd pair of U & N estimates above is based on J & S from Seidelmann (1992).

    FormAI pol!shh, IT’s missUN no linksnow:

    This is a diagnostic strategy precisely exposing an exponential in clear conflict with a double exponential. A hair-splitting subtle difference in this case is period-doubling.

  47. Paul Vaughan says:

    Scene IC Route to Secure IT Count Sol

    4370 = 2*5*19*23

    “√(Φ-φ) lost my keys
    √(Φ-φ) can’t C strait anymore
    What’s go.in.on.on the floor?
    Spin that record Babe” — Lady Gaga Just D-ants

    Φ(2) = 1
    Φ(5) = 4
    Φ(19) = 18
    Φ(23) = 22

    Euler’s totient function is a multiplicative function, meaning that if two numbers m and n are relatively prime, then φ(mn) = φ(m)φ(n). This function gives the order of the multiplicative group of integers modulo n […]”

    Φ(3) = 2
    Φ(37) = 36

    s(4370) = 4270
    “Then one foggy Christmas eve…” — Rue Dolph.in Red Knows Reign D-a√(Φ-φ)r
    5106 = 4270 + 836 = 2*3*23*37

  48. Paul Vaughan says:

    1584 Miscellaneous Tips from General Ramanujan’s Travel Guide

    Today’s trip begins with ET (Euler’s Totient).

    count: 40
    lowest:
    Φ(1679) = 1584
    the next 2:
    Φ(1691) = 1584
    Φ(1985) = 1584
    double:
    Φ(3382) = 1584
    Φ(3970) = 1584
    highest:
    Φ(6210) = 1584

    Having boldly left a little mystery, we rightly move on not icing hawk key:

    Φ(4370) = 1584
    Φ(5106) = 1584

    It is possible for an infinite set of integers to be pairwise coprime. Notable examples include the set of all prime numbers, the set of elements in Sylvester’s sequence, and the set of all Fermat numbers.

    Φ(4270+836) = 1584
    Φ(s(4370)+836) = 1584

    Recall that 836 is untouchable, meaning it’s not an s(n) for any n — i.e. there is NO number with aliquot sum = 836.

    A comparative view of Jovian order is shaping up:
    global exponential constraint with no local adjustment
    • global order subtly-retuned (prime-D-iffew prefer) 4 fit with double-exponential local greed

    104.000034332275 = ⌊(e^√58π)^(1/2)⌉^2 – e^√58π = 156816^2 – e^√58π
    104.001742574386 = ⌊(e^√22π)^(1/2)⌉^2 – e^√22π = 1584^2 – e^√22π
    104.000034332275 = ⌊(e^√58π)^(1/4)⌉^4 – e^√58π = 396^4 – e^√58π

    396 = √156816
    1584 / 396 = 4 = 396 / 99
    156816 / 1584 = 99
    √(58-22) = √36 = 6 = s(6) = s(25)

    d(p,r,k)
    d = difference
    p = power
    r = root
    k = count or level in docks
    152 = d(2,1/2,25) – d(1,1/2,25)
    152 = (d(3,1/2,25) – d(1,1/2,25)-8744)/2 = (9048-8744)/2 = 304/2

    review:
    levels k=10,13,18,22 converge on 104
    quadruple those
    levels k=40,52,72,88 converge on 8744

    4370 = (84*104+8744)/4
    We explore these things in stages. Each stage brings more clarity.

    Landscape ecology hierarchy theory and years of field experience using detailed botany taxonomy keys helped base a mindset to orient in rich territory.

    Conventional linear “thinking” about what’s “best” doesn’t help see (sort and classify) how every thing naturally fits in God’s creative design.

    s((84*104+8744)/4) = 4270
    Φ(s((84*104+8744)/4)+836) = 1584

    Discrete-continuous relations are full of nested, recursive, hierarchical structure. Western math education is fatally deficient. What do you think bad elite (to be neither confused nor conflated with good elite, if such a thing exists) can do with ignorance of trade secrets?

    Given a chance to speak (remember what Conway & Norton said) to the pope about luck D-own west turn spread of severe financial terror past just “rogue” nations to “count less” millions of homeland western citizens, what might I say?

    “and if(IT)s reel then √(Φ-φ) don’t want tune O
    √(Φ-φ) no. ya reel good”
    — “don’t speak” no. doubt

    139560 = d(3,1/2,58) – d(1,1/2,58)
    70^2 = (196560-139560-8744+744)/10 = 73500/15 = 2*36750/15
    review: 744 = d(1,1/2,k) – d(3,1/2,k) for k = 43-24, 43, 43+24

    836 is the next-lowest weird number after 70. 836 is untouchable. 70 is not. “70p is weird for all primes p ≥ 149

    A ware no. word scan do, weave numbers.

  49. Paul Vaughan says:

    CorrectS[hh!]UN[8]O[R]well.in.sight

    The tech “know”so-called CR at IC “eXpeRts” never advise-D of period-dub e ln SIM MET try.

    “the bloop ill O-pens euRise
    is theRe a bet te[a]R weigh?
    a knew Religion pRess scRibed
    to those without the faith
    is IT 2 late 2 go back?
    is IT 2 late 2 go?
    theRe’s No. won heaR
    and peep e/11 eveRy waRe:
    yeaR .oN euR O-won” — Queens of the Stone Age

    Won step back.
    73500 = 3*(196883-4270/5-29)/8
    2 steps 4 word.

    489425.981694385 = 1 / ( -300J+904S-301U-301N )
    489426 = 2*3*7*43*271 = 1806*271
    -0.000003740221 = % error

    where ABCD estimate JSUN using Ramanujan’s Perfect Superstring:
    245061.761049564 = 1 / ( -300A+904B-301C-301D )
    245061.75 = (10*196560-4270-836)/8
    0.000004508890 = % error

    Recall: s(4370) = 4270

    First-order model’s base-D on the baby monster.
    Leech precisely IDs the point of symmetry that’s well-corrected by Sylvester’s double-exponential greed.

    We’ll take the seen IC route away from luck D-own weather Joe Trump or Don Biden.
    1 = 1/2 + 1/2 be 4+after USelection: know matter what curry US IT obstructs, free dem. from eur. increasingly savage financial terror schemes directed not only at “rogue” nations abroad but “count less” millions of homeland western citizens left unable to pay rent and buy food since they were CANCELED in a west turn tournament left by incompetent organizers with no consolation round.

    PC’s plan’s UNstable. Withdrawn BE support can only be restored under more favorable circumstances should they stably develop to sustain trust — including: no more Boris $ under mind “free” doom for a monarchy planning a reign of widespread homeland financial torment with no consolation.

  50. Paul Vaughan says:

    Ramanujan’s the man who knew in.ph.in.IT’s perfect superstring:

    22.1392314983836 = 1/(3V-5E+2J)
    22.1392315068494 = (φ^22+1/11)^(e/11+1/22)
    11 = d(2,1/2,6) – d(1,1/2,6) = 59mod24 = Φ(71mod24)/2 = Φ(47mod24)/2 = Φ(23)/2
    6 = s(6) = s(25) = Φ(7) = Φ(9) = Φ(14) = Φ(18) = Φ(Φ(19))
    6 = Φ(s(2*Φ(Φ(22)))) = Φ(s(s(s(s(22))))) = Φ(s(s(Φ(22))))
    6 = Φ(s(2*Φ(Φ(Φ(23))))) = Φ(s(s(s(s(Φ(23)))))) = Φ(s(s(Φ(Φ(23)))))
    Φ(23)
    83.9908695437061 = Φ( d(4,1/2,6) – d(1,1/2,6) )/2 – d(1,1/2,6)
    83.9908202014941 = 1/(J^2-S^2)/2
    104 = d(1,1/2,s(d(4,1/2,6)-d(1,1/2,6)))-d(2,1/2,s(d(4,1/2,6)-d(1,1/2,6)))

    Note We/11:
    lim s→∞
    (φ^(2s)+1/s)^(e/s+1/(2s))
    = φ(φφ)^e

  51. Paul Vaughan says:

    Can’s e/11-D: mmmIC Dawn a11-D’s Perfect Hire Arch IC AI Notice

    225 = d(1,1/2,9) – d(3,1/2,9) = 15^2
    71 = d(1,1/2,9) – d(2,1/2,9)
    Φ(71) = 70
    70^2 = 1^2+2^2+3^2+…+22^2+23^2+24^2

    73500 = ( d(1,1/2,9) – d(3,1/2,9) )^(1/2)*Φ( d(1,1/2,9) – d(2,1/2,9) )^2
    36750 = ( d(1,1/2,9) – d(3,1/2,9) )^(1/2)*Φ( d(1,1/2,9) – d(2,1/2,9) )^2/2

    “√(Φ-φ)awe sum things√(Φ-φ)thought.in√(Φ-φ)verse awe
    EU’ve gotta whole.in U.in.N√(Φ-φ)verse awe”
    — Queens of the Stone Age “Monster’s.in the Paris Sol”

    18 = d(2,1/5,Φ(Φ(163))) – d(1,1/5,Φ(Φ(163))) = Φ(Φ(Φ(163))) = Φ(19)
    22 = Φ(Φ( d(1,1/5,Φ(Φ(163))) – d(5,1/5,Φ(Φ(163))) )) = Φ(23)

    47 = d(1,1/5,Φ(Φ(163))) – d(5,1/5,Φ(Φ(163)))
    59 = 2*d(1,1/5,Φ(Φ(163))) – d(3,1/5,Φ(Φ(163))) – d(5,1/5,Φ(Φ(163)))
    71 = d(1,1/5,Φ(Φ(163))) – d(3,1/5,Φ(Φ(163))) = s(s(Φ(163)))

    225 = d(4,1/5,Φ(Φ(163))) – d(1,1/5,Φ(Φ(163)))
    342 = d(1,1/5,Φ(Φ(163))) – d(6,1/5,Φ(Φ(163))) = 2*171

    73500 = ( d(4,1/5,Φ(Φ(163))) – d(1,1/5,Φ(Φ(163))) )^2 * Φ( d(1,1/5,Φ(Φ(163))) – d(3,1/5,Φ(Φ(163))) )^2
    36750 = ( d(4,1/5,Φ(Φ(163))) – d(1,1/5,Φ(Φ(163))) )^2 * Φ( d(1,1/5,Φ(Φ(163))) – d(3,1/5,Φ(Φ(163))) )^2/2

    Φ(Φ( 2*d(1,1/5,Φ(Φ(163))) – d(3,1/5,Φ(Φ(163))) – d(5,1/5,Φ(Φ(163))) )) =
    28 = s(28) = Φ(58) = Φ(Φ(59))
    6 = s(6) = Φ(18) = Φ(Φ(54)) = Φ(Φ(Φ(162))) = Φ(Φ(Φ(Φ(163)))) = Φ(Φ(19))

  52. Paul Vaughan says:

    IC a few obviOus typos: IT’s no. thing per sun AI on the C[ENSO]Rship.

  53. Paul Vaughan says:

    In Julian years:

    1.00001743371442 = 1/E

    1.00001743390371 = (1-(1/240)^1)^(0/1)/(1-(1/240)^2)^(2/2)/(1-(1/240)^3)^(3/3)/(1-(1/240)^4)^(2/4)/(1-(1/240)^5)^(5/5)/(1-(1/240)^6)^(1/6)

    *365.25 gives days:

    365.256367733331 = 1/E
    365.256367664193
    0.000000018929 = % error
    That’s 0.597351554461056 seconds per century.

  54. Paul Vaughan says:

    1.61803398874989 = φ
    0.618033988749895 = Φ

    1 = φ – Φ
    Φ = 1 / φ

    2.71828182845905 = e

    2.71828182845905 = (1-Φ^1)^(0/1)/(1-Φ^2)^(2/2)/(1-Φ^3)^(3/3)/(1-Φ^4)^(2/4)/(1-Φ^5)^(5/5)/(1-Φ^6)^(1/6)/(1-Φ^7)^(7/7)/(1-Φ^8)^(4/8)/(1-Φ^9)^(6/9)/(1-Φ^10)^(3/10)/(1-Φ^11)^(11/11)/(1-Φ^12)^(4/12)/(1-Φ^13)^(13/13)/(1-Φ^14)^(5/14)/(1-Φ^15)^(7/15)/(1-Φ^16)^(8/16)/(1-Φ^17)^(17/17)/(1-Φ^18)^(6/18)/(1-Φ^19)^(19/19)/(1-Φ^20)^(8/20)/(1-Φ^21)^(11/21)/(1-Φ^22)^(9/22)/(1-Φ^23)^(23/23)/(1-Φ^24)^(8/24)/(1-Φ^25)^(20/25)/(1-Φ^26)^(11/26)/(1-Φ^27)^(18/27)/(1-Φ^28)^(12/28)/(1-Φ^29)^(29/29)/(1-Φ^30)^(9/30)/(1-Φ^31)^(31/31)/(1-Φ^32)^(16/32)/(1-Φ^33)^(19/33)/(1-Φ^34)^(15/34)/(1-Φ^35)^(23/35)/(1-Φ^36)^(12/36)/(1-Φ^37)^(37/37)/(1-Φ^38)^(17/38)/(1-Φ^39)^(23/39)/(1-Φ^40)^(16/40)/(1-Φ^41)^(41/41)/(1-Φ^42)^(13/42)/(1-Φ^43)^(43/43)/(1-Φ^44)^(20/44)/(1-Φ^45)^(24/45)/(1-Φ^46)^(21/46)/(1-Φ^47)^(47/47)/(1-Φ^48)^(16/48)/(1-Φ^49)^(42/49)/(1-Φ^50)^(20/50)/(1-Φ^51)^(31/51)/(1-Φ^52)^(24/52)/(1-Φ^53)^(53/53)/(1-Φ^54)^(18/54)/(1-Φ^55)^(39/55)/(1-Φ^56)^(24/56)/(1-Φ^57)^(35/57)/(1-Φ^58)^(27/58)/(1-Φ^59)^(59/59)/(1-Φ^60)^(16/60)/(1-Φ^61)^(61/61)/(1-Φ^62)^(29/62)/(1-Φ^63)^(36/63)/(1-Φ^64)^(32/64)/(1-Φ^65)^(47/65)/(1-Φ^66)^(21/66)/(1-Φ^67)^(67/67)/(1-Φ^68)^(32/68)/(1-Φ^69)^(43/69)/(1-Φ^70)^(25/70)/(1-Φ^71)^(71/71)/(1-Φ^72)^(24/72)/(1-Φ^73)^(73/73)/(1-Φ^74)^(35/74)/(1-Φ^75)^(40/75)

  55. Paul Vaughan says:

    Moderators: Goal done approximation caught in the filter natural e starts on the left, extending infinitely far right to phinally reach strict mathematical proof.

  56. Paul Vaughan says:

    Perfect He’er

    M the orrery: 496 = s(496) = 2/3*744
    He’er.ET.IC string theory comm. pact iff IC k? shh! UN.

    Knock, knock.
    WHO’s there?
    J.in.very.ant.
    Jane VI rant WHO??
    j-invariant 2pi/3 is anyon nome?

    194 = average(225,163)
    172 = 194-22 = 9+163
    171 = 9+ET(163)

  57. tallbloke says:

    Woah! There’s been a lot of work going on in here.

  58. Paul Vaughan says:

    Rog, the tide was flooding as I paddled across the narrows.
    A long, black helicopter flew over very low, commanding attention.
    After it passed, something falling on what appeared to be a small parachute caught my eye.
    I watched the gentle landing on the water’s surface.
    I altered course to inspect.
    It was not a parachute but a pink heart-shaped helium balloon, with a draping note including the word “fun” and a black pentagon with points numbered 1 to 5 in white.
    Δ(√(73500/15)) = Δ(70) = 25 = 5^2

  59. Paul Vaughan says:

    Notation Alert

    I’m refining notation — going back to R() for Ramanujan and d() is the difference of a pair of R()s that gives a positive integer. Power p=1 (the fraction) is always a member of the differencing pair.

    I’ll illustrate with an example:

    R(1,1/5,54) = -0.00832631991693233 = ⌊(e^π*54^(1/5))^(1/1)⌉^1 – e^π*54^(1/5)

    R(3,1/5,54) = -71.0083263199169 = ⌊(e^π*54^(1/5))^(1/3)⌉^3 – e^π*54^(1/5)
    R(5,1/5,54) = -47.0083263199169 = ⌊(e^π*54^(1/5))^(1/5)⌉^5 – e^π*54^(1/5)

    71 = d(3,1/5,54) = R(1,1/5,54) – R(3,1/5,54)
    47 = d(5,1/5,54) = R(1,1/5,54) – R(5,1/5,54)

    59 = average(d(3,1/5,54),d(5,1/5,54))

    The convention change makes notes more terse.

  60. Paul Vaughan says:

    Juncture 162

    With UN a poll lag(ET)IC mnemonic PRloom.inairai$e We beg.in IT’s j(U-N)k shh!invaryant.

    At a juncture deciding weather a new day dawns, recalling Polya’s prescient advice to solve a simpler problem, let’s simply consider:

    1. inverse totients (upscale branches).
    2. totient sequences (downscale stem shared by upscale branches).

    What cross-diss up plan eerie curry O’s IT?

    There’s a cureOus lack of conventional mainstream reference to totient sequences. IT’s the “tie UN” USelection 4 “tie = won” if the method of loci (mind palace memory method) works well enough.

    Wikipedia links to the concept aliquot sequence. The concept is used to define perfect numbers. Recall that n is perfect if n = s(n). For example, a chain we’ve noted:

    s(652) = 496 = s(496) = 49+149+298 = (2/3)*744

    652 = 4*163 gives a path to 496, meaning 496 is touchable (what they call it). Once the stem reaches 496, it ends. Perfect numbers are aliquot sequence termini.

    Notably 28 is untouchable. If you start an aliquot sequence at 28 it stays there (i.e. 28 is perfect), but since 28’s untouchable you can’t use any aliquot sequence from anywhere else to get to 28.

    The analogies for aliquot’s touchable and untouchable in ET (Euler Totient) context are totient and nontotient. We’re building a thesaurus to help us navigate cross-disciplinary communication.

    Having appreciated the fundamental utility of aliquot sequences as a sorting and classification guide, I quite naturally, instinctively, and enthusiastically catalogued ET sequences and began looking for analogies. When I write notes I’m calling the ET sequences totient sequences. As you’ll see they’re a moonlight key to monstrous sorting and classification.

    There’s UN other type O seek wins I’m explore ring 4 EU tell IT as a fun dem. meant AI sort & class iff IC case sun guide, D-note-D by the small greek letter D-e/11-TA, but 4 now let’s just note key junction 162:

    Φ(163) = 162 —– 163 is prime, nontotient, and (TAKE NOTE) the largest of the 9 Heegner numbers
    Φ(243) = 162 = Φ(3^5) ———– 243 is nontotient
    Φ(326) = 162 = Φ(2*163) ———– 326 is nontotient
    Φ(486) = 162 = Φ(2*3^5) —– 486 = 2*243 is totient (points up for example to 729 = 3^3^2)
    ^ upscale ^
    ————–
    v downscale v
    Φ(163) = 162
    Φ(162) = 54
    Φ(54) = 18
    Φ(18) = 6
    Φ(6) = 2
    Φ(2) = 1
    Φ(1) = 1

    Discrete functions I developed (generalizing a now infinitely less “mysterious” Ramanujan “trick”) IDed almost-integers as spikes that sort by modulus. Now I’m tracing spike origins to orient in conventional paradigms.

    Context: Here we’re off in a corner focused on e and pi (special values assigned to 2 variables), but stay aware that the method generalizes several-fold.

    Note the 54 in the downscale totient chain above. By now some readers may be thinking: “There are so many numbers. How can I keep track of them all? How can so many be important?” Suggestion: As with specialized members playing different roles on a team, big-picture fans note well where and how pieces fit the whole. We can see de Vries in discrete context without which de Vries mnemonICally wood knot B de Vries.

    Let’s note 1 more branch to 54:

    Φ(81) = 54 = Φ(162/2)
    Φ(54) = 18
    Φ(18) = 6
    Φ(6) = 2

    Note that the branches above 54 from 162 & 81 (half of 162) share a common stem from 54 down.

    I write left neither 4 USelection rig nor well divided audience no. win next comm. on C[ENSO]Rship (weather easy or not) will hinge on “Junk Sure” Φ(163).

    God bless you all — weather left, right, Chinese, Russian, North Korean, Persian, American, OR CZECH (U. Charles!) — as with General Ramanujan We D-tour threw monstrous moonshine on an otherwise hazard US trail.

    Here I leave 1 mysterious note:
    Φ(4374) = 1458

  61. Paul Vaughan says:

    Won Course Braiding Streams from Left and Right

    This comment will be left deliberately terse to rightly invite nonpartisan appreciation of nature.

    M rep review:
    196883 = 47*59*71 = (59-12)*(59)*(59+12)

    Φ(118) = 58 —————————– 118 is nontotient (means no branches upward from there)
    Φ(58) = 28 = s(28) = Φ(59) = Φ(29) ———– 29 & 59 nontotient
    Φ(28) = 12
    Φ(12) = 4
    Φ(4) = 2
    Φ(2) = 1
    Φ(1) = 1

    Recall 28 = s(28) means 28 is “perfect” (in number theory lingo).

    We’ve looked at totient sequences passing through 54 & 58. Here’s the intermediate chain passing through 56 = average(58,54) = 2*28 = 2*s(28) :

    Φ(236) = 116 = Φ(177) = Φ(354) — 177, 236, 354 nontotient (meaning capped — i.e…)
    Φ(116) = 56 = Φ(87) = Φ(174) — 87, 174 nontotient (no branches up from stem below go higher)
    Φ(56) = 24
    Φ(24) = 8
    Φ(8) = 4
    Φ(4) = 2
    Φ(2) = 1
    Φ(1) = 1

    Summary

    163 – 59 = 104 IDs del pezzo surface structure.
    A del Pezzo surf ace circuit starts and ends with 59:

    Φ(59) = 58
    d(2,1/2,58)=d(4,1/2,58)=104
    163=59+104
    Φ(163) = 162
    Φ(162) = 54
    59=average(d(3,1/5,54),d(5,1/5,54))=average(71,47)

    The perfect connection:
    Φ(58) = 28 = s(28) = average(58,54)/2

  62. Paul Vaughan says:

    ZZ$ox: Sharp Dressed Mayan$Top

    Note well Heegner M-ET del PeZZo surf ace on the C[ENSO]Rship:

    E6 (degree 3): 28 = 27+(6-5) = 27+(4-3) = 27+1 (line)
    E7 (degree 2): 58 = 56+(7-5) = 56+(4-2) = 56+2 (bitangent)
    E8 (degree 1): 243 = 240+(8-5) = 240+(4-1) = 240+3 (tritangent)

    the mob comm.s.CR.awe.11.in/out — Queens of the Stone Age

    My[UN]key orientsuper string he’er[ET(IC)]: Table 1 p.19.

    1728 = 2*744+240
    8128 = 7920+208 = s(8128) —- 8128 is perfect; 7920 is M11 order

    Monde stir seats high above left Field’shhall O[We/11]play in boss tune: my M?(y/n)-dis up hear on UNother wave, cove.ReD.in he’er” — Queens of the Stone Age “Monsters.in the Paris Sol.”

  63. Paul Vaughan says:

    Ally Caught Up in Bushy Taled Top O Log IC AI Totient Trees

    So far weave no.comm.on.tory.on the 3rd root, which first sprouts 59 trees in a maze sun 4 est con text:

    59 = average(47,71)
    196883 = 47*59*71 = (59-12)*(59)*(59+12) ————– M rep
    196560 ———————————————————— Leech rep

    71 = s(s(162)) = s(s(Φ(163)))

    323 = d(2,1/3,59) = 196883-196560 = 298 + Δ(70)
    s(323) = 37
    d(2,1/2,37) = 104

    493 = d(3,1/3,59)
    s(493) = 47

    At canopy level won squirrel isn’t lost in academic publication floods algebra storm.

    Φ(59) = 58
    d(2,1/2,58) = d(4,1/2,58) = 104

    Φ(58) = 28 ————- remember 28 = s(28) means 28’s perfect
    d(4,1/2,28) = 196585 ——- perfect number 28 gives a perfect split
    298 = 196883-196585 = 323 – Δ(70)
    25 = 196585-196560 = Δ(70) = Φ(70) – μ(70) = 323 – 298

    de vries first order quantum hale effect jovian circuit anyon?
    7920+208=8128 ——– 7920 is M11 order; 8128 is perfect
    8128-3700=4428 —— 37 is sharp 104-yielding level
    4428-58=4370 ———- 58 is sharp 104-yielding level; 4428 defines JS relations to first order
    4370=2*5*19*23 ——– s(4370) = 4270 ties UN to JS (first order); 4370 is B rep
    7920=5*Φ(5)*Φ(19)*Φ(23) —— yet another bilateral-pentagonal 5/2 key ET IDs

    Chased by owls while looking for food and shelter just watch an amazed squirrel run through the circuit tree.

  64. Paul Vaughan says:

    Route 5 Steps to Fifth Root.in Sight

    N8O worry O[r] left.in on dawns golf righter keys:

    “V(ic)tory i$ Mayan — Putt, tee $myth “The War e ^ O[r] ”

    Polya: “Solve a simpler problem.”

    66, 71, 76 —– 2*76 = 152 = 128+24
    54, 59, 64 —– 2*64 = 128 = average(152,104) = 152-24 = 104+24
    42, 47, 52 —– 2*52 = 104 = 128-24

    Phew! C[eNSO]Rship nearly had US divided and conquered ITself! be 4 share ring nonpartisan finds comm. on! ground.

    That was the look forward to a simple ending.
    Now let’s back track.

    54 is the big fifth-root spike.
    47 & 71 are 2 of the noteworthy products.

    Their average 59 is the big third-root spike.
    Let’s simply explore fifth-root connect sun with square-root.

    59-5 = 54 = s(42)
    59+5 = 64

    71+5 = 76
    2*76 = 152

    47+5 = 52
    2*52 = 104

    (71+5)+(47+5) = 128 = 2*(59+5)
    152+104 = 256 = 4*64

    47-5 = 42
    256+42 = 298 ————– JS first order (review)

    A Maze UN: D-air Rat “Miss Fear” IC Look Do[w]n

    Something te/11s US this isn’t what climb ET “super” comm. pew dare sore luck kin fear.

    fall O mystery O jung11chi11ed…

    Caught “ally”: wood knot no. by curry US IT what’s UN known?

    s(42) = 54 ———- 42,
    s(54) = 66 ———- 54, &
    s(66) = 78 ———- 66 on the same branch share ring aliquot sea stem
    s(78) = 90
    s(90) = 144
    s(144) = 259
    s(259) = 45
    s(45) = 33
    s(33) = 15
    s(15) = 9
    s(9) = 4
    s(4) = 3
    s(3) = 1
    s(1) = 0

    We a11 no. IT‘s knot right 2 look left.in 2 blew collar find $UN ants air red quest yen.

    Putt turn wreck cog niche$yen.in phi sox red: “green” monde stir B(est) tune f(UN) way.

    “WHO’s the hunter. Whose the game?”

    s(5#) = 42 ——————— antswore: BUNT! (left field shh! all O)

    “I feel the beat call Eur. name
    I haled EU close.in victory
    I don’t want too tame Eur. animal style
    EU won’t be caged.in the call of the will.ed” — Patty Smyth “The Warrior”

    Red door well note won mystery 66 left unadduressed 4 coverage UN[he!ll]O-there day with $O asin(O[r]We’ll).

  65. Paul Vaughan says:

    Hale 240

    2, 12, & 240 are both highly and sparsely totient.

    1/H = 1/(3V-5E+2J) = 22.1392315068494 = (φ^22+1/11)^(e/11+1/22)
    Solving for 1/V:
    1/V = 1/((1/3)H+(5/3)E-(2/3)J) ———————— 1/J
    0.615197263582188
    0.615197263396975 = 1/V (Seidelmann 1992)
    0.000000030106 = % error

    lim s→∞
    (φ^(2s)+1/s)^(e/s+1/(2s))
    = φ(φφ)^e

  66. Paul Vaughan says:

    e-Luke Shh(Own) Guard Do[w]ns Perfect Links !Sol Look! Shone

    27, 81, & 243 are perfect totient numbers:

    ΣΦ(54) = ΣΦ(27) = 27 = 18+6+2+1
    ΣΦ(162) = ΣΦ(81) = 81 = 54+18+6+2+1
    ΣΦ(163) = ΣΦ(243) = 243 = 162+54+18+6+2+1

    https://oeis.org/A082897

    Φ(163) = 162 = Φ(243)
    Φ(162) = 54 = Φ(81)
    Φ(54) = 18 = Φ(27)
    Φ(18) = 6
    Φ(6) = 2
    Φ(2) = 1

    163-104=59
    Φ(59) = 58
    Φ(58) = 28 = Φ(29)
    Φ(28) = 12
    Φ(12) = 4
    Φ(4) = 2
    Φ(2) = 1

    Hale O Monster USlink B(ET) we’en 2 stems:
    104 = d(2,1/2,s(196883-196560)) = d(2,1/2,58) = d(4,1/2,58)

    Just little sure pries: All Lie !Caught! Hide U-N Truth

    “Sometimes the faster IT gets, the less EU need tune OTHseasurfacetemperatch[ET]O[r]well

  67. Paul Vaughan says:

    B(est) Tune Perfect Green CO2shh!UN Monster Level 7 in f(U-N weigh)

    φ(8128) = 4096
    φ(4096) = 2048
    φ(2048) = 1024
    φ(1024) = 512
    φ(512) = 256 = φ(496)
    φ(256) = 128
    φ(128) = 64
    φ(64) = 32
    φ(32) = 16 = φ(28)
    φ(16) = 8
    φ(8) = 4
    φ(4) = 2
    φ(2) = 1
    φ(1) = 0

    R(k,1/2,7) = 24.067904774739 = ⌊(e^√7π)^(1/k)⌉^k – e^√7π for k=2,3,4,6,12
    Calculating ⌊(e^√7π)^(1/k)⌉^k :
    k=2: 64^2 = 4096 = φ(8128)
    k=3: 16^3 = 4096 = φ(8128)
    k=4: 8^4 = 4096 = φ(8128)
    k=6: 4^6 = 4096 = φ(8128)
    k=12: 2^12 = 4096 = φ(8128)
    Notice:
    R(k,1/2,7) = 24.067904774739 = φ(8128) – e^√7π for k=2,3,4,6,12

    Again recall 28=s(28), 496=s(496), & 8128=s(8128) are perfect numbers.

    Best tune crash-course threw number theory seas solar system order UN dare green monster in fun ways left field.

    “Mathematics is the queen of the sciences—and number theory is the queen of mathematics.” — Gauss

    Left field creatures: Nature’s beauties there to sea (surf ace stem peer ashore) if Eur. will.int(el)look past the C[e^(NSO)]Rship. (IT AImost divided and conquered US.)

  68. Paul Vaughan says:

    Room O[r]s of short circuit in.wash.un.tune

    Mayan D-Palace readers remember SUM thing CO2shh!UN reversing the tale end of bollinger’s symbol IC 1952 pentagon vis IT. From “ok” LA home a rumor doesn’t eve UN have to be true or well 4 students 2 remember? The tale ends with a baby.

    You may 4 get 2 individuals, but not how they add up.in education AI controversy.

    Φ(69) = 44
    φ(69) = 25

    Political junkies love D-bait.in rumors weather true or not.

    Definition of cototient:
    Φ(n)+φ(n)=n

    Bollinger’s paper about a 44 year cycle was published in 1952 when the Pentagon’s focus was the Korean War. Let’s look at the pieces.

    The totient of 19 is 18.
    52 is half of 104 and written backwards 5^2 = 25.
    Half of 44 is 22 and that is the totient of 23.

    4370 = 19*5*2*23

    Script-writers, Bollinger’s theme song is Lady Gaga’s “Just D-ants”.
    Ally caught 4370 is 4270, pointing straight to U-N.

    Even worse, IT’s reverse land.in the river times galaxy:

    Take the 9 out of 1952. You have 152. Use the 9 with half of 22 (as did Ramanujan) to key Figure$16 tames 9*11=IX*XI 4×4 times just Rome UN no. morals.

    Of course Ramanujan knew 1584 is the totient of 4370 and every hitchhiker nos. 42+104+152=298 phinally ties US to jovian Hale weather through Junction 162 from “Ramanujan’s constant” to totient 58 via 59 or scaled perfectly threw ally caught 37=s((59-12)*59*(59+12)-196560) in 11 each lettuce.

    Really, is there any point in B-labor.in UNcloaked narrative any further at this point? That’s enough creative writing room or wells.in wash.un.tune. AIron court.in MSM D-baits key word door.in monster US KOre[]on wa[B] peek r&B baby [r]oomer fall O-win pan tug gone 69 CO2$UN symbol IC a11y !caught!

    Nons[ENS]e^Poll[O]see Puttin’ on monster US “green” links? ITees the treat e We sign-D.

  69. Paul Vaughan says:

    Five

    Constant new tettitory is a costly source of cross-disciplinary delight.

    I’ve decided number theory should be a staple of math education for all fields. It wasn’t a part of any program I ever studied, so I’m learning as I explore.

    Here I leave another note knowing specialists should be able to move much faster with general orientation.

    φ(70) = 46
    φ(46) = 24 = Φ(70) = φ(φ(70))

    70’s totient Φ(70) equals its double-cototient φ(φ(70)). That wasn’t covered in my ecology courses, but I now know its relevant to Earth’s stability.

    √Δ(70) = √(Φ(70)-μ(70)) = √(φ(φ(70))-μ(70)) = φ(25) = 5
    Δ(70) = Φ(70)-μ(70) = φ(φ(70))-μ(70) = φ(25)^2 = 5^2 = 25

    70 = √(73500/15)

    5 = √Δ(√(73500/15)) = √(Φ(√(73500/15))-μ(√(73500/15))) = √(φ(φ(√(73500/15)))-μ(√(73500/15)))
    = φ(Δ(√(73500/15))) = φ(Φ(√(73500/15))-μ(√(73500/15))) = φ(φ(φ(√(73500/15)))-μ(√(73500/15)))

    Alert readers: Watch for typos.

    Would naive hobbyists looking at Titius-Bode “Law” focus on how modular forms fit together stably? With 2020 hindsight our observations suggest no at step 1.

  70. Paul Vaughan says:

    Review and Consolidate

    This note gives technical clarification where there’s considerable risk of misunderstanding (below) after further elucidating a “perfect totient” structure introduced above.

    Notation: Φ() for ET and φ() for cototient.
    A subset of perfect totient stems alternates or zippers perfectly with cototient stems.

    Φ(729) = 486 = 2*φ(729)
    Φ(486) = 162 = 2*φ(243)
    Φ(162) = 54 = 2*φ(81)
    Φ(54) = 18 = 2*φ(27)
    Φ(18) = 6 = 2*φ(9)
    Φ(6) = 2 = 2*φ(3)
    Φ(2) = 1
    Φ(1) = 1

    φ(729) = 243 = Φ(729)/2
    φ(243) = 81 = Φ(486)/2
    φ(81) = 27 = Φ(162)/2
    φ(27) = 9 = Φ(54)/2
    φ(9) = 3 = Φ(18)/2
    φ(3) = 1 = Φ(6)/2
    φ(1) = 0

    There’s an analogy with part of this note on Heegner numbers:
    =
    The Heegner numbers greater than 3 can also be found using the Kronecker symbol, as follows: A number k [greater than] 3 is a Heeger number if and only if s = Sum_{j = 1..k} j * (j|k) is prime, which happens to be negative, where (x|y) is the Kronecker symbol. Also note for these results s = -k. But if s = -k is used as the selection condition (instead of primality), then the cubes of {7, 11, 19, 43, 67, 163} are also selected, followed by these same numbers to 9th power (and presumably followed by the 27th or 81st power). – Richard R. Forberg, Jul 18 2016
    =
    https://oeis.org/A003173

    The backbone is powers of 3. Complimentary stems are thus half or double opposing zipper teeth (or 3/2 or 2/3 looking the other way).

    What I’m outlining here is equivalent to what’s illuminated in MaKay & He (2015).

    Above I noted 27+1=28, 56+2=58, 240+3=243. 243 is on the zipper, anchored to the largest Heegner number 163 via 162. 58 & 28 are not on the zipper. Some readers may get confused because 54 & 27 are on the zipper.

    To avoid confusion: separate the two 27s conceptually by pathway.

    One is on the zipper and the other is derived from a peripheral structure that ties 163 to 59 through 104 (detailed in a few earlier comments). Go down the totient chain from 59 to 58 to 28 to get 58-2(bitangent)=56 and 28-1(line)=27.

    27=27 no doubt, but keep the pathways sorted when parameterizing topological course. Looking at the wrong stem leads to an incorrectly parameterized model.

    Go up the zipper from 162 to 243 to get 243-3(tritangent)=240. In the other direction that chain goes down through 54 & 27 — and that path should NOT be confused with the path through 59, 58, and 28 on another stem (a separate structure adjoined by a 104 bridge).

    Crystallized awareness of links and routes should (not trying to be naive about how 5 politics equals 2+2 in hard D-Orwellian times) eliminate the risk of misunderstanding.

    String.in the links back so far covered in he’er:
    perfect totient
    59 stem
    162 stem — tied up with top Heegner 163 & ally caught 4 further perfection s(652) lead.in he’er ET IC to 496
    ● Most won’t be ready for top-level Heegner perspective …with type O’s left.in to folly tempt UN fact-chuckers dare UN 2 fix major typos USing later “notation alert” (find.in maybe stew arts O-port-tune-IT 4 OBsurveysUN….)

    The term “totient” was coined by Sylvester. How fitting that his greedy double-exponential sequence broke the perfect solar system superstring symmetry with period doubling symmetry (more on U-N in the Jovian reference frame soon….including a 0% error).

  71. Paul Vaughan says:

    D-visor Some Luck D-own Less UN

    Things you might never stop to realize:
    √Δ(√(73500/15)) = √Δ(70) = √25 = 5
    σ(σ(σ(σ(Δ(√(73500/15)))))) = 104
    σ(σ(σ(σ(σ(Δ(√(73500/15))))))) = 210 = 7#
    σ(σ(σ(σ(σ(σ(Δ(√(73500/15)))))))) = 24^2 = 576
    σ(σ(σ(104/2))) = 260 = σ(171)
    σ(104/4) = 42
    σ(260/4) = 84 = σ(44)
    σ(66) = σ(70) = 12^2 = 144
    σ(σ(836/4)) = σ(240) = 744
    σ(4270/2/5) = σ(s(4370)/2/5) = 496
    Weather UNeven nor not even Joe Biden nos. 2+2 climate politics are about √Δ(√(73500/15)).

  72. Paul Vaughan says:

    MacK Sum U-N PRai$e^(Year Comp. Aim)

    1/11/500/2
    11500
    111.5
    11/2 = 5.5
    1 / 11500 ~= 111.5(U-N)-5.5(J+S)

    Thus:
    171.406220509617 = 111.5/(5.5*(J+S)+1/11500)
    171.406220601552 = 1/(U-N)
    -0.000000053635 = % error

    1 / 489426 = 1/2/3/7/43/271 ~= -300J+904S-301U-301N
    Recall 489426 =2*3*7*43*271
    1204 = 2*(301+301) = 2*2*7*43
    = 1806-301-301 = 2*3*7*43 – 2*7*43 = 300+904

    29.3625733662892 = 1/((J+S-1/2/3/7/43/271)/1204+S)
    29.3625733662893 = 4/(J+S+U+N) = jovian harmonic mean
    0.000000000000 = % error

    Sylvester seek wins!

    55.6462717641964 = 1/(4*((J+S-1/2/3/7/43/271)/1204+S)-J-S)
    55.6462717641972 = 1/(U-N)
    -0.000000000001 = % error

    84.0168459111161 = 2/(4*((J+S-1/2/3/7/43/271)/1204+S)-J-S+(5.5*(J+S)+1/11500)/111.5)
    84.016845922161 = 1/U
    -0.000000013146 = % error

    164.791315682562 = 2/(4*((J+S-1/2/3/7/43/271)/1204+S)-J-S-(5.5*(J+S)+1/11500)/111.5)
    164.791315640078 = 1/N
    0.000000025781 = % error

    With complements, join $11 each turn.in a round at:
    271 = 1000-729

  73. Paul Vaughan says:

    59 and 1728 j-invariant

    59’s at the core of more than i$ geneR(ally known) to war e ^ Or[well]s in Mayan Victory.
    Simple beginnings: at r=1/3, 59’s the level of the big spike.

    M rep’s anchored productively ± 12:
    196883=47*59*71=(59-12)*(59)*(59+12)=59*(59^2-12^2)
    12 = Φ(Φ(Φ(59))) = φ(φ(φ(Φ(59)))) and thus:
    196883=59*(59^2-Φ(Φ(Φ(59)))^2)=59*(59^2-φ(φ(φ(Φ(59))))^2)

    This one is fantastic.
    j-invariant hookup swings simply ± 5:

    27=1728/(59+5)
    32=1728/(59-5)
    59=27+32

    54=2*1728/(59+5) —— recall 54 points directly to 47 & 71 at r=1/5
    64=2*1728/(59-5)
    59=(54+64)/2

    1728=27*64
    1728=32*54

    48=AVERAGE(32,64)
    36=HARMEAN(27,54)
    1728=AVERAGE(32,64)*HARMEAN(27,54)

    Compassionate Lefties: Do you imagine 1728=AVERAGE(32,64)*HARMEAN(27,54) as a response if you probe thoughts on climate? Sensible private thoughts are thus silent: “don’t speak” no. doubt. GCD(left,right)=1 signals no common factor in climate dialogue. It doesn’t mean you have to viciously hate and savagely push terrorizing financial lockdown into homelessness.

    Φ(59) = 58
    Δ(58) = 27
    s(58) = 32
    59 = Δ(Φ(59)) + s(Φ(59))

    No. brainer under present circumstances: The New Zealand National Party should NOT have campaigned to lower taxes. A prequisite is a generous campaign for superior math education — a noble imperative for civilizations facing challenges which can not be solved with politics.

    George Polya’s advice: “Solve a simpler problem.” Done:

    1728 = 2*Δ(Φ(59))*s(Φ(59))

  74. Paul Vaughan says:

    O-Bay AI lie CAUGHT

    Aliquot.in moonlight sights tree-top baby monde stirrin’ minimal faithful representation.

    Hunted by U-No. they’re predatorial owl, the cute test right fly yen squirrel $scrambled further up the tree…

    4270=2*2135=s(4370) ——– 4370 is B rep
    s(2135) = 841

    Knew wall along far left “insight” looked extreme mist (the major west stern fault). Count O[r]link.in project seas comm. pound.in e^(XpeRt) curry US IT “nos. B(est)” what-the-ignorant missed: UN dare “Bei. area” guide D-ants.

    841=836+5 —— another member of the +5 pattern (review)
    s(836+5) = 5# ——– recall that 836 is the smallest untouchable weird number

    s(841) = 30
    s(30) = 42
    s(42) = 54
    s(54) = 66 — seq.U-Ns continues as “ayaM” conΦDOn’t EU AI red e^no.?

    Doubly greedy Sylvester’s (double exponential) Sequence precisely marks period doubling symmetry from perfect jovian superstring symmetry. The subtle difference is several orders of magnitude smaller than the dominant marks of B, M, & Leech and thus it motivates careful contemplation of model detail diagnostics.

    B, M, & Leech in solar system order are crystal clear. On this trail an investigator is tripping over strictly equal fits that keep piling up (thousands of noteworthy things). Someone will discover a terse way to encode a monstrous array of fits …or if you prefer cynicism, that’s already a trade secret and there are difficult people prepared to go to dangerous lengths to keep a precious gem hidden.

  75. Paul Vaughan says:

    West Turn Security Makes 298 BIG Investments in Better Math Education

    Homeland innocents learned
    to crave unilateral freedom simply because
    on CR Ave., U-N ill at Orwell free doom be cause.

    Multilateralism — a word with rapidly increasing negative connotations — means bad dictators “representing” several billion savagely bully millions of good folks in consequently faltering democracies, just like they do to their “own”.

    In 1984 UN e^(thick hale) nos. antonym 4:2020 hindsight.

    Φ(11) = 10 = Φ(22)
    Φ(19) = 18
    Φ(23) = 22
    s(323) = 37 = Φ(149)/4 = Φ(298)/4 = s(298+5^2)
    s(152) = Φ(149) = Φ(298) = 148 = 4*37 O No!

    Emerge.in-C-meet-UN secure ET count sol: western citizens might use math to explore elements sov. climb ET ….if they no. ET = Euler’s Totient.

    d(2,1/2,10) = d(2,1/2,13) = d(2,1/2,22) = d(2,1/2,18) = d(2,1/2,37) = d(4,1/2,58) =
    d(2,1/2,Φ(22)) = d(2,1/2,Φ(11)) = d(2,1/2,104/8) = d(2,1/2,Φ(23)) = d(2,1/2,Φ(19)) =
    d(2,1/2,s(323)) = d(1,1/2,Φ(59)) = d(2,1/2,Φ(149)/4) = d(2,1/2,Φ(298)/4) = 104

    s(652)=s(4*163)=496=s(496) is perfect
    163-104=59 trade perfect seek routes indeed!
    Φ(Φ(59)) = Φ(58) = Φ(29) = 28 = s(28)
    Just divide 4 UNother puzzle piece:

    R(1,1/2,7) = 0.0679047747389632 = ⌊(e^√7π)^(1/1)⌉^1 – e^√7π = 4072 – 4071.93209522526
    298 = 4370 – ⌊e^√7π⌉ = 4370 – 4072 = 2*5*19*23-ROUND(EXP(7^(1/2)*PI()),0)

    With a maze UN guide dense IT’s AI a cure rat must fear. IC WHO invited sci11UNs bullies to demonstate the US mull tee laud orwell “leadership” “change”.

    If I were Donald Biden or Joe Trump I would NOT debate but simply state:

    1. Pandemic: I will not lock you down. Without delay we will stop lockdown freaks from causing you to go broke and become homeless. We don’t want you living in terror that they’re going to pull some kind of trick — which predators surely do, availed opportunity.

    2. Climate: Superior math learning outcomes in homeland math education are a security imperative. We dare not face the monstrous cost of keeping 4/5 of homeland citizens in the dark.

    1984 = 4*s(4*163) = 4*496 = 8/3 * 744 = 1728 + 104 + 152
    You really think George Orwell didn’t No.?
    “don’t speak” no. doubt

    70p is weird for all primes p ≥ 149

    1728=2*(836+28) ——– 28 is lowest perfect; 836 is lowest weird untouchable; 1728 of j-invariant
    1728=4*496-104-152=8*744/3-104-152=1984-104-152=8128-6400
    298=104+152+42=2/3*744-149-49=496-149-49
    =4*496-1728+42=8*744/3-1728+42=1984-1728+42 hitchhiker guides orwell too jovi-invariant

  76. Paul Vaughan says:

    D-anger
    Steep Cliffs
    Stay on the Trail
    Φ(85) = 64
    Φ(128) = 64
    Φ(136) = 64
    Φ(160) = 64
    Φ(170) = 64
    Φ(192) = 64
    Φ(204) = 64
    Φ(240) = 64 ~~~~~~~ “I’m goin’ 2 Whishh! IT awe” — The White Stripes
    ^v^v^v^v^v^v
    Φ(64) = 32
    Φ(32) = 16
    Φ(16) = 8
    Φ(8) = 4
    Φ(4) = 2
    Φ(2) = 1
    Φ(1) = 1
    240 = 1728 mod 744
    “make the sweat drip out of every pore”
    271 = 1729 mod 729
    Φ(241) = 240
    Φ(287) = 240
    Φ(305) = 240
    Φ(325) = 240
    Φ(369) = 240
    Φ(385) = 240
    Φ(429) = 240
    Φ(465) = 240
    Φ(482) = 240
    Φ(488) = 240
    Φ(495) = 240
    Φ(496) = 240 ~~~~~~~~~~~~ “a 7 ACE shh!U-N…” – TWS
    Φ(525) = 240
    Φ(572) = 240
    Φ(574) = 240
    Φ(610) = 240
    Φ(616) = 240
    Φ(620) = 240
    Φ(650) = 240
    Φ(700) = 240
    Φ(732) = 240
    Φ(738) = 240
    Φ(744) = 240 ~~~~~~~~~~~~~ “cou dn’s top…” ———————–
    Φ(770) = 240
    Φ(792) = 240
    Φ(858) = 240
    Φ(900) = 240
    Φ(924) = 240
    Φ(930) = 240
    Φ(990) = 240
    Φ(1050) = 240
    ^v^v^v^v^v^v^v
    Φ(240) = 64
    Φ(64) = 32
    Φ(32) = 16
    Φ(16) = 8
    Φ(8) = 4
    Φ(4) = 2
    Φ(2) = 1
    Φ(1) = 1
    “…and dem. B[]den right be 4 the lord” — Jack Black

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