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A143324 Table T(n,k) by antidiagonals. T(n,k) is the number of length n primitive (=aperiodic or period n) k-ary words (n,k >= 1).

Original entry on oeis.org

1, 2, 0, 3, 2, 0, 4, 6, 6, 0, 5, 12, 24, 12, 0, 6, 20, 60, 72, 30, 0, 7, 30, 120, 240, 240, 54, 0, 8, 42, 210, 600, 1020, 696, 126, 0, 9, 56, 336, 1260, 3120, 4020, 2184, 240, 0, 10, 72, 504, 2352, 7770, 15480, 16380, 6480, 504, 0, 11, 90, 720, 4032, 16800, 46410, 78120, 65280, 19656, 990, 0
Offset: 1

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Author

Alois P. Heinz, Aug 07 2008

Keywords

Comments

Column k is Dirichlet convolution of mu(n) with k^n.
The coefficients of the polynomial of row n are given by the n-th row of triangle A054525; for example row 4 has polynomial -k^2+k^4.

Examples

			T(2,3)=6, because there are 6 primitive words of length 2 over 3-letter alphabet {a,b,c}: ab, ac, ba, bc, ca, cb; note that the non-primitive words aa, bb and cc don't belong to the list; secondly note that the words in the list need not be Lyndon words, for example ba can be derived from ab by a cyclic rotation of the positions.
Table begins:
  1,  2,   3,    4,    5, ...
  0,  2,   6,   12,   20, ...
  0,  6,  24,   60,  120, ...
  0, 12,  72,  240,  600, ...
  0, 30, 240, 1020, 3120, ...
		

Crossrefs

Rows n=1-10 give: A000027, A002378(k-1), A007531(k+1), A047928(k+1), A061167, A218130, A133499, A218131, A218132, A218133.
Main diagonal gives A252764.

Programs

  • Maple
    with(numtheory): f0:= proc(n) option remember; unapply(k^n-add(f0(d)(k), d=divisors(n)minus{n}), k) end; T:= (n,k)-> f0(n)(k); seq(seq(T(n, 1+d-n), n=1..d), d=1..12);
  • Mathematica
    f0[n_] := f0[n] = Function [k, k^n - Sum[f0[d][k], {d, Complement[Divisors[n], {n}]}]]; t[n_, k_] := f0[n][k]; Table[Table[t[n, 1 + d - n], {n, 1, d}], {d, 1, 12}] // Flatten (* Jean-François Alcover, Dec 12 2013, translated from Maple *)

Formula

T(n,k) = Sum_{d|n} k^d * mu(n/d).
T(n,k) = k^n - Sum_{d
T(n,k) = A143325(n,k) * k.
T(n,k) = A074650(n,k) * n.
So Sum_{d|n} k^d * mu(n/d) == 0 (mod n), this is a generalization of Fermat's little theorem k^p - k == 0 (mod p) for primes p to an arbitrary modulus n (see the Smyth link). - Franz Vrabec, Feb 09 2021