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A265026 First differences of A048701.

%I A265026 #27 Dec 26 2022 13:16:18
%S A265026 3,6,6,18,12,6,12,66,24,12,24,6,24,12,24,258,48,24,48,12,48,24,48,6,
%T A265026 48,24,48,12,48,24,48,1026,96,48,96,24,96,48,96,12,96,48,96,24,96,48,
%U A265026 96,6,96,48,96,24,96,48,96,12,96,48,96,24,96,48
%N A265026 First differences of A048701.
%C A265026 Comment from _Altug Alkan_, Dec 03 2015: (Start) Except for 3, all terms are divisible by 6 (cf. A048702, A265027).
%C A265026 Proof: Binary palindromes of even length (A048701) are odd for n > 0. So A048701(n) - A048701(n-1) is an even number for n > 1. Because the length is even and palindromic numbers are symmetric, for any digit “1” that is related with 2^n in its expansion which n is even, there are another digit “1” that is related with 2^m in its expansion which m is odd. 2^n+2^m is always divisible by 3 if n is even and m is odd. Therefore A048701(n) is divisible by 3, so A048701(n) - A048701(n-1) is divisible by 3 for n > 0. In conclusion, A048701(n) - A048701(n-1) is always divisible by 6 for n > 1. (End)
%H A265026 N. J. A. Sloane, <a href="/A265026/b265026.txt">Table of n, a(n) for n = 1..32767</a>
%F A265026 a(n) = A048701(n) - A048701(n-1). - _Altug Alkan_, Dec 03 2015
%o A265026 (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 A265026 vector(100, n, (a048701(n) - a048701(n-1))) \\ _Altug Alkan_, Dec 03 2015
%Y A265026 Cf. A048701, A048702, A265027.
%K A265026 nonn
%O A265026 1,1
%A A265026 _N. J. A. Sloane_, Nov 30 2015