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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

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

From Vladimir Shevelev, Apr 23 2011: (Start)

Also number of non-equivalent necklaces of 5 beads each of them painted by one of n colors.

The sequence solves the so-called Reis problem about convex k-gons in case k=5. The full solution was given by H. Gupta (1979); I gave a short proof of Gupta's result and showed an equivalence of this problem and every one of the following problems: enumerating the bracelets of n beads of 2 colors, k of them black, and enumerating the necklaces of k beads each of them painted by one of n colors.

a(n) is an essentially unimprovable upper estimate for the number of distinct values of the permanent in (0,1)-circulants of order n with five 1's in every row. (End)

a(n+5) is the number of symmetry-allowed, linearly-independent terms at n-th order in the series expansion of the T_1 X h vibronic perturbation matrix, H(Q) (cf. Dunn & Bates). - Bradley Klee, Jul 20 2015

From Petros Hadjicostas, Jul 17 2018: (Start)

Let (c(n): n >= 1) be a sequence of nonnegative integers and let C(x) = Sum_{n>=1} c(n)*x^n be its g.f. Let k be a positive integer. Let a_k = (a_k(n): n >= 1) be the output sequence of the DIK[k] transform of sequence (c(n): n >= 1), and let A_k(x) = Sum_{n>=1} a_k(n)*x^n be its g.f. See Christian G. Bower's web link below. It can be proved that, when k is odd, A_k(x) = ((1/k)*Sum_{d|k} phi(d)*C(x^d)^(k/d) + C(x^2)^((k-1)/2)*C(x))/2.

For this sequence, k = 5, c(n) = 1 for all n >= 1, and C(x) = x/(1-x). Thus, a(n) = a_5(n) for all n >= 1. Since a_k(n) = 0 for 1 <= n <= k-1, the offset of this sequence is n = k = 5. Applying the formula for the g.f. of DIK[5] of (c(n): n >= 1) with C(x) = x/(1-x) and k = 5, we get A(x) = A_5(x) = x^5*((1/5)*Sum_{d|5} phi(d)*(1-x^d)^(-5/d) + (1+x)/(1-x^2)^3)/2, which obviously equals the g.f. in the formula section below.

The g.f. is also a special case of Herbert Kociemba's formula that is valid for both even and odd k: A_k(x) = x^k*((1/k)*Sum_{d|k} phi(d)*(1-x^d)^(-k/d) + (1+x)/(1-x^2)^Floor[(k+2)/2])/2.

Here, a(n) is defined to be the number of n-bead bracelets of two colors with 5 black beads and n-5 white beads. But it is also the number of dihedral compositions of n with 5 positive parts. (This statement is equivalent to Vladimir Shevelev's statement above that a(n) is the "number of non-equivalent necklaces of 5 beads each of them painted by one of n colors." By "necklaces" he means "turnover necklaces". See paragraph (2) of Section 2 in his 2004 paper in the Indian Journal of Pure and Applied Mathematics.)

Two cyclic compositions of n (with k = 5 parts) belong to the same equivalence class corresponding to a dihedral composition of n if and only if one can be obtained from the other by a rotation or reversal of order. (End)

The row polynomials P(n,x) := Sum_{k=0..p(n)-1} a(n,k)*x^k, n >= 1, appear in the numerator of the g.f. G(p(n),x) for the numbers N(p(n),m) of inequivalent m-bead necklaces of two colors with p(n) beads of one color and m-p(n) beads of the other color. Here p(n)=A000040(n) (prime numbers). Equivalently, N(p(n),m) counts inequivalent necklaces with p(n) beads which are labeled with nonnegative numbers, such that the sum of the labels is m. For a proof of this equivalent formulation see a comment in A032191. Inequivalence is meant with respect to the cyclic group C_p(n).

This necklace g.f. is G(p(n),x) = P(n,x)/((1-x^p(n))*(1-x)^(p(n)-1)), n >= 1. The row polynomials P(n,x) are defined above. This g.f. is Z(C_p(n),x), the two variable (x[1] and x[p(n)]) cycle index polynomial for the cyclic group of prime order p(n), with substitution x[1]->1/(1-x^1)and x[p(n)]->1/(1-x^p(n)). This follows by Polya enumeration if the above mentioned labeled necklace problem is solved.

The row length sequence for this array a(n,k) is A000040(n) (n-th prime number), [2, 3, 5, 7, 11, 13, 17, 19, 23, 29, ...].

The rows of this signed array are symmetric: a(n,k) = a(n,p(n)-1-k), n >= 2, k = 0..(p(n)-1)/2. See the explicit formula below.

The formulas for a(n,k), given below, produces in fact integers.

G.f. for column k, k>=0 (without leading zeros): sum(A103718(k,m)*p(n)^m,m=0..k)/k! produces for all n> pi(n) integers, where pi(n):=A000720(n), primes not exceeeding n.

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)

Row less the last element is palindrome for n=odd or n=power of 2 where n is the row number (observation-conjecture).

From Petros Hadjicostas, Jul 13 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).

According to the definition of this sequence, for 1 <= k <= n, T(n, k) 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 = 0 (mod n). The proof of Lemma 5.1 in Barnes (1959) implies that T(n, k) = (1/n) * Sum_{s | gcd(n, k)} (-1)^(k - (k/s)) * phi(s) * binomial(n/s, k/s) for 1 <= k <= n.

For fixed k >= 1, the g.f. of the column (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), which generalizes Herbert Kociemba's formula from A032801.

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) = A054535(s, v) = Sum_{d | gcd(s,v)} d * Moebius(s/d) is Ramanujan's sum (even though it was first discovered 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) for 1 <= k <= n.

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=4, we have R(n, k=4, v=0) = A032801(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.

The reason we have T(2*m+1, k) = A037306(2*m+1, k) = A047996(2*m+1, k) for m >= 0 and k >= 1 is the following. When n = 2*m + 1, all divisors s of gcd(n, k) are odd. In such a case, k - (k/s) is even for all k >= 1, and thus (-1)^(k - (k/s)) = 1, and thus T(n = 2*m+1, k) = (1/n) * Sum_{s | gcd(n, k)} phi(s) * binomial(n/s, k/s) = A037306(2*m+1, k) = A047996(2*m+1, k).

By summing the products of the g.f. of column k times y^k from k = 1 to k = infinity, we get the bivariate g.f. for the array: Sum_{n, k >= 1} T(n, k)*x^n*y^k = Sum_{s >= 1} (phi(s)/s) * log((1 - x^s + (-x*y)^s)/(1 - x^s)) = -x/(1 - x) - Sum_{s >= 1} (phi(s)/s) * log(1 - x^s + (-x*y)^s).

Letting y = 1 in the above bivariate g.f., we get the g.f. of the sequence (Sum_{1 <= k <= n} T(n, k): n >= 1) is -x/(1 - x) - Sum_{s >= 1} (phi(s)/s) * log(1 - x^s + (-x)^s) = -x/(1 - x) + Sum_{m >= 0} (phi(2*m + 1)/(2*m + 1)) * log(1 - 2*x^(2*m+1)), which is the g.f. of sequence A082550. Hence, sequence A082550 consists of the row sums.

There is another important interpretation of this array T(n, k) which is related to some of the references for sequence A047996, but because the discussion is too lengthy, we omit the details.

(End)