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%I #18 Dec 22 2018 16:39:04
%S 38,2478782
%N Positive integers k for which 1 + 6*2^(k+2) divides the Fermat number 1 + 2^2^k.
%C On Keller's linked page, to find the terms, you run through the tables and find all rows with k = 3 and with n exactly 3 greater than m, then that m belongs to this sequence. - _Jeppe Stig Nielsen_, Dec 04 2018
%H Wilfrid Keller, <a href="http://www.prothsearch.com/fermat.html">Prime factors k*2^n + 1 of Fermat numbers F_m</a>
%e a(1)=38 because 38 is the smallest positive integer k for which 1 + 6*2^(k+2) divides the Fermat number 1 + 2^2^k.
%t aQ[n_] := PowerMod[2, 2^n, 1 + 6*2^(n+2)] == 6*2^(n+2); Select[Range[3000000], aQ] (* _Amiram Eldar_, Dec 04 2018 *)
%o (PARI) isOK(n) = Mod(2, 1+3*2^(n+3))^(2^n) + 1 == 0 \\ _Jeppe Stig Nielsen_, Dec 03 2018
%Y Cf. A103477, A103478.
%K nonn,bref,hard,more
%O 1,1
%A Serhat Sevki Dincer (mesti_mudam(AT)yahoo.com), Feb 07 2005
%E Sequence name trimmed by _Jeppe Stig Nielsen_, Dec 03 2018