A369925 Number of uniform circular words of length n with adjacent elements unequal using an infinite alphabet up to permutations of the alphabet.

1, 0, 1, 1, 2, 1, 6, 1, 33, 23, 295, 1, 4877, 1, 44191, 141210, 749316, 1, 31762349, 1, 309754506, 3980911205, 4704612121, 1, 1303743206944, 55279816357, 2737023412201, 343866841144704, 564548508168226, 1, 145630899385513158, 1, 2359434158555273239

Offset

0,5

Comments

A word is uniform here if each symbol that occurs in the word occurs with the same frequency.

a(n) is the number of ways to partition [n] into parts of equal size and no part containing values that differ by 1 modulo n.

Links

Andrew Howroyd, Table of n, a(n) for n = 0..200

Formula

a(n) = Sum_{d|n} A369923(d, n/d} for n > 0.

a(p) = 1 for prime p.

Example

a(1) = 0 because the symbol 'a' is considered to be adjacent to itself in a circular word. The set partition {{1}} is also excluded because 1 == 1 + 1 (mod 1).
The a(6) = 6 words are ababab, abacbc, abcabc, abcacb, abcbac, abcdef.
The corresponding a(6) = 6 set partitions are:
   {{1,3,5},{2,4,6}},
   {{1,3},{2,5},{4,6}},
   {{1,4},{2,5},{3,6}},
   {{1,4},{2,6},{3,5}},
   {{1,5},{2,4},{3,6}},
   {{1},{2},{3},{4},{5},{6}}.

Prog

(PARI) \\ Needs T(n, k) from A369923.
a(n) = {if(n==0, 1, sumdiv(n, d, T(d, n/d)))}

Crossrefs

The case for adjacent elements possibly equal is A038041.

Cf. A369923, A369924 (linear words).

Sequence in context: A277440 A321597 A083720 * A055878 A331654 A346864

Adjacent sequences: A369922 A369923 A369924 * A369926 A369927 A369928

Keyword

nonn

Author

Andrew Howroyd, Feb 06 2024

Status

approved