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Equivalently, T(n,k) = number of necklaces with k black beads and n-k white beads (binary necklaces of weight k).

The same sequence arises if we take the table U(n,k) = number of necklaces with n black beads and k white beads and read it by antidiagonals (cf. A241926). - Franklin T. Adams-Watters, May 02 2014

U(n,k) is also equal to the number of ways to express 0 as a sum of k elements in Z/nZ. - Jens Voß, Franklin T. Adams-Watters, N. J. A. Sloane, Apr 30 2014 - May 05 2014. See link ("A Note on Modular Partitions and Necklaces") for proof.

The generating function for column k is given by the substitution x_j -> x^j/(1-x^j) in the cycle index of the Symmetric Group of order k. - R. J. Mathar, Nov 15 2018

From Petros Hadjicostas, Jul 12 2019: (Start)

Regarding the comments above by Voss, Adams-Watters, and Sloane, note that Fredman (1975) proved that the number S(n, k, v) of vectors (a_0, ..., a_{n-1}) of nonnegative integer components that satisfy a_0 + ... + a_{n-1} = k and Sum_{i=0..n-1} i*a_i = v (mod n) is given by S(n, k, v) = (1/(n + k)) * Sum_{d | gcd(n, k)} A054535(d, v) * binomial((n + k)/d, k/d) = S(k, n, v).

This result was also proved by Elashvili et al. (1999), who also proved that S(n, k, v) = Sum_{d | gcd(n, k, v)} S(n/d, k/d, 1). Here, S(n, k, 0) = A241926(n, k) = U(n, k) = T(n + k, k) (where T(n, k) is the current array). Also, S(n, k, 1) = A245558(n, k). See also Panyushev (2011) for more general results and for generating functions.

Finally, note that A054535(d, v) = c_d(v) = Sum_{s | gcd(d,v)} s*Moebius(d/s). These are the Ramanujan sums, which equal also the von Sterneck function c_d(v) = phi(d) * Moebius(d/gcd(d, v))/phi(d/gcd(d, v)). We have A054535(d, v) = A054534(v, d).

It would be interesting to see whether there is a proof of the results by Fredman (1975), Elashvili et al. (1999), and Panyushev (2011) for a general v using Molien series as it is done by Sloane (2014) for the case v = 0 (in which case, A054535(d, 0) = phi(d)). (Even though the columns of array A054535(d, v) start at v = 1, we may start the array at column v = 0 as well.)

(End)

U(n, k) is the number of equivalence classes of k-tuples of residues modulo n, identifying those that differ componentwise by a constant and those that differ by a permutation. - Álvar Ibeas, Sep 21 2021

Also structured triakis octahedral numbers (vertex structure 9) (Cf. A100171 = alternate vertex); and structured heptagonal anti-prism numbers (Cf. A100185 = structured anti-prisms).

If Y is a 2-subset of a 2n-set X then, for n>=2, a(n-1) is the number of 4-subsets of X intersecting Y. - Milan Janjic, Nov 18 2007

Let M(2n-1) be a (2n-1)x(2n-1) matrix whose (i,j)-entry equals i^2/(i^2+sqrt(-1)) if i=j and equals 1 otherwise. Then a(n) equals (-1)^(n+1) times the real part of prod(k^2+sqrt(-1),k=1...2n-1) times the determinant of M(2n-1). - John M. Campbell, Sep 07 2011

Principal diagonal of the convolution array A213752. - Clark Kimberling, Jun 20 2012

The Fuss-Catalan numbers are Cat(d,k)= [1/(k*(d-1)+1)]*binomial(k*d,k) and enumerate the number of (d+1)-gon partitions of a (k*(d-1)+2)-gon (cf. Whieldon and Schuetz link). a(n)= Cat(n,4), so enumerates the number of (n+1)-gon partitions of a (4*(n-1)+2)-gon. Analogous series are A000326 (k=3) and A234043 (k=5). Also, a(n)= A006918(4n+1) = A008610(4n+1) = A053307(4n+1) with offset=0. - Tom Copeland, Oct 05 2014

Triangle obtained from A047996 by dropping the first column (k=0), and row (n=0).

T(n, k) = number of different ways the number n can be expressed as ordered sums of k positive integers, counting only once those ordered sums that can be transformed into each other by a cyclic permutation.

These might be described as cyclic compositions, or more loosely as cyclic partitions. - N. J. A. Sloane, Sep 05 2012

The g.f. is Z(C_6,x)/x^6, the 6-variate cycle index polynomial for the cyclic group C_6, with substitution x[i]->1/(1-x^i), i=1,...,6. Therefore by Polya enumeration a(n+6) is the number of cyclically inequivalent 6-necklaces whose 6 beads are labeled with nonnegative integers such that the sum of labels is n, for n=0,1,2,... See A102190 for Z(C_6,x). Note the equivalence of this formulation with the one given as this sequence's name: start with a black 6-necklace (all 6 beads have labels 0). Insert after each of the 6 black beads k white ones if the label was k and then disregard the labels. - Wolfdieter Lang, Feb 15 2005

The g.f. of the CIK[k] transform of the sequence (b(n): n>=1), which has g.f. B(x) = Sum_{n>=1} b(n)*x^n, is CIK[k](x) = (1/k)*Sum_{d|k} phi(d)*B(x^d)^{k/d}. Here, k = 6, b(n) = 1 for all n >= 1, and B(x) = x/(1-x), from which we get another proof of the g.f.s given below. - Petros Hadjicostas, Jan 07 2018

"CIK[ 7 ]" (necklace, indistinct, unlabeled, 7 parts) transform of 1, 1, 1, 1, ...

The g.f. is Z(C_7,x)/x^7, the 7-variate cycle index polynomial for the cyclic group C_7, with substitution x[i]->1/(1-x^i), i=1,...,7. Therefore by Polya enumeration a(n+7) is the number of cyclically inequivalent 7-necklaces whose 7 beads are labeled with nonnegative integers such that the sum of labels is n, for n=0,1,2,... See A102190 for Z(C_7,x) and the comment in A032191 on the equivalence of this problem with the one given in the 'Name' line. - Wolfdieter Lang, Feb 15 2005

From Petros Hadjicostas, Dec 08 2017: (Start)

For p prime, if a_p(n) is the number of necklaces with p black beads and n-p white beads, then (a_p(n): n>=1) = CIK[p](1, 1, 1, 1, ...). Since CIK[k](B(x)) = (1/k)*Sum_{d|k} phi(d)*B(x^d)^{k/d} with k = p and B(x) = x + x^2 + x^3 + ... = x/(1-x), we get Sum_{n>=1} a_p(n)*x^n = ((p-1)/(1 - x^p) + 1/(1 - x)^p)*x^p/p, which is Herbert Kociemba's general formula for the g.f. when p is prime.

We immediately get a_p(n) = ((p-1)/p)*I(p|n) + (1/p)*C(n-1,p-1) = ((p-1)/p)*I(p|n) + (1/n)*C(n,p) = ceiling(C(n,p)/n), which is a generalization of the conjecture made by N. J. A. Sloane and Wolfdieter Lang. (Here, I(condition) = 1 if the condition holds, and 0 otherwise. Also, as usual, for integers n and k, C(n,k) = 0 if 0 <= n < k.)

Since the sequence (a_p(n): n>=1) is column k = p of A047996(n,k) = T(n,k), we get from the documentation of the latter sequence that a_p(n) = T(n, p) = (1/n)*Sum_{d|gcd(n,p)} phi(d)*C(n/d, p/d), from which we get another proof of the formulae for a_p(n).

(End)

The g.f. is Z(C_8,x)/x^8, the 8-variate cycle index polynomial for the cyclic group C_8, with substitution x[i]->1/(1-x^i), i=1,...,8. Therefore by Polya enumeration a(n+8) is the number of cyclically inequivalent 8-necklaces whose 8 beads are labeled with nonnegative integers such that the sum of labels is n, for n=0,1,2,... See A102190 for Z(C_8,x). See the comment in A032191 on the equivalence of this problem with the one given in the `Name' line. - Wolfdieter Lang, Feb 15 2005

From Petros Hadjicostas, Aug 31 2018: (Start)

The CIK[k] transform of sequence (c(n): n>=1) has generating function A_k(x) = (1/k)*Sum_{d|k} phi(d)*C(x^d)^{k/d}, where C(x) = Sum_{n>=1} c(n)*x^n is the g.f. of (c(n): n>=1).

When c(n) = 1 for all n >= 1, we get C(x) = x/(1-x) and A_k(x) = (x^k/k)*Sum_{d|k} phi(d)*(1-x^d)^{-k/d}, which is the g.f. of the number a_k(n) of necklaces of n beads of 2 colors with k of them black and n-k of them white.

Using Taylor expansions, we can easily prove that a_k(n) = (1/k)*Sum_{d|gcd(n,k)} phi(d)*binomial(n/d - 1, k/d - 1) = (1/n)*Sum_{d|gcd(n,k)} phi(d)*binomial(n/d, k/d), which is Robert A. Russell's formula in the Mathematica code below.

For this sequence k = 8, and thus we get the formulae below.

(End)

The g.f. is Z(C_9,x)/x^9, the 9-variate cycle index polynomial for the cyclic group C_9, with substitution x[i]->1/(1-x^i), i=1,...,9. Therefore by Polya enumeration a(n+9) is the number of cyclically inequivalent 9-necklaces whose 9 beads are labeled with nonnegative integers such that the sum of labels is n, for n=0,1,2,... See A102190 for Z(C_9,x). See the comment in A032191 on the equivalence of this problem with the one given in the `Name' line. - Wolfdieter Lang, Feb 15 2005

From Petros Hadjicostas, Jul 12 2019: (Start)

By reading carefully the proof of Lemma 5.1 (pp. 65-66) in Barnes (1959), we see that he actually proved a general result (even though he does not state it in the lemma). For 1 <= k <= n, let T(n, k) be the number of unordered sets b_1, b_2, ..., b_k of k distinct integers from 1..n such that b_1 + b_2 + ... + b_k = 0 (mod n). The proof of Lemma 5.1 in the paper implies that T(n, k) = (1/n) * Sum_{s | gcd(n, k)} (-1)^(k - (k/s)) * phi(s) * binomial(n/s, k/s).

For fixed k >= 1, the g.f. of the sequence (T(n, k): n >= 1) (with T(n, k) = 0 for 1 <= n < k) is (x^k/k) * Sum_{s|k} phi(s) * (-1)^(k - (k/s)) / (1 - x^s)^(k/s).

For k = 4, we get T(n, k=4) = (1/n) * Sum_{d | gcd(n, 4)} (-1)^(4/s) * phi(d) * binomial(n/d, 4/d), which agrees with Barnes' 3-part formula in Lemma 5.1 and with the formula in N. J. A. Sloane's Maple program below. It also agrees with Colin Barker's formula below.

For k = 4, the g.f. is (x^4/4) * Sum_{s|4} phi(s) * (-1)^(4/s) /(1 - x^s)^(4/s), which agrees with Herbert Kociemba's g.f. below.

Barnes' (1959) formula is a special case of Theorem 4 (p. 66) in Ramanathan (1944). If R(n, k, v) is the number of unordered sets b_1, b_2, ..., b_k of k distinct integers from 1..n such that b_1 + b_2 + ... + b_k = v (mod n), then he proved that R(n, k, v) = (1/n) * Sum_{s | gcd(n,k)} (-1)^(k - (k/s)) * binomial(n/s, k/s) * C_s(v), where C_s(v) = A054533(s, v) is Ramanujan's sum (even though it was discovered first around 1900 by the Austrian mathematician R. D. von Sterneck).

Because C_s(v = 0) = phi(s), we get Barnes' (implicit) result; i.e., R(n, k, v=0) = T(n, k) and a(n) = R(n, k=4, v=0) = T(n, k=4).

For k=2, we have R(n, k=2, v=0) = T(n, k=2) = A004526(n-1) for n >= 1. For k=3, we have R(n, k=3, v=0) = T(n, k=3) = A058212(n) for n >= 1. For k=5, we have R(n, k=5, v=0) = T(n, k=5) = A008646(n-5) for n >= 5.

(End)

Von Sterneck (1902, 1903) dealt with this problem. In his notation, we need to find (n)i, the number of integer solutions of the congruence n = x_1 + ... + x_i (mod M) such that 0 <= x_1 < x_2 < ... < x_i < M, where (his n) = 0, (his M) = (our n), and (his i) = 4. He gave several formulas for solving this problem for various cases of his M (M = prime, M = product of two primes, M = power of 2, etc.). - _Petros Hadjicostas, Aug 20 2019

Columns can be computed by a method analogous to that of Fripertinger for isometry classes of linear codes, disallowing scalar transformation of individual coordinates.

Regarding the formula for column k = 1, note that A241926(q - 1, n) counts, up to coordinate permutation, one-dimensional subspaces of (F_q)^n generated by a vector with no zero component.