A265027 - OEIS

%I #28 Dec 26 2022 13:18:12

%S 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 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 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 First differences of A048701 divided by 6.

%C 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 N. J. A. Sloane, <a href="/A265027/b265027.txt">Table of n, a(n) for n = 2..32767</a>

%F a(n) = A265026(n) / 6, for n > 1. - _Altug Alkan_, Dec 03 2015

%t Differences@ Select[Range@ 12000, Reverse@ # == # && EvenQ@ Length@ # &@ IntegerDigits[#, 2] &]/6 (* _Michael De Vlieger_, Dec 04 2015 *)

%o (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 vector(100, n, (a048701(n+1) - a048701(n)) / 6) \\ _Altug Alkan_, Dec 03 2015

%Y Cf. A042950, A048701, A070875, A123760, A265026.

%K nonn

%O 2,3

%A _N. J. A. Sloane_, Nov 30 2015