[math-fun] claim of proof for No Odd Perfect Numbers
Paper (*cross-listing*): hep-th/0401052 From: Simon Davis Dr <davis@math.uni-potsdam.de> Date: Thu, 8 Jan 2004 23:56:42 GMT (9kb) Title: A Proof of the Odd Perfect Number Conjecture Authors: Simon Davis Comments: TeX, 13 pages Report-no: RFSC 04-01 Subj-class: High Energy Physics - Theory; Number Theory The finiteness of the number of solutions to the rationality condition for the existence of odd perfect numbers is deduced from the prime decomposition of the product of repunits defined by the sum of the divisors, $\sigma(N)$. All of the solutions to this rationality condition satisfy the inequality ${{\sigma(N)}\over N}\ne 2$. This technique is then used to demonstrate that there are no odd perfect numbers $N=(4k+1)^{4m+1} \prod_{i=1}^\ell q_i^{2\alpha_i}$. http://arXiv.org/abs/hep-th/0401052 13 pages ---------------------- I looked at this, but haven't puzzled through it yet. A crank? Putting the paper in hep-th is curious, and the author affiliation is unusual. The surface markers look mostly reasonable: real references, real notation. The logic is abbreviated, and it's hard to tell if it's merely explanatory. Page 1 leaps immediately into complicated algebra, eschewing background. There are well known required shapes for hypothetical odd perfect numbers. (These arise from equating 2N with the formula for the sum of divisors of N, and analyzing the even/odd properties of base P repunits. "Elementary" number theory.) One requirement is that all but one prime divisor have even exponent, making a square of a prime or prime power. One part of Davis's argument is that the divisor sum term for these primes (a factor of P^2 leads to a divisor-sum factor of P^2 + P + 1, a base P repunit 111) leads to new primes > P, which must then appear as new divisors of the hypothetical perfect number. For example, 7^2 -> 7^2 + 7 + 1 = 57 = 3 * 19 giving the new prime 19. Continuing the argument, 19^2 -> 381 = 3 * 127, and so on. If this always led to new larger primes, it would prove NoOPN. But it doesn't always give new primes (67^2 -> 4557 = 3 * 7 * 7 * 31 and 79^2 -> 6321 = 3 * 7 * 7 * 43), and I think his argument is concerned with handling the exceptional cases. Rich rcs@cs.arizona.edu
This suggests something that appears not yet to appear in OEIS (that was a good talk you gave in Phoenix, Neil) 67, 79, 137, 149, 163, 181, 191, 211, 229, 263, 269, 277, 313, 373, 431, 439, 499 [E&OE] Primes p such that all prime divisors of p^2 + p + 1 are less than p. The corresp seq for p^2 + 1 is A073501. R. On Wed, 14 Jan 2004, Richard Schroeppel wrote:
Paper (*cross-listing*): hep-th/0401052 From: Simon Davis Dr <davis@math.uni-potsdam.de> Date: Thu, 8 Jan 2004 23:56:42 GMT (9kb)
Title: A Proof of the Odd Perfect Number Conjecture Authors: Simon Davis Comments: TeX, 13 pages Report-no: RFSC 04-01 Subj-class: High Energy Physics - Theory; Number Theory
The finiteness of the number of solutions to the rationality condition for the existence of odd perfect numbers is deduced from the prime decomposition of the product of repunits defined by the sum of the divisors, $\sigma(N)$. All of the solutions to this rationality condition satisfy the inequality ${{\sigma(N)}\over N}\ne 2$. This technique is then used to demonstrate that there are no odd perfect numbers $N=(4k+1)^{4m+1} \prod_{i=1}^\ell q_i^{2\alpha_i}$.
http://arXiv.org/abs/hep-th/0401052 13 pages
----------------------
I looked at this, but haven't puzzled through it yet. A crank? Putting the paper in hep-th is curious, and the author affiliation is unusual. The surface markers look mostly reasonable: real references, real notation. The logic is abbreviated, and it's hard to tell if it's merely explanatory. Page 1 leaps immediately into complicated algebra, eschewing background.
There are well known required shapes for hypothetical odd perfect numbers. (These arise from equating 2N with the formula for the sum of divisors of N, and analyzing the even/odd properties of base P repunits. "Elementary" number theory.) One requirement is that all but one prime divisor have even exponent, making a square of a prime or prime power. One part of Davis's argument is that the divisor sum term for these primes (a factor of P^2 leads to a divisor-sum factor of P^2 + P + 1, a base P repunit 111) leads to new primes > P, which must then appear as new divisors of the hypothetical perfect number. For example, 7^2 -> 7^2 + 7 + 1 = 57 = 3 * 19 giving the new prime 19. Continuing the argument, 19^2 -> 381 = 3 * 127, and so on. If this always led to new larger primes, it would prove NoOPN. But it doesn't always give new primes (67^2 -> 4557 = 3 * 7 * 7 * 31 and 79^2 -> 6321 = 3 * 7 * 7 * 43), and I think his argument is concerned with handling the exceptional cases.
Rich rcs@cs.arizona.edu
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While in New Zealand I met Eamonn O'Brien, who's one author of a paper that, among other things, tabulates the number of groups of each order up to 2000. This has led me to find quite q few interesting sequences, and given quite a lot of fun. First, some comments on the numbers. For n = 1 to 32, the number of groups of order n is 1,1,1,2,1,2,1,5,2,2,1,5,1,2,1,14,1,5,1,5,2,2,1,15,2,2,5,4,1,4,1,51. Certain numbers (such as 1,2,5,15) are much more common than others. The numbers of groups of orders 1,2,4,8,16,32,64,128,256,512,1024 are 1,1,2,5,14,51,267,2328,56092,10494213,49487365422. The number of groups of order at most 2000 but not 1024 is only 423164062, so the ones of order 1024 outnumber the others 10-to-1! Remarkably, the number of groups of order p^n (p prime) is known quite closely - it has the form p^{ (2n^3/27) + O(n^{5/2}) }, the 5/2 here improving on an earlier bound of 8/3. The smallest N for which the number of groups of order N is 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 is 1 4 75 28 8 42 375 510 308 90 140 88 56 16 24 100 675 156 1029 820 1875 For n = 22, we must have N > 2000. This raises the interesting question of whether every positive integer is the number of groups of some order? O'Beirne told me that Keith Devlin has worked on this, but I didn't get the impression he'd got very far. Lenstra and I pushed it quite a long way, though with some mistakes that I've since been rectifying. I'll report on this later. I propose to use GROUPS(N) for the number of groups of order N, but will abbreviate it to G(N) in this letter. Let's call N group-deficient or group-abundant according as G(N) < N or G(N) > N, and group-perfect if G(N) = N. Then it seems almost certain that 1 is the only group-perfect number, and that almost all numbers are group-deficient. The best we've proved in this direction, though, is that all squarefree numbers (and so most numbers) are group-deficient. Here's the elegant proof that was found by Lenstra and simplified a bit by me: Any group of squarefree order N has a presentation of the form 1 = a^d = b^(N/d), ab = ba^k, where d is some divisor of n, and k is prime to d. So the number of such groups is at most the sum of phi(d) over d dividing N, which is well-known to be N. However, if N > 1 this counts the cyclic group twice (for d = 1 and d = N), and so then the number of groups of order N is strictly less than N. One of the conjectures above can be put in form A(N)/N -> 0, where A(N) is the number of group-abundant numbers up to N. However, the convergence is slow. For example, if N = 49487365422 and n = 1024, then every multiple of n below N is group-abundant, and so A(N)/N exceeds 1/n. We can replace n by any 2^m > 16 and N by GROUPS(n), which is approximately 2^{(2n^3)/27}, to see that this argument proves that we have 1 A(N)/N > ------------------------- roughly, 3 times cuberoot(logN /2) for an infinity of N, the logarithm being taken to base 2. Here's how the sequence of group-abundant numbers starts: 1,32,48,64,96,128,144,160,192,256,288,320,384,432,448,480,512,576,640,648, 672,704,720,768,800,832,864,896,960,1024,1088,1152,1216,1248,1280,1296, 1344,1408,1440,1458,1536,1600,1664,1728,1792,1920,1944 Lenstra's theorem tells us that none of these can be squarefree - in fact so far they've always been divisible by some fourth power, which suggests that perhaps that theorem can be improved. Most of my thinking has been on the universality question - "is every positive integer the number of groups of some order?". As I said, I'll report on that later. It gives rise to some more interesting sequences by the way, including 7,11,19,29,31,47,49,... . REgards to all - JHC
--- John Conway <conway@Math.Princeton.EDU> wrote:
While in New Zealand I met Eamonn O'Brien, who's one author of a paper that, among other things, tabulates the number of groups of each order up to 2000. This has led me to find quite q few interesting sequences, and given quite a lot of fun.
It's worth noting that all of this deals with Sequence #1, in the OEIS heirarchy. http://www.research.att.com/cgi-bin/access.cgi/as/njas/sequences/eisA.cgi?An... I give a picture in my recent MAA column, Sequence Pictures. http://www.maa.org/editorial/mathgames/mathgames_12_08_03.html H.-U. Besche had the first 2000 terms in the Small Groups Library, which now seems to be broken. Here are the first 1000 terms. 1, 1, 1, 2, 1, 2, 1, 5, 2, 2, 1, 5, 1, 2, 1, 14, 1, 5, 1, 5, 2, 2, 1, 15, 2, 2, 5, 4, 1, 4, 1, 51, 1, 2, 1, 14, 1, 2, 2, 14, 1, 6, 1, 4, 2, 2, 1, 52, 2, 5, 1, 5, 1, 15, 2, 13, 2, 2, 1, 13, 1, 2, 4, 267, 1, 4, 1, 5, 1, 4, 1, 50, 1, 2, 3, 4, 1, 6, 1, 52, 15, 2, 1, 15, 1, 2, 1, 12, 1, 10, 1, 4, 2, 2, 1, 231, 1, 5, 2, 16, 1, 4, 1, 14, 2, 2, 1, 45, 1, 6, 2, 43, 1, 6, 1, 5, 4, 2, 1, 47, 2, 2, 1, 4, 5, 16, 1, 2328, 2, 4, 1, 10, 1, 2, 5, 15, 1, 4, 1, 11, 1, 2, 1, 197, 1, 2, 6, 5, 1, 13, 1, 12, 2, 4, 2, 18, 1, 2, 1, 238, 1, 55, 1, 5, 2, 2, 1, 57, 2, 4, 5, 4, 1, 4, 2, 42, 1, 2, 1, 37, 1, 4, 2, 12, 1, 6, 1, 4, 13, 4, 1, 1543, 1, 2, 2, 12, 1, 10, 1, 52, 2, 2, 2, 12, 2, 2, 2, 51, 1, 12, 1, 5, 1, 2, 1, 177, 1, 2, 2, 15, 1, 6, 1, 197, 6, 2, 1, 15, 1, 4, 2, 14, 1, 16, 1, 4, 2, 4, 1, 208, 1, 5, 67, 5, 2, 4, 1, 12, 1, 15, 1, 46, 2, 2, 1, 56092, 1, 6, 1, 15, 2, 2, 1, 39, 1, 4, 1, 4, 1, 30, 1, 54, 5, 2, 4, 10, 1, 2, 4, 40, 1, 4, 1, 4, 2, 4, 1, 1045, 2, 4, 2, 5, 1, 23, 1, 14, 5, 2, 1, 49, 2, 2, 1, 42, 2, 10, 1, 9, 2, 6, 1, 61, 1, 2, 4, 4, 1, 4, 1, 1640, 1, 4, 1, 176, 2, 2, 2, 15, 1, 12, 1, 4, 5, 2, 1, 228, 1, 5, 1, 15, 1, 18, 5, 12, 1, 2, 1, 12, 1, 10, 14, 195, 1, 4, 2, 5, 2, 2, 1, 162, 2, 2, 3, 11, 1, 6, 1, 42, 2, 4, 1, 15, 1, 4, 7, 12, 1, 60, 1, 11, 2, 2, 1, 20169, 2, 2, 4, 5, 1, 12, 1, 44, 1, 2, 1, 30, 1, 2, 5, 221, 1, 6, 1, 5, 16, 6, 1, 46, 1, 6, 1, 4, 1, 10, 1, 235, 2, 4, 1, 41, 1, 2, 2, 14, 2, 4, 1, 4, 2, 4, 1, 775, 1, 4, 1, 5, 1, 6, 1, 51, 13, 4, 1, 18, 1, 2, 1, 1396, 1, 34, 1, 5, 2, 2, 1, 54, 1, 2, 5, 11, 1, 12, 1, 51, 4, 2, 1, 55, 1, 4, 2, 12, 1, 6, 2, 11, 2, 2, 1, 1213, 1, 2, 2, 12, 1, 261, 1, 14, 2, 10, 1, 12, 1, 4, 4, 42, 2, 4, 1, 56, 1, 2, 1, 202, 2, 6, 6, 4, 1, 8, 1, 10494213, 15, 2, 1, 15, 1, 4, 1, 49, 1, 10, 1, 4, 6, 2, 1, 170, 2, 4, 2, 9, 1, 4, 1, 12, 1, 2, 2, 119, 1, 2, 2, 246, 1, 24, 1, 5, 4, 16, 1, 39, 1, 2, 2, 4, 1, 16, 1, 180, 1, 2, 1, 10, 1, 2, 49, 12, 1, 12, 1, 11, 1, 4, 2, 8681, 1, 5, 2, 15, 1, 6, 1, 15, 4, 2, 1, 66, 1, 4, 1, 51, 1, 30, 1, 5, 2, 4, 1, 205, 1, 6, 4, 4, 7, 4, 1, 195, 3, 6, 1, 36, 1, 2, 2, 35, 1, 6, 1, 15, 5, 2, 1, 260, 15, 2, 2, 5, 1, 32, 1, 12, 2, 2, 1, 12, 2, 4, 2, 21541, 1, 4, 1, 9, 2, 4, 1, 757, 1, 10, 5, 4, 1, 6, 2, 53, 5, 4, 1, 40, 1, 2, 2, 12, 1, 18, 1, 4, 2, 4, 1, 1280, 1, 2, 17, 16, 1, 4, 1, 53, 1, 4, 1, 51, 1, 15, 2, 42, 2, 8, 1, 5, 4, 2, 1, 44, 1, 2, 1, 36, 1, 62, 1, 1387, 1, 2, 1, 10, 1, 6, 4, 15, 1, 12, 2, 4, 1, 2, 1, 840, 1, 5, 2, 5, 2, 13, 1, 40, 504, 4, 1, 18, 1, 2, 6, 195, 2, 10, 1, 15, 5, 4, 1, 54, 1, 2, 2, 11, 1, 39, 1, 42, 1, 4, 2, 189, 1, 2, 2, 39, 1, 6, 1, 4, 2, 2, 1, 1090235, 1, 12, 1, 5, 1, 16, 4, 15, 5, 2, 1, 53, 1, 4, 5, 172, 1, 4, 1, 5, 1, 4, 2, 137, 1, 2, 1, 4, 1, 24, 1 --Ed Pegg Jr
participants (4)
-
Ed Pegg Jr -
John Conway -
Richard Guy -
Richard Schroeppel