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%I #19 Feb 01 2019 20:08:49
%S 563,1291,1733,1907,2477,2609,2693,2837,3533,3677,4157,4517,5693,
%T 12809,15077,19997,25603,28517,29573,29837,31517,32237,32717,34949,
%U 37277,43613,43973,44453,50333,52253,62477,68213,69197,72893,74717
%N Integers k such that 2^(k-1) == 1 (mod k) and 2^(m-1) == 1 (mod m), where m is defined as m = k*(A000265(k-1) - 1) + 1.
%C The first condition of the definition means terms k are either primes or Fermat pseudoprimes to base 2 (see A001567). If the parameter m is also prime or Fermat pseudoprime to base 2, then k belongs to this sequence.
%C Composites in this sequence: 1771946607940820033, 14356915031659973281, ... - _Max Alekseyev_
%C If k = 1 and p is a composite number, then p == 0 (mod 3). (Is this heuristics or strict? - _R. J. Mathar_, Apr 04 2011) Terms to illustrate these cases:
%C k = 1291 = 2 * 645 + 1; p = 645 = 215 * 3.
%C k = 25603 = 2 * 12801 + 1; p = 12801 = 4267 * 3.
%C k = 424843 = 2 * 212421 + 1; p = 212421 = 70807 * 3.
%C k = 579883 = 2 * 289941 + 1; p = 289941 = 96647 * 3.
%C k = 4325443 = 2 * 2162721 + 1; p = 2162721 = 720907 * 3.
%H Alzhekeyev Ascar M, <a href="/A187849/b187849.txt">Table of n, a(n) for n = 1..2297</a>
%p isA187849 := proc(n) local redn,k,p,m; if modp(2^(n-1),n) = 1 then redn := n-1 ; k := A007814(redn) ; p := (n-1)/2^k ; m := n*(p-1)+1 ; is( modp(2^(m-1),m) = 1 ); else false; end if; end proc:
%p for n from 1 do if isA187849(n) then print(n); end if; end do: # _R. J. Mathar_, Mar 30 2011
%K nonn
%O 1,1
%A _Alzhekeyev Ascar M_, Mar 14 2011
%E Edited by _Max Alekseyev_, Jun 04 2011
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