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Integers of the form (2^k + 1)/(2k + 1).
2

%I #19 Nov 28 2014 22:46:04

%S 1,2,3,5,27,565,7085,48771,1266205,9099507,17602325,128207979,

%T 26494256091,11147523830125,84179432287299,165269711096165,

%U 281629680514649643,4246732448623781667,126774939137440139965,1925041114036033717685,14833445639443302757131

%N Integers of the form (2^k + 1)/(2k + 1).

%C a(A103579(n)) is a subsequence.

%C Numbers n such that 2n + 1 divides 2^n + 1: 0, 1, 2, 5, 6, 9, 14, 18, 21, 26, 29, 30, 33, 41, 50, 53, ...

%H Colin Barker, <a href="/A247094/b247094.txt">Table of n, a(n) for n = 1..400</a>

%e 1 is in this sequence because (2^1 + 1)/(2*1 + 1) = 1,

%e 2 is in this sequence because (2^0 + 1)/(2*0 + 1) = 2,

%e 3 is in this sequence because (2^5 + 1)/(2*5 + 1) = 3.

%o (PARI) s=[]; for(k=0, 100, t=(2^k + 1)/(2*k + 1); if(type(t)=="t_INT", s=concat(s, t))); s=vecsort(s,,8) \\ _Colin Barker_, Nov 18 2014

%Y Cf. A081856, A081858, A103579, A224486, A247132.

%K nonn

%O 1,2

%A _Juri-Stepan Gerasimov_, Nov 18 2014

%E a(19) corrected by _Colin Barker_, Nov 18 2014