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%I #27 Sep 21 2021 21:31:16
%S 1,80519,107663,1284113,1510313,3933023,4557713,24849833,71871113,
%T 80646143,98058097,104832833,106694033,146987033,168204191,188997463,
%U 205428713,332693873,333681761,336327863,380284847,533039513,552913169,711999113,725943719,805031663,1000519033,1069441313,1476327353,1610020913
%N Numbers k such that 2^(2*k-1) == 1 (mod k).
%C Odd numbers k such that ord(2,k) divides 2*k-1, where ord(2,k) is the multiplicative order of 2 modulo k.
%C Numbers k such that 2*k is in A006935. For k > 1, k is a term if and only if 2*k is an even pseudoprime to base 2.
%C Odd terms in A130421. Complement of A347908 in A130421.
%C Terms > 1 must be composite, since for odd primes p we have 2^(2*p-1) == 2 (mod p). If k > 1 is a term, then 2*k-1 must also be composite, since ord(2,k) | (2*k-1) and ord(2,k) <= eulerphi(k) <= k-1 < 2*k-1.
%C If k > 1 is a term, then (2^(2*k-1) - 1)/k is composite. Proof: since 2*k-1 is composite, write 2*k-1 = u*v, u >= v > 1, then (2^(2*k-1) - 1)/k = (2^u - 1)*(2^(u*(v-1)) + ... + 2^u + 1)/k. Since k | 2^(2*k-1) - 1, there exist positive integers a,b such that a*b = k and that a | 2^u - 1 and b | 2^(u*(v-1)) + ... + 2^u + 1. Note that (2^u - 1)/a, (2^(u*(v-1)) + ... + 2^u + 1)/b >= (2^u - 1)/k >= (2^sqrt(2*k-1) - 1)/k > 1, so (2^(2*k-1) - 1)/k is the product of two integers > 1, so it is composite.
%C 2^t - 1 is a term if and only if 2^(t+1) == 3 (mod t) (t = 1, 111481, 465793, ... in A296370).
%H Jianing Song, <a href="/A347906/b347906.txt">Table of n, a(n) for n = 1..1319</a> (contains all terms below 10^15; based on Max Alekseyev's b-file for A006935)
%F a(n) = A006935(n)/2.
%e 80519 is a term since 80519 divides 2^161037 - 1 (the multiplicative order of 2 modulo 80519 is 261, which is a divisor of 161037). Note that 2 * 80519 = 161038 = A006935(2) is the smallest even pseudoprime to base 2.
%o (PARI) isA347906(k) = if(k%2 && !isprime(k), Mod(2, k)^(2*k-1)==1, 0)
%Y Cf. A006935, A130421, A347908, A296370.
%Y Cf. A347907 (a similar sequence).
%K nonn
%O 1,2
%A _Jianing Song_, Sep 18 2021
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