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A341658 Primes p such that p^2 - 1 has 32 divisors. 5
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 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
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.
LINKS
EXAMPLE
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
MATHEMATICA
Select[Range[6000], PrimeQ[#] && DivisorSigma[0, #^2 - 1] == 32 &] (* Amiram Eldar, Feb 26 2021 *)
PROG
(PARI) isok(p) = isprime(p) && (numdiv(p^2-1) == 32); \\ Michel Marcus, Feb 26 2021
CROSSREFS
Sequence in context: A066502 A125870 A076439 * A168474 A341666 A247896
KEYWORD
nonn
AUTHOR
Jon E. Schoenfield, Feb 26 2021
STATUS
approved

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Last modified April 25 03:15 EDT 2024. Contains 371964 sequences. (Running on oeis4.)