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%I #22 Dec 22 2018 16:38:48
%S 5,23,73,125,1945,23471
%N Positive integers k for which 1 + 5*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 = 5 and with n exactly 2 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)=5 because 5 is the smallest positive integer k for which 1 + 5*2^(k+2) divides the Fermat number 1 + 2^2^k.
%t Select[Range[1, 2000], Mod[1 + PowerMod[2, 2^#, 1 + 5*2^(# + 2)], 1 + 5*2^(# + 2)] == 0 &] (* _Julien Kluge_, Jul 08 2016 *)
%o (PARI) isok(n) = Mod(2, 1+5*2^(n+2))^(2^n) + 1 == 0; \\ _Michel Marcus_, Apr 29 2016
%Y Cf. A000215, A083575, A103477, A103479.
%K nonn,more,hard
%O 1,1
%A Serhat Sevki Dincer (mesti_mudam(AT)yahoo.com), Feb 07 2005
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