A337848 Odd integers k>=5 such that 2^((k-1)/2)-1 == 0 (mod k*(k-3)/2).

73, 241, 2593, 5113, 8713, 18433, 53593, 55681, 86113, 102241, 126337, 127873, 158113, 181721, 184369, 186049, 208393, 219313, 221537, 241921, 262657, 267913, 282313, 314161, 314401, 341641, 362521, 398441, 415873, 450913, 534241, 619921, 651169, 731881, 953473, 1045801, 1153441, 1294177, 1554281, 2023921, 2162401, 2345401, 2533681

Offset: 1

Author

Benoit Cloitre, Sep 26 2020

Keywords

• nonn

Comments

Computed terms are prime. Is it always the case? If not it would be interesting to compute the pseudoprimes.

1234125721 = 24841*49681, 4294901761 = 193*22253377, 6602556241 = 57457*114913 are composite counterexamples to the assumption that all terms are prime. - Hugo Pfoertner, Sep 26 2020

These are a(420), a(705) and a(830). Together with a(956) = 10025492401 = 101 * 701 * 141601 they are the first 4 composite terms. - Amiram Eldar, Jun 17 2022

Links

• Amiram Eldar, Table of n, a(n) for n = 1..1000

Crossrefs

Cf. A337818.

Programs

Mathematica

Select[Range[5, 10^6, 2], PowerMod[2, (# - 1)/2, #*(# - 3)/2] == 1 &] (* Amiram Eldar, Sep 26 2020 *)

PARI

is(n) = n%2 && n>=3 && Mod(2, n*(n-3)/2)^((n-1)/2) ==1