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%I #6 Aug 20 2023 16:57:40
%S 0,1,2,4,8,16,25,32,45,55,64,91,95,99,125,128,135,143,153,155,161,175,
%T 187,225,235,245,247,256,261,273,275,279,285,289,297,319,329,333,335,
%U 355,363
%N Numbers k such that ceiling(2^(2^k mod k)/3) is odd.
%C Conjecture: the odd values of ceiling(2^(2^k mod k)/3) are Jacobsthal numbers (and the even values are 1 plus a Jacobsthal number).
%Y Cf. A005578, A001045.
%K easy,nonn
%O 1,3
%A _Paul Barry_, Oct 08 2005
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