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Numbers k with property that 2^(k-1) == 1 (mod k) and 2^((3*k-1)/2) - 2^((k-1)/2) + 1 == 0 (mod k).
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%I #18 Oct 13 2020 05:45:24

%S 3,5,11,13,19,29,37,43,53,59,61,67,83,101,107,109,131,139,149,157,163,

%T 173,179,181,197,211,227,229,251,269,277,283,293,307,317,331,347,349,

%U 373,379,389,397,419,421,443,461,467,491,499,509,523,541,547,557,563

%N Numbers k with property that 2^(k-1) == 1 (mod k) and 2^((3*k-1)/2) - 2^((k-1)/2) + 1 == 0 (mod k).

%C All composites in this sequence are 2-pseudoprimes, see A001567.

%C The subsequence of composites begins: 3277, 29341, 49141, 80581, 88357, 104653, 196093, 314821, 458989, 476971, 489997, ..., . - _Robert G. Wilson v_, Oct 02 2010

%C The sequence includes all the primes of A003629. - _Alzhekeyev Ascar M_, Mar 09 2011

%C If we consider the composites in this sequence which are in the modulo classes == 3 (mod 8) or == 5 (mod 8), they are moreover strong pseudoprimes to base 2 (see A001262). - _Alzhekeyev Ascar M_, Mar 09 2011

%C Are there any composites in this sequence which are *not* in the two modulo classes == {3,5} (mod 8)? - _R. J. Mathar_, Mar 29 2011

%H Amiram Eldar, <a href="/A175865/b175865.txt">Table of n, a(n) for n = 1..10000</a>

%e 3 is a term since 2^(3-1)-1 = 3 is divisible by 3, and 2^((3*3-1)/2) - 2^((3-1)/2) + 1 = 15 is divisible by 3.

%t fQ[n_] := PowerMod[2, n - 1, n] == 1 && Mod[ PowerMod[2, (3 n - 1)/2, n] - PowerMod[2, (n - 1)/2, n], n] == n - 1; Select[ Range@ 570, fQ] (* _Robert G. Wilson v_, Oct 02 2010 *)

%Y Cf. A001262, A001567, A003629.

%K nonn

%O 1,1

%A _Alzhekeyev Ascar M_, Sep 30 2010

%E More terms from _Robert G. Wilson v_, Oct 02 2010