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A374841 Numbers k such that 2^(2^k-2) == 1 (mod k^2).

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%I A374841 #32 Jan 05 2025 09:05:14
%S A374841 3,7,19,43,73,127,163,337,341,379,487,601,881,883,937,1387,1459,1801,
%T A374841 2593,2647,2857,3079,3529,3673,3943,4057,4201,4681,5419,5461,5881,
%U A374841 6121,6481,6529,6553,6571,6841,7481,7993,8233,8911,9001,9199,9241,9721,10261,10657,11161,11827,12241
%N A374841 Numbers k such that 2^(2^k-2) == 1 (mod k^2).
%C A374841 If p is an odd prime and 2^(2^p-2) == 1 (mod p), then 2^(2^p-2) == 1 (mod p^2).
%C A374841 If 2^(k-1) == 1 (mod k) and 2^(2^k-2) == 1 (mod k), then 2^(2^k-2) == 1 (mod k^2).
%C A374841 Composite terms that are not Fermat pseudoprimes to base 2 are 66709, 951481, ...
%C A374841 Note that 66709 = 19*3511 and 951481 = 271*3511, where 3511 is a Wieferich prime.
%H A374841 Amiram Eldar, <a href="/A374841/b374841.txt">Table of n, a(n) for n = 1..10000</a> (terms 1..250 from Michel Marcus)
%e A374841 3 is a term, because 3^2 divides 2^(2^3-2) - 1 = 2^6 - 1 = 63.
%t A374841 Select[Range[12500], PowerMod[2, 2^# - 2, #^2] == 1 &] (* _Amiram Eldar_, Jul 22 2024 *)
%o A374841 (PARI) isok(k) = Mod(2, k^2)^(2^k-2) == 1; \\ _Michel Marcus_, Jan 05 2025
%Y A374841 Cf. A001220, A001567, A069051 (> 2 is a subsequence), A217468 (subsequence).
%K A374841 nonn
%O A374841 1,1
%A A374841 _Thomas Ordowski_, Jul 22 2024
%E A374841 More terms from _Amiram Eldar_ Jul 22 2024