A341658 Primes p such that p^2 - 1 has 32 divisors.

29, 43, 53, 59, 61, 67, 83, 107, 157, 173, 193, 227, 277, 283, 317, 347, 563, 653, 733, 787, 877, 907, 997, 1213, 1237, 1283, 1307, 1523, 1867, 2083, 2693, 2797, 2803, 3253, 3413, 3517, 3643, 3677, 3733, 3803, 4253, 4363, 4547, 4723, 5387, 5443, 5483, 5717

Offset: 1

Author

Jon E. Schoenfield, Feb 26 2021

Keywords

• nonn

Comments

Conjecture: sequence is infinite.

All terms are primes p such that p^2 - 1 is of the form 24*q*r = 2^3 * 3 * q * r (where q and r are distinct primes), with only three exceptions: 53, 107, and 193 (see Example section).

For primes p > 3, p^2 - 1 = (p-1)*(p+1) will have p-1 and p+1 as consecutive even numbers (so one of them is divisible by 4, so their product is divisible by 8), and one of p-1 and p+1 will be divisible by 3. For each term other than 53, 107, and 193, the factors p-1 and p+1 are, in some order, numbers of the forms 2*q and 12*r or 4*q and 6*r.

Examples

			      p =               factorization
   n  a(n)  p^2 - 1      of (p^2 - 1)
  --  ----  -------  -------------------
   1   29      840   2^3 * 3   *  5 *  7
   2   43     1848   2^3 * 3   *  7 * 11
   3   53     2808   2^3 * 3^3 * 13
   4   59     3480   2^3 * 3   *  5 * 29
   5   61     3720   2^3 * 3   *  5 * 31
   6   67     4488   2^3 * 3   * 11 * 17
   7   83     6888   2^3 * 3   *  7 * 41
   8  107    11448   2^3 * 3^3 * 53
   9  157    24648   2^3 * 3   * 13 * 79
  10  173    29928   2^3 * 3   * 29 * 43
  11  193    37248   2^7 * 3   * 97
		

Crossrefs

Cf. A000005, A000040, A309906, A341655.

Programs

Mathematica

Select[Range[6000], PrimeQ[#] && DivisorSigma[0, #^2 - 1] == 32 &] (* Amiram Eldar, Feb 26 2021 *)

PARI

isok(p) = isprime(p) && (numdiv(p^2-1) == 32); \\ Michel Marcus, Feb 26 2021