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A265027 First differences of A048701 divided by 6.

%I A265027 #28 Dec 26 2022 13:18:12
%S A265027 1,1,3,2,1,2,11,4,2,4,1,4,2,4,43,8,4,8,2,8,4,8,1,8,4,8,2,8,4,8,171,16,
%T A265027 8,16,4,16,8,16,2,16,8,16,4,16,8,16,1,16,8,16,4,16,8,16,2,16,8,16,4,
%U A265027 16,8,16,683,32,16,32,8,32,16,32,4,32,16,32,8,32,16,32,2,32,16,32,8,32,16,32,4,32,16,32,8
%N A265027 First differences of A048701 divided by 6.
%C A265027 Indices n such that a(n) = 1 are equal to row sums of Lucas triangle. In other words, a(A042950(n)) = 1. Additionally, a(A070875(n)) = 2 and a(A123760(n)) = 4. - _Altug Alkan_, Dec 04 2015
%H A265027 N. J. A. Sloane, <a href="/A265027/b265027.txt">Table of n, a(n) for n = 2..32767</a>
%F A265027 a(n) = A265026(n) / 6, for n > 1. - _Altug Alkan_, Dec 03 2015
%t A265027 Differences@ Select[Range@ 12000, Reverse@ # == # && EvenQ@ Length@ # &@ IntegerDigits[#, 2] &]/6 (* _Michael De Vlieger_, Dec 04 2015 *)
%o A265027 (PARI) a048701(n) = my(f); f = length(binary(n)) - 1; 2^(f+1)*n + sum(i=0, f, bittest(n, i) * 2^(f-i));
%o A265027 vector(100, n, (a048701(n+1) - a048701(n)) / 6) \\ _Altug Alkan_, Dec 03 2015
%Y A265027 Cf. A042950, A048701, A070875, A123760, A265026.
%K A265027 nonn
%O A265027 2,3
%A A265027 _N. J. A. Sloane_, Nov 30 2015