A152061 Counts of unique periodic binary strings of length n.
0, 0, 2, 2, 4, 2, 10, 2, 16, 8, 34, 2, 76, 2, 130, 38, 256, 2, 568, 2, 1036, 134, 2050, 2, 4336, 32, 8194, 512, 16396, 2, 33814, 2, 65536, 2054, 131074, 158, 266176, 2, 524290, 8198, 1048816, 2, 2113462, 2, 4194316, 33272, 8388610, 2, 16842496, 128, 33555424
Offset: 0
Keywords
Examples
a(3) = 2 = |{ 000, 111 }|, a(4) = 4 = |{ 0000, 1111, 0101, 1010 }|.
Links
- Alois P. Heinz, Table of n, a(n) for n = 0..2000
- Achilles A. Beros, Bjørn Kjos-Hanssen, and Daylan Kaui Yogi, Planar digraphs for automatic complexity, arXiv:1902.00812 [cs.FL], 2019.
Crossrefs
Row sums of A050870.
A050871 is bisection (even part). - R. J. Mathar, Sep 24 2011
Programs
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Maple
with(numtheory): a:= n-> `if`(n=0, 0, 2^n -add(mobius(n/d)*2^d, d=divisors(n))): seq(a(n), n=0..100); # Alois P. Heinz, Sep 26 2011
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Mathematica
a[0] = 0; a[n_] := 2^n - Sum[MoebiusMu[n/d]*2^d, {d, Divisors[n]}]; Table[a[n], {n, 0, 50}] (* Jean-François Alcover, Jul 04 2019 *) -
Python
from sympy import mobius, divisors def A152061(n): return -sum(mobius(n//d)<
Chai Wah Wu, Sep 21 2024
Formula
a(n) = 2^n - A001037(n) * n for n>0, a(0) = 0.
a(n) = 2^n - A027375(n) for n>0, a(0) = 0.
a(n) = 2^n - Sum_{d|n} mu(n/d) 2^d for n>0, a(0) = 0.
a(n) = 2^n - A143324(n,2).
a(n) = 2 * A178472(n) for n > 0. - Alois P. Heinz, Jul 04 2019
Comments