

A265026


First differences of A048701.


3



3, 6, 6, 18, 12, 6, 12, 66, 24, 12, 24, 6, 24, 12, 24, 258, 48, 24, 48, 12, 48, 24, 48, 6, 48, 24, 48, 12, 48, 24, 48, 1026, 96, 48, 96, 24, 96, 48, 96, 12, 96, 48, 96, 24, 96, 48, 96, 6, 96, 48, 96, 24, 96, 48, 96, 12, 96, 48, 96, 24, 96, 48
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OFFSET

1,1


COMMENTS

Comment from Altug Alkan, Dec 03 2015: (Start) Except for 3, all terms are divisible by 6 (cf. A048702, A265027).
Proof: Binary palindromes of even length (A048701) are odd for n > 1. So A048701(n+1)  A048701(n) 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+1)  A048701(n) is divisible by 3 for n > 0. In conclusion, A048701(n+1)  A048701(n) is always divisible by 6 for n > 1. (End)


LINKS

N. J. A. Sloane, Table of n, a(n) for n = 1..32767


FORMULA

a(n) = A048701(n+1)  A048701(n).  Altug Alkan, Dec 03 2015


PROG

(PARI) a048701(n) = my(f); f = length(binary(n1))  1; 2^(f+1)*(n1) + sum(i=0, f, bittest(n1, i) * 2^(fi));
vector(100, n, (a048701(n+1)  a048701(n))) \\ Altug Alkan, Dec 03 2015


CROSSREFS

Cf. A048701, A048702, A265027.
Sequence in context: A269525 A036252 A103463 * A304335 A223048 A319446
Adjacent sequences: A265023 A265024 A265025 * A265027 A265028 A265029


KEYWORD

nonn


AUTHOR

N. J. A. Sloane, Nov 30 2015


STATUS

approved



