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%I #18 Sep 20 2021 17:42:33
%S 2,14,1022,20066,485918,2531678,3677198,8277458,8893262,21122318,
%T 26358638,39852014,42448478,76712318,131492498,144322478,164360606,
%U 175126478,176647378,196705598,249126626,306789074,317051378,438023138,497041538,696970718,996520658
%N Even numbers k such that 2^(2*k) == 2 (mod k).
%C Numbers of the form 2*t where 2^(4*t-1) == 1 (mod t).
%C Even terms in A130421. Complement of A347906 in A130421.
%C If k > 14 is a term, then k/2 must be composite, since for odd primes p we have 2^(4*p) == 16 (mod 2*p). If k = 2*t > 2 is a term, then 2*k-1 = 4*t-1 must also be composite, since ord(2,t) | (4*k-1) and ord(2,t) <= eulerphi(t) <= t-1 < 4*t-1.
%C If k = 2*t > 2 is a term, then (2^(2*k) - 2)/k = (2^(4*k-1) - 1)/t is composite. See A347907 for a proof.
%C 2*(2^t - 1) is a term if and only if 2^(t+2) == 5 (mod t) (t = 1, 3, 9, 871, 2043, 2119, 8769, ...).
%H Jianing Song, <a href="/A347908/b347908.txt">Table of n, a(n) for n = 1..1298</a> (contains all terms below 10^15; based on Max Alekseyev's b-file for A130421)
%F a(n) = A347907(n)*2.
%e 14 is a term since 14 divides 2^28 - 2.
%o (PARI) isA347908(k) = if(k%4==2, k=k>>1; if(isprime(k) && k!=7, 0, Mod(2, k)^(4*k-1)==1), 0)
%Y Cf. A130421, A347906, A347907.
%K nonn
%O 1,1
%A _Jianing Song_, Sep 18 2021
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