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A301453 a(n) is the number of ways of writing the binary expansion of n as a concatenation of nonempty substrings with no two consecutive equal substrings. 6

%I #47 Apr 15 2018 15:03:06

%S 1,1,2,1,3,4,3,3,6,7,7,6,5,6,6,4,10,13,14,11,11,14,14,12,9,11,11,9,9,

%T 12,10,7,17,23,26,20,20,26,25,21,23,26,28,22,22,27,26,20,16,20,22,17,

%U 17,22,20,18,18,21,23,18,16,20,17,14,31,40,46,36,39,49

%N a(n) is the number of ways of writing the binary expansion of n as a concatenation of nonempty substrings with no two consecutive equal substrings.

%C Leading zeros in the binary expansion of n are ignored.

%C The value a(0) = 1 corresponds to the empty concatenation.

%C The following sequences f correspond to the numbers of ways of writing the binary expansion of a number as a concatenation of substrings with some specific features:

%C f f(2^n-1) Features

%C ------- -------- --------

%C A215244 A011782 Substrings are palindromes.

%C A301453 A003242 This sequence; no two consecutive equal substrings.

%C A302395 A032020 All substrings are distinct.

%C A302436 A000012 Substrings with Hamming weight at most 1.

%C A302437 A000045 Substrings with Hamming weight at most 2.

%C A302439 A000012 Substrings are aperiodic.

%C For any such sequence f, the function n -> f(2^n-1) corresponds to a composition of n.

%H Rémy Sigrist, <a href="/A301453/b301453.txt">Table of n, a(n) for n = 0..10000</a>

%H Rémy Sigrist, <a href="/A301453/a301453.png">Scatterplot of the second ordinal transform of the first 1000000 terms</a>

%H <a href="/index/Bi#binary">Index entries for sequences related to binary expansion of n</a>

%F a(2^n - 1) = A003242(n) for any n >= 0.

%e For n = 19: the binary expansion of 19, "10011", can be split in 11 ways into nonempty substrings with no two consecutive equal substrings:

%e - (10011),

%e - (1001)(1),

%e - (100)(11),

%e - (10)(011),

%e - (10)(01)(1),

%e - (10)(0)(11),

%e - (1)(0011),

%e - (1)(001)(1),

%e - (1)(00)(11),

%e - (1)(0)(011),

%e - (1)(0)(01)(1).

%e Hence a(19) = 11.

%o (PARI) a(n{, pp=0}) = if (n==0, return (1), my (v=0, p=1); while (n, p=(p*2) + (n%2); n\=2; if (p!=pp, v+=a(n, p))); return (v))

%Y Cf. A000012, A000045, A003242, A011782, A032020, A215244, A301453, A302395, A302436, A302437, A302439.

%K nonn,base

%O 0,3

%A _Rémy Sigrist_, Apr 08 2018

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