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Integers k>=3 such that 2^k == 2 (mod k*(k-1)*(k-2)/6).
1

%I #30 Mar 27 2024 21:12:48

%S 3,5,37,101,44101,3766141,8122501,18671941,35772661,36969661,

%T 208168381,425420101,725862061,778003381,818423101,1269342901,

%U 9049716901,27221068981,60138957061,125980182901,137330493301,314912454781,315322826869,478543291381,667933881301

%N Integers k>=3 such that 2^k == 2 (mod k*(k-1)*(k-2)/6).

%C Computed terms are prime. Is it always the case? If not it would be interesting to compute the smallest pseudoprime.

%C It seems that all larger terms are of the form 180*k + 1, starting at a(5) = 44101 = 180*245 + 1. - _Hugo Pfoertner_, Sep 27 2020

%C Other terms of the form 180*k+1 (which are all prime): 60138957061, 125980182901, 137330493301, 478543291381, 667933881301, 700212813301, 701030830501, 720023604301, 766514618101, 778382658901. - _Chai Wah Wu_, Oct 06 2020

%H Delbert L. Johnson, <a href="/A337858/b337858.txt">Table of n, a(n) for n = 1..28</a>

%t Select[Range[3, 10^7], PowerMod[2, #, #*(# - 1)*(# - 2)/6] == 2 &] (* _Amiram Eldar_, Sep 27 2020 *)

%o (PARI) is(n) = n>=3 && Mod(2, n*(n-1)*(n-2)/6)^n ==2

%Y Cf. A337818, A337846.

%K nonn

%O 1,1

%A _Benoit Cloitre_, Sep 26 2020

%E a(12)-a(18) from _Amiram Eldar_, Sep 27 2020

%E a(19)-a(25) from _Delbert L. Johnson_, Mar 27 2024