cp's OEIS Frontend

From Richard L. Ollerton, May 07 2021: (Start)

Here, as in A000031, turning over is not allowed.

(1/n) * Dirichlet convolution of phi(n) and k^n. (End)

For prime rows, these appear to be evaluations of Moreau's necklace polynomials at the integers with several combinatorial interpretations (see Wikipedia link). - Tom Copeland, Oct 20 2014

From Petros Hadjicostas, Nov 05 2017: (Start)

The g.f. for column k follows easily from I. Gessel's formulas for this sequence. Since S(1,k) = k-1, we have T(1,k) = k + S(1,k) - (k - 1). Thus, Sum_{n >= 1} T(n,k)*x^n = k*x + Sum_{n >= 1} (1/n)*Sum_{d|n} (k - 1)^d*phi(n/d)*x^n - Sum_{s=0} (k-1)*x^{2*s+1}. Letting m = n/d, we get that (column k g.f.) = k*x - (k - 1)*x/(1 -x^2) + Sum_{m >= 1} (phi(m)/m)*Sum_{d >= 1}((k - 1)*x^m)^d/d. But Sum_{d>=1} z^d/d = -log(1 - z), and so (column k g.f.) = k*x - (k - 1)*x/(1 - x^2) - Sum_{m >= 1} (phi(m)/m)*log(1 - (k - 1)*x^m).

The other formula for the g.f. of column k of this sequence follows from the formula Sum_{n >= 1} (phi(n)/n)*log(1 + t^n) = t/(1 - t^2), which in turn follows from the well-known series Sum_{n >= 1} phi(n)*t^n/(1 + t^n) = t*(1 + t^2)/(1 - t^2)^2.

The extra term k*x in the g.f. for column k is due to the fact that we conventionally assume that a necklace with only one bead, colored with one of the k colors available, is such that there are "no adjacent beads having the same color" (even though, strictly speaking, a single bead is adjacent to itself when we go around the circle of the necklace).

One can use the g.f. for column k to derive the so-called "Empirical for row n" formulae that are denoted by a(k) and given in the formula section below (from n = 1 to n = 7). For example, for n = 3, a(k) = a(k, x=0), where a(k, x) = (1/3!)*d^3/dx^3 (column k g.f.). Here, d^3/dx^3 stands for "third derivative w.r.t. x". If we let f(x) = x/(1 - x^2) and g(x, m) = -log(1 - (k - 1)*x^m), then f^{(3)}(0) = 6, while g^{(3)}(0,m) = 2*(k - 1)^3 for m = 1, 0 for m=2, 6*(k - 1) for m = 3, and 0 for m >= 4. Then, a(k) = (1/6)*(-6*(k - 1) + 2*(k - 1)^3 + (2/3)*6*(k - 1)) = (1/3)*k^3 - k^2 + (2/3)*k. Using this method, one can derive an "Empirical for row n" formula for a(k) for any positive integer n. (End)

T(n,k) is the number of n-bead necklaces with up to k different colored beads. - Yves-Loic Martin, Sep 29 2020

Number of ways to color vertices of a hexagon using <= n colors, allowing rotations and reflections.

Equivalently, the number of distinct hexagons that can be tiled using equilateral triangles of n different colors. - Lekraj Beedassy, Dec 29 2007

Number of ways to color slots of a 2 X 3 matrix with the respective symmetric groups S_2 and S_3 acting on the rows / columns. - Marko Riedel, Jan 26 2017

For the base case of m=2 the sequence counts circulant digraphs up to Cayley isomorphism. Two circulant graphs are Cayley isomorphic if there is a d, which is necessarily prime to n, that transforms through multiplication modulo n the step values of one graph into those of the other. For squarefree n this is the only way that two circulant graphs can be isomorphic. (See Liskovets reference for a proof.)

Alternatively, the number of mappings with domain {1..n-1} and codomain {1..m} up to equivalence. Mappings A and B are equivalent if there is a d, prime to n, such that A(i) = B(i*d mod n) for i in {1..n-1}. This sequence differs from A132191 only in that sequence also includes 0 in the domain which introduces an extra factor of m into the results since zero multiplied by anything is zero.

All column sequences are polynomials of order n-1 and these are the cycle index polynomials.

This sequence is also related to A075195(n, m) which counts necklaces and A285548(m, n) which is the sequence described in the Titsworth reference. In particular, A075195 is the analogous array with equivalence determined through the additive group instead of by multiplication whereas A285548 allows for both addition and multiplication.