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Visual Prime Pattern identified |
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| May3-11, 01:56 PM | #69 |
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Visual Prime Pattern identified
Raphie,
Thanks for the wealth of information. Your posts really help me to see things I've never thought about before. Sorry for the delay but I've been trying to formalize my ideas before I get to lost exploring other areas. I have attached a rough draft of what I've been working on. I'd love any feedback you can offer. Jeremy |
| May5-11, 11:36 AM | #70 |
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I fixed a few mistakes.
http://www.tubeglow.com/test/Pythagorean%20lattice.pdf |
| May7-11, 02:03 PM | #71 |
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i wish i was smart
amazing article and amazing animtion I find so infinetly interesting... can you explain me shortly the animation? i dont really understand what it means. I am doing 1st year eng. so i dont know that much math |
| May8-11, 04:38 AM | #72 |
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VAR(X) = E[X]^2 - E[X^2] Also note the following symmetrical equation form: x^2 + 2xy + y^2 = z. RELATED LINKS: Variance http://en.wikipedia.org/wiki/Variance The Expectation Operator http://arnoldkling.com/apstats/expect.html E(X+Y)^2 = E(X^2 + Y^2 + 2XY) Expected value http://en.wikipedia.org/wiki/Expected_value Best, RF |
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| May11-11, 04:36 PM | #73 |
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very interesting patterns if I use a little recursion. Its in 3D best viewed in 1900*1200 resolution. Keep the mouse off of the page while it loads and it will plot in 2d first. source code is available. http://www.tubeglow.com/test/PL3D/P_Lattice_3D.html
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| May11-11, 05:57 PM | #74 |
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Keep in mind, ala Marcus du Sautoy, that every time you add in a new prime you are, in effect, adding in another variable that will create "waves" and "ripples" interacting with all the other waves and ripples created by the other primes, meaning that the shapes you model will morph endlessly as you go further and further down the number line, or, rather, further out from the origin along the surface of your Minkowski-esque "prime lattice light cone." But, just because you know that it will "morph," this in no way excludes the possibility, even perhaps likelihood, of regularities in relation to the manner of "timing" by which new variables are introduced, because each and every new variable is recursively made possible only by the "multiplicative failure" of the primes that preceded it to fully "cover" "number space". Also keep in mind that variables equate with dimensions: Variable == Parameter == Dimension Best, RF |
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| May12-11, 08:46 PM | #75 |
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edit: I attached a better image and its projection. this is a kind of root system right? http://en.wikipedia.org/wiki/Root_system http://upload.wikimedia.org/wikipedi...ystems.svg.png |
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| May12-11, 10:28 PM | #76 |
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More "timing" links. Natural squares and partition numbers maybe? this is based on 12:
http://1.bp.blogspot.com/_u6-6d4_gsS...1600/roots.PNG |
| May13-11, 03:41 AM | #77 |
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And in relation to the number 12... n(n+ceiling(2^n/12)) http://oeis.org/A029929 The first 8, but only 8 numbers in this series are proven (lattice) kissing numbers. - RF RELATED LINK Leech lattice http://en.wikipedia.org/wiki/Leech_lattice The Leech lattice is also a 12-dimensional lattice over the Eisenstein integers. This is known as the complex Leech lattice, and is isomorphic to the 24-dimensional real Leech lattice... |
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| May13-11, 09:55 AM | #78 |
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e.g. 40 - 24 = 4^2 40 + 24 = 4^3 One can use this mathematical fact to easily obtain integer solutions to the following: Period^2 = 4*pi^2/GM * Distance^3 (Kepler's 3rd Law) e.g. (40 + 24)^2 = (40 - 24)^3 = 4^6 = 4096 (6 + 2)^2 = (6 - 2)^3 = 2^6 = 64 (= sqrt 4096) The Pentagonal Pyramid numbers, of course, are the summation of the Pentagonal numbers, which are already well-known to be related to the "timing" and/or "tuning" of the primes. p^2 - 1 == 1 mod (24) for all p > 3 (p^2 - 1)/24 is Pentagonal for all p > 3. And, also, as I mentioned previously, 24 s^2 is the Period^2 one obtains if one replaces L/g in the formula for a pendulum with zeta(2)^-2 = (pi^2/6)^-2, where (the reciprocal of) zeta(2) gives the probability of two randomly selected integers being relatively prime. - RF Note: Pentagonal Pyramid Numbers have a very easy to remember formula n*T_n = (+) Pentagonal Pyramid # and n*T_-n = (-) Pentagonal Pyramid #, for T_n a Triangular Number. |
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| May14-11, 12:42 PM | #79 |
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Raphie,
Thanks! I'm working on some projections.I definitly see an Eisenstein integer connection. edit: I'm working on the pendulum |
| May14-11, 08:31 PM | #80 |
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A060967 Number of prime squares <= 2^n. http://oeis.org/A060967 0, 0, 1, 1, 2, 3, 4, 5, 6, 8, 11, 14, 18, 24... RULE: SUBTRACT 1 -1, -1, 0, 0, 1, 2, 3, 4, 5, 9, 10, 13, 17, 23... RULE: ITERATE INTO THE "QRIME" NUMBER SEQUENCE {0, 1 U Primes} indexed from -1; p'_(n-1) 0, 0, 1, 1, 2, 3, 5, 7, 11, 17, 29, 41, 59, 83... RULE: ADD 1 1, 1, 2, 2, 3, 04, 06, 08, 12, 018, 030, 042, 060, 084... RULE: |MULTIPLY BY (n-2)| 2, 1, 0, 2, 6, 12, 24, 40, 72, 126, 240, 378, 600, 924... In formula form... K_(n-2) = (n-2) * (1 + p'_(-1 + COUNT[Number of prime squares <= 2^n])) for n = 2 --> 10 The 3rd through 11th values are the (proven lattice) Kissing Numbers up to Dimension 8, the very same ones you get by inserting n into the formula: n(n+ceiling(2^n/12)). And the 2nd and 12th values? T_1^3 = 1 and T_3^3 = 378 (& 600 = 2*T_24, while 924 is a Central Binomial Coefficient, the sum of proper divisors of which = 1764 == 42^2) 378 - totient (107) = 272 = K_9; 107 = p'_28 = p'_(T_7) = p'_(T_(Lucas_4) 001 - totient (002) = 000 = K_0; 002 = p'_01 = p'_(T_1) = p'_(T_(Lucas_1) 2 is the 1st Mersenne Prime Exponent, and 107 the 11th (1 = Lucas_1, 11 = Lucas_5, and 1 and 28 are both k-Perfect Numbers). These two numbers also have the property, that, when triangulated, you get a Lucas Number. There are only 3 Triangular Lucas Numbers: 1, 3 and 5778. (See: Lucas Number http://mathworld.wolfram.com/LucasNumber.html ) 0001 = T_001 = Lucas_01 = Lucas_(Lucas_1) = Lucas_(Lucas_(T_1)) = Lucas_(1*T_1) 5778 = T_107 = Lucas_18 = Lucas_(Lucas_6) = Lucas_(Lucas_(T_3)) = Lucas_(2*T_2) (1 and 6 are k-Perfect Numbers, 1,6 & 18 are the first 3 Pentagonal Pyramid Numbers) Also, see.... k*Lucas_n + 1 is a prime of Lucas Number Index http://www.physicsforums.com/showthread.php?t=497766 - RF |
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| May14-11, 10:15 PM | #81 |
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A POSSIBLY RELATED SEQUENCE Suppose the sum of the digits of prime(n) and prime(n+1) divides prime(n) + prime(n+1). Sequence gives prime(n). http://oeis.org/A127272 2, 3, 5, 7, 11, 17, 29, 41, 43, 71, 79, 97, 101, 107... e.g. (2 + 3)/(2+3) = 1 (3+5)/(3+5) = 1 (5+7)/(5+7) = 1 (7+11)/(7+(1+1)) = 2 (11+13/((1+1)+(1+3)) = 4 (17+19/((1+7)+(1+9)) = 2 (29+31/((2+9)+(3+1)) = 4 (41+43/((4+1) + (4+3)) = 7 (43+47/((4+3)+(4+7)) = 5 (71+73)/((7+1)+(7+3)) = 8 (79+83)/((7+9)+(9+7)) = 5 (97+101)/((9+7)+(1+0+1)) = 11 (101+103)/((1+0+1) + (1+0+3) = 34 (107+109)/((1+0+7)+(1+0+9) = 12 ALSO... Numbers n such that 1 plus the sum of the first n primes is divisible by n+1. http://oeis.org/A158682 2, 6, 224, 486, 734, 50046, 142834, 170208, 249654, 316585342, 374788042, 2460457826, 2803329304, 6860334656, 65397031524, 78658228038 002 - 002 = 000 = K_00 012 - 006 = 006 = K_02 (Max) 600 - 224 = 336 = K_10 (Lattice Max known) 924 - 486 = 438 = K_11 (Lattice Max known) 6/(5+1) = 1 42/(6+1) = 6 143100/(224+1) = 636 775304/(486+1) = 1592 Like I said, especially given that these two progressions are ones I came across in the process of writing that last post to you, "hmmmm..." RELATED PROGRESSIONS Integer averages of first n noncomposites for some n. http://oeis.org/A179860 1, 2, 6, 636, 1592, 2574, 292656, 917042, 1108972, 1678508, 3334890730, 3981285760, 28567166356, 32739591796, 83332116034 a(n) is the sum of the first A179859(n) noncomposites. http://oeis.org/A179861 1, 6, 42, 143100, 775304, 1891890, 14646554832, 130985694070, 188757015148, 419047914740, 1055777525624570390, 1492138298614167680, 70288308055831268412, 91779857115464381780, 571686203669195590338 Numbers n that divide the sum of the first n noncomposites. http://oeis.org/A179859 1, 3, 7, 225, 487, 735, 50047, 142835, 170209, 249655, 316585343, 374788043, 2460457827, 2803329305, 6860334657 This number, in particular, I find interesting... 142835 = 5*7^2*11*53 = (142857 - par_8) = (142857 - 22) vs. 1/7 = .142857 (repeating) Indexing from 0, 142857 is the 24th Kaprekar Number 1, 3, 7 and 225, the 1st 4 terms in that last sequence above == (2^1 - 1)^1, (2^2 - 1)^1, (2^3 - 1)^1, (2^4 - 1)^2. - RF |
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| May14-11, 11:18 PM | #82 |
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Jeremy, as an FYI, and by way of giving another example, if one desires to mathematically derive, say, the Dimension 10 Lattice Kissing Number from a convolution of primes and partition numbers, a far simpler way to do it is as follows:
p'_((par_n - 1) * p'_(par_(n-1) - 1) p'_(1-1) -1 = 0 p'_(1-1) -1 = 0 p'_(2-1) -1 = 1 p'_(3-1) -1 = 2 p'_(5-1) -1 = 6 p'_(7-1) -1 = 12 p'_(11-1) -1 = 28 0*0 = 0 = K_0 0*1 = 0 = K_0 1*2 = 2 = K_1 2*6 = 12 = K_3 6*12 = 72 = K_6 12*28 = 336 = K_10 In order, that formula returns maximal (proven except for Dimension 10) lattice sphere packings for Dimensions equal to 6 consecutive Triangular Numbers: T_-1, T_0, T_1, T_2, T_3, T_4 On the other hand, if you simply add 1 to the first 7 partition numbers, and multiply by n... (1+1)*0 = 0 = K_0 (1+1)*1 = 2 = K_1 (2+1)*2 = 6 = K_2 (3+1)*3 = 12 = K_3 (5+1)*4 = 24 = K_4 (7+1)*5 = 40 = K_5 (11+1)*6 = 72 = K_6 ... then you get Maximal (proven) Lattice Sphere packings to dimension 6. Best, RF |
| May15-11, 10:48 AM | #83 |
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I'm going through you posts now. I reworked the visual a little. Click any where on the page after it loads the black back ground then press:
1 = normal growth of the equation. after it builds for a while you can notice the pattern and the timing. Seems to be timed like a pendulum. or 2 = normal "inverse growth". or 3 = fractal pattern generation up/down arrrow = zoom in out left/right arrow = fractal limit increase/decrease. d = 3d on/off http://www.tubeglow.com/test/PL3D2/P_Lattice_3D_2.html |
| May16-11, 10:17 AM | #84 |
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Jeremy, firstly, the page you linked to doesn't seem to work with my system.
Secondly, I wouldn't read too much in to any single example I might give. It's all of the examples, taken together, and the picture they are seeming to paint (or the tune they are seeming to play) that I find most interesting. Thirdly, a critic would reasonably note that the indices of the prime numbers I am giving are all quite small. And that's a fair point. But then one has to explain away as "coincidence" relationships such as the following: For 1, 2, 3, 4 and 6 the solutions to the Crystallographic Restriction Theorem, then consider lattices in the following Dimensions: (1 - 1)^2 + 1 - totient (1) = 0 (2 - 1)^2 + 1 - totient (2) = 1 (3 - 1)^2 + 1 - totient (3) = 3 (4 - 1)^2 + 1 - totient (4) = 8 (6 - 1)^2 + 1 - totient (6) = 24 Dimensions {0 & 24} Union {1, 3, 8}, the dimensions associated with the Standard Model of Physics = SU(3)×SU(2)×U(1) Then, for F_n a Fibonacci Number and T_n a Triangular Number... And for... 2, 4, 6, 10, 22 == totient (1st 5 safe "qrimes") == 2 * (1, 2, 3, 5, 11) where... 01 = p'_(1 - 1) = par_1 02 = p'_(2 - 1) = par_2 03 = p'_(3 - 1) = par_3 05 = p'_(4 - 1) = par_4 11 = p'_(6 - 1) = par_6 Then... p_00001 - p_01 = p_F_02 - p_((F_0)*(T_(pi(pi(01) + 1))) + 1) = 000002 - 002 = 000002 = K_0 p_00003 - p_02 = p_F_04 - p_((F_1)*(T_(pi(pi(02) + 1))) + 1) = 000005 - 003 = 000002 = K_1 p_00008 - p_04 = p_F_06 - p_((F_2)*(T_(pi(pi(03) + 1))) + 1) = 000019 - 007 = 000012 = K_3 p_00055 - p_07 = p_F_10 - p_((F_3)*(T_(pi(pi(05) + 1))) + 1) = 000257 - 017 = 000240 = K_8 p_17711 - p_31 = p_F_22 - p_((F_4)*(T_(pi(pi(13) + 1))) + 1) = 196687 - 127 = 196560 = K_24 Note: 1, 2, 3, 5 & 13 are the Prime Numbers | (2^n - 1) is Twice Triangular (aka "The Ramanujan-Nagell Pronic Numbers"). And 2, 3, 5, 17 and 257 are all Fermat Primes, while 2, 3, 7, 17 and 127 (and also 19) are all Mersenne Prime Exponents, the 1st, 2nd, 4th, 6th and 12th (19 is the 7th). p'_1 - 1 = 02 - 1 = 1 p'_2 - 1 = 03 - 1 = 2 p'_3 - 1 = 05 - 1 = 4 p'_4 - 1 = 07 - 1 = 6 p'_6 - 1 = 13 - 1 = 12 The condensed way to state the above is as follows: -------------------------------------------------------------------------------- for... K_n = n-th Kissing Number p'_(n-1) = n-th n in N | -1 < d(n) < 3 --> {0,1,2,3,5,7,11,13...} c_(n-1) = n-th n in N | -1 < totient(n) < 3 --> {0,1,2,3,4,6} E_n = n-th Mersenne Prime Exponent F_n = n-th Fibonacci Number then for range n = 0 --> 4... FORMULA K_((c - 1)^2 + 1 - totient (c)) = (p'_(F_(2(p'_(c - 1))))) - (E_(p'_c - 1)) -------------------------------------------------------------------------------- 2, 3, 5, 7, 13 [= {p_c} == {n in N | d(p_c - 1) = c}], as well as being the first 5 Mersenne Prime Exponents, are also the unique prime divisors of the Leech Lattice: K_24 = 196560 And, as I believe you may already know, this particular set of primes has been associated with anomaly cancellations in 26 Dimensional Bosonic String Theory by Frampton and Kephart: Mersenne Primes, Polygonal Anomalies and String Theory Classification http://arxiv.org/abs/hep-th/9904212 Best, RF ============================================ Also... 00 = p'_-1 = (1 - 1) ------------------------ 01 = p'_(p'_-1) = (2 - 1) 02 = p'_(p'_(p'_-1)) = (3 - 1) 03 = p'_(p'_(p'_(p'_-1))) = (4 - 1) 05 = p'_(p'_(p'_(p'_(p'_-1)))) = (6 - 1) 11 = p'_(p'_(p'_(p'_(p'_(p'_-1))))) = (12 - 1) for 1, 2, 3, 4, 6, 12 --> the divisors of 12 And also... (01 * 0) + 3 - d(01) = 02 = p'_01 --> 01st Mersenne Prime Exponent (02 * 2) + 3 - d(02) = 05 = p'_03 --> 03rd Mersenne Prime Exponent (03 * 4) + 3 - d(03) = 13 = p'_06 --> 05th Mersenne Prime Exponent (05 * 6) + 3 - d(05) = 31 = p'_11 --> 08th Mersenne Prime Exponent (11 * 8) + 3 - d(11) = 89 = p'_24 --> 10th Mersenne Prime Exponent for 1, 3, 5, 8, 10 --> Sum of Divisors (SUM d(n)) for n = 1 through 5 Here are the 1st 14 Mersenne Prime Exponents (inclusive of 1)... 1, 2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521 (Range = Lucas_1 --> Lucas_13) As you may or may not have noticed, in the last few posts I've referenced every one of these excepting 61 and 521 (= Lucas_13 = Lucas_(p'_(sigma_5)), indexing from 0, the 13th Mersenne Prime Exponent). ((61-1)*11) = totient (p'_11^2) = 660 = (T_36 - sqrt (36)), by the way, is a simple group that, musically speaking, is one-perfect 5th above A-440 and its my prediction for the maximal Kissing Number in 11 dimensions (+ or - 12). "Coincidentally," 660 - 12 = 648, is the maximal known lattice sphere packing in 12 dimensions (= 2*18^2 = Lucas_0*Lucas_(sigma_(5))^2 = p'_(Lucas_5)^2 - p'_(sigma_5)), while 36 (=2*18) is the totient of the 12th prime number, 37. Finally, in the interests of clarity, note that the below are all just absurdly long-winded, even if contextually relevant, ways of stating: 0, 1, 2, 2, 4: (pi(pi(01) + 1)) == d(pi (01)); 01 --> (1-1)th Mersenne Prime Exp == pi (pi (02)); 02 --> 1st Mersenne Prime Exp = 0 (pi(pi(02) + 1)) == d(pi (02)); 02 --> (2-1)th Mersenne Prime Exp == pi (pi (03)); 03 --> 2nd Mersenne Prime Exp = 1 (pi(pi(03) + 1)) == d(pi (03)); 03 --> (3-1)th Mersenne Prime Exp == pi (pi (05)); 05 --> 3rd Mersenne Prime Exp = 2 (pi(pi(05) + 1)) == d(pi (05)); 05 --> (4-1)th Mersenne Prime Exp == pi (pi (07)); 07 --> 4th Mersenne Prime Exp = 2 (pi(pi(13) + 1)) == d(pi (13)); 13 --> (6-1)th Mersenne Prime Exp == pi (pi (17)); 17 --> 6th Mersenne Prime Exp = 4 And finally, finally... a statement such as p_((F_4)*(T_(pi(pi(13) + 1))) + 1) is an even more long-winded way of stating: p_31, which is the 12th Mersenne Prime Exponent and/or the iterated 8th Mersenne Prime Exponent. As such, I went back in the post and included the condensed formula... |
| May16-11, 08:07 PM | #85 |
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for... c_(n-1) --> {0,1,2,3,4,6} == pi {1, 2, 3, 5, 7, 13} --> Divisors of Prime Divisors of Leech Lattice --> Integers with Totient < 3 --> n in N | d(p_c - 1) = c --> Solutions to 2*cos (2*pi/(n + 1 - sgn(n)) is in N E_(n-1) --> {0,1,2,3,5,7,13,17,19...} --> 0, 1 Union Mersenne Prime Exponents then... K_((c_n - 1)^2 + 1 - totient (c_n)) = (2^E_(c_n - 1)) - 2)*((E_(c_n - 1) - 1)*totient (c_(n-1))) Expansion: K_00 = (2^01 - 2) * ((01 - 1)*totient (0)) = 0000 * 00 = 0 K_01 = (2^02 - 2) * ((02 - 1)*totient (1)) = 0002 * 01 = 2 K_03 = (2^03 - 2) * ((03 - 1)*totient (2)) = 0006 * 02 = 12 K_08 = (2^05 - 2) * ((05 - 1)*totient (3)) = 0030 * 08 = 240 K_24 = (2^13 - 2) * ((13 - 1)*totient (4)) = 8190 * 24 = 196560 --------------------------------------------------------------------------------- POSSIBLY RELATED PROGRESSION y such that y^2=C(x,0)+C(x,1)+C(x,2)+C(x,3) is soluble 0, 1, 2, 8, 24, 260, 8672 R. K. Guy, Unsolved Problems in Number Theory, Section D3. http://oeis.org/A047695 --------------------------------------------------------------------------------- c_n & E_(c_n - 1) - 1) can be linked in the following manner: Denote i-phi(x)_n as the n-th integers with a totient of x (The "Inverted Totient Function") Denote s(x) as the number of Solutions to i-phi(x) Denote J_n as the y solutions to 2^y - 1 is Triangular (Ramanujan-Nagell Triangular Numbers) Then... i-phi(J) --> ------------------ i-phi(00) --> 00; ------- Solutions = 1 (Mathematica Definition) i-phi(01) --> 01 02; ------- Solutions = 2 i-phi(02) --> 03 04 06; ------- Solutions = 3 i-phi(04) --> 05 08 10 12; ------- Solutions = 4 i-phi(12) --> 13 21 26 28 36 42; ------- Solutions = 6 00 = 2T_pi(01) = 2T_d(01-1) = 2T_0 02 = 2T_pi(02) = 2T_d(02-1) = 2T_1 06 = 2T_pi(03) = 2T_d(03-1) = 2T_2 12 = 2T_pi(05) = 2T_d(05-1) = 2T_3 42 = 2T_pi(13) = 2T_d(13-1) = 2T_6 And right there you've got yourself, potentially, a nice clean bridge between (just for starters...), Ramanujan-Nagell, the Solutions to the Crystallographic Restriction Theorem and the Divisors of the Leech Lattice/Frampton-Kephart Primes. s(J) = c i-phi(J)_1 = (J-1) + totient (c) i-phi(J)_c = 2T_(d(J)) i-phi(J)_c = 2T_(pi(J+1)) and... Delta (i-phi(J)_1, i-phi(J)_c) = p'_(J - 2) = 0, 1, 3, 7, 29 Thus, for instance... K_(totient(s(J))J) = (2^J + 1)*(totient(s(J))J) K_00 = (2^(00+1) - 2) * (00*1) = 0 K_01 = (2^(01+1) - 2) * (01*1) = 2 K_04 = (2^(02+1) - 2) * (02*2) = 24 K_08 = (2^(04+1) - 2) * (04*2) = 240 K_24 = (2^(12+1) - 2) * (12*2) = 196560 Hard to get more simple than that. With the nice little bonus that... pi (2^01) = 0001 = 2^00 + 0 = 2^(T_-1) + 0 pi (2^02) = 0002 = 2^00 + 1 = 2^(T_0) + 1 pi (2^03) = 0004 = 2^01 + 2 = 2^(T_1) + 2 pi (2^05) = 0011 = 2^03 + 3 = 2^(T_2) + 3 pi (2^13) = 1028 = 2^10 + 4 = 2^(T_4) + 4 Thus... pi (2^J+1) = 2^(T_(s(J) - 2)) + n If that and the other relationships presented in this post don't make you and/or anyone else who comes across this go "hmmmm," then I really don't know what will. Go back to the beginning of this post and you'll see you can make both formulas relating to the previous post far, far simpler by substitution. But it still doesn't change the nature of the relationships. Rather, all that changes is the apparent simplicity of the relationships. e.g. K_(n(mod 5)+0) = (s(J)_(n-1) + 0*6)(n+0)^0 = 00, 02, 06, 012, 024 K_(n(mod 5)+4) = (s(J)_(n-1) + 1*6)(n+4)^1 = 24, 40, 72, 126, 240 Best, RF |
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