A296975 Number of aperiodic normal sequences of length n.

1, 2, 12, 72, 540, 4668, 47292, 545760, 7087248, 102247020, 1622632572, 28091562840, 526858348380, 10641342923148, 230283190977300, 5315654681435520, 130370767029135900, 3385534663249753392, 92801587319328411132, 2677687796244281955480, 81124824998504073834516

Offset: 1

Author

Gus Wiseman, Dec 22 2017

Keywords

• nonn

Comments

A finite sequence is normal if it spans an initial interval of positive integers. It is aperiodic if every cyclic rotation is different.

Examples

			The a(3) = 12 aperiodic normal sequences are 112, 121, 122, 123, 132, 211, 212, 213, 221, 231, 312, 321.
The 15 non-aperiodic normal sequences of length 6 are: 111111, 112112, 121121, 121212, 122122, 123123, 132132, 211211, 212121, 212212, 213213, 221221, 231231, 312312, 321321.
		

Links

• Andrew Howroyd, Table of n, a(n) for n = 1..200

Crossrefs

Cf. A000670, A000740, A001037, A019536, A027375, A060223, A095684, A185700, A296976, A296977, A296978.

Programs

Mathematica

Table[DivisorSum[n,MoebiusMu[n/#]*Sum[k!*StirlingS2[#,k],{k,#}]&],{n,25}]

PARI

\\ here b(n) is A000670.
b(n)={polcoef(serlaplace(1/(2-exp(x+O(x*x^n)))),n)}
a(n)={sumdiv(n, d, moebius(d)*b(n/d))} \\ Andrew Howroyd, Aug 29 2018

Formula

a(n) = n * A060223(n) = Sum_{d|n} mu(d) * A000670(n/d).