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%I #27 Apr 10 2017 12:24:33
%S 3,14,52,184,656,34688,2118656,134438912,537346048,9007202811510784,
%T 2417851639318318791262208,633825300114170432793740312576,
%U 2535301200456572518883997515776
%N Numbers of the form 2^n*(2^(n+1)+2n+1) where 2^(n+1)+2n+1 is prime.
%C This sequence is a subsequence of A083874 (see A083874 & A105330).
%C This is because these numbers satisfy tau(n) + sigma(n) = 2n when n = 2^k * p with p is prime; for instance tau(14) + sigma(14) = 4 + 24 = 28 = 2 x 14. [See References.] - _Bernard Schott_, Apr 07 2017
%D J.-M. De Koninck and A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Ellipses, Problème 723, page 93.
%F a(n) = 2^A105330(n)*(2^(A105330(n)+1) + 2*A105330(n) + 1). - _Bernard Schott_, Apr 07 2017
%e 9007202811510784 is in the sequence because 9007202811510784 = 2^26*(2^27 + 2*26 + 1) and 2^27 + 2*26 + 1 is prime.
%t Do[If[PrimeQ[2^(m + 1) + 2m + 1], Print[2^m(2^(m + 1) + 2m + 1)]], {m, 0, 110}]
%t 2^# (2^(#+1)+2#+1)&/@Select[Range[0,100],PrimeQ[2^(#+1)+2#+1]&] (* _Harvey P. Dale_, Nov 13 2012 *)
%Y Cf. A007503, A083874, A105330.
%K nice,nonn
%O 1,1
%A _Farideh Firoozbakht_, Apr 28 2005
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