cp's OEIS Frontend

A350780 Numbers that are the number of divisors of p^2 - 1 for some prime p.

%I A350780 #37 Feb 16 2025 22:24:24
%S A350780 2,4,8,10,16,18,20,24,28,30,32,36,40,42,48,54,56,60,64,70,72,80,84,88,
%T A350780 90,96,100,104,108,112,120,126,128,132,136,140,144,150,152,156,160,
%U A350780 162,168,176,180,182,184,192,196,198,200,204,208,210,216,220,224,228
%N A350780 Numbers that are the number of divisors of p^2 - 1 for some prime p.
%C A350780 For all primes p > 73, tau(p^2 - 1) >= A309906(2) = 32.
%H A350780 Jianing Song, <a href="/A350780/b350780.txt">Table of n, a(n) for n = 1..192</a>
%H A350780 Jianing Song, <a href="/A350780/a350780.pdf">Notes on A350780 and A358881 (1)</a>
%H A350780 Jianing Song, <a href="/A350780/a350780_1.pdf">Notes on A350780 and A358881 (2)</a>
%H A350780 Jianing Song, <a href="/A350780/a350780_1.txt">PARI program for A350780 and A358881</a>
%e A350780 184 is a term: p = 111149057 is a prime, and p^2 - 1 = (p-1)*(p+1) = 2^22 * 3 * 53 * 18524843, which has 23*2*2*2 = 184 divisors.
%e A350780 190 is not a term: 190 = 2 * 5 * 19, so a number with 190 divisors must be of the form q^189, q^94 * r, q^37 * r^4, q^18 * r^9, or q^18 * r^4 * s, and for every prime p > 3, p^2 - 1 is a multiple of 24 = 2^3 * 3, so all the forms with 190 divisors are easily ruled out except for q^18 * r^4 * s. If p^2 - 1 = q^18 * r^4 * s, then it's one of the products 2^18 * 3^4 * s, 2^18 * r^4 * 3, 3^18 * 2^4 * s, or q^18 * 2^4 * 3. Each of these can be shown to be impossible by examining all possible ways of factoring the product into two even factors (p-1 and p+1) that differ by exactly two.
%e A350780 From _Jianing Song_, Feb 11 2025: (Start)
%e A350780 Let Omega = A001222, k be an even number and p be a prime.
%e A350780  - Omega(k) <= 2. If odd p != 3, 7 (not necessarily prime) satisfies tau(p^2 - 1) = k = 2q for prime q, then p^2 - 1 = 2^(q-1)*P for some prime P, so (p-1,p+1) = (2^(q-2),2P) or (2P,2^(q-2)), which means that P = 2^(q-3) +- 1. Note that "-" is impossible since q-3 is even, so we have q = 2^r + 3, P = 2^2^r + 1, and p = 2^(2^r+1) + 1 for some r. In particular, p must be divisible by 3, so p cannot be prime.
%e A350780  - Omega(k) = 3, or k = 16, 24, 36, or 54. Then tau(p^2 - 1) = k has finitely many solutions p == 1, 5 (mod 6) (not necessarily prime). See my first pdf link in the Links section for a proof. In fact, it seems that if we require p to be prime, then k <= 518, and the complete list of (k,p), Omega(k) = 3 is (k,p) = (8,5), (18,17), (20,23), (28,31), (30,73), (42,97), (70,2593), (182,1492993), and (518,4803028329503971873=32*3^36+1).
%e A350780  - If k/2 has only prime factors congruent to 1 modulo 4, then tau(p^2 - 1) = k has no solutions for odd p. See my second pdf link in the Links section for a proof.
%e A350780  - If k/2 has only prime factors congruent to 1 modulo 2*r for some odd r >= 3, then tau(p^2 - 1) = k for odd p implies that p is of the form p = 2^(2*r-1)*M^(2*r) + 1 for some M.
%e A350780  - In general, if d(x) = k, then the largest exponent in the canonical factorization of x must be at least gpf(k)-1, where gpf = A006530 is the largest prime factor function. So if d(p^2 - 1) = k, then one of p-1 and p+1 must be divisible by M^(gpf(k)-1) for some odd prime M or by 2^(gpf(k)-2).
%e A350780 Conjecture: if Omega(k) >= 4, k != 16, 24, 36, or 54, and k/2 has a prime factor not congruent to 1 modulo 4, then tau(p^2 - 1) = k has infinitely many solutions. (End)
%o A350780 (PARI) \\ See Links. _Jianing Song_, Feb 16 2025
%Y A350780 Cf. A000005, A309906, A341660, A358881.
%K A350780 nonn,hard
%O A350780 1,1
%A A350780 _Jon E. Schoenfield_, May 02 2022