A350780 Numbers that are the number of divisors of p^2 - 1 for some prime p.
2, 4, 8, 10, 16, 18, 20, 24, 28, 30, 32, 36, 40, 42, 48, 54, 56, 60, 64, 70, 72, 80, 84, 88, 90, 96, 100, 104, 108, 112, 120, 126, 128, 132, 136, 140, 144, 150, 152, 156, 160, 162, 168, 176, 180, 182, 184, 192, 196, 198, 200, 204, 208, 210, 216, 220, 224, 228
Offset: 1
Examples
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. 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. From _Jianing Song_, Feb 11 2025: (Start) Let Omega = A001222, k be an even number and p be a prime. - 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. - 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). - 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. - 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. - 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). 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)
Links
- Jianing Song, Table of n, a(n) for n = 1..192
- Jianing Song, Notes on A350780 and A358881 (1)
- Jianing Song, Notes on A350780 and A358881 (2)
- Jianing Song, PARI program for A350780 and A358881
Programs
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PARI
\\ See Links. Jianing Song, Feb 16 2025
Comments