cp's OEIS Frontend

Variant of A047812 (Parker's partition triangle).

Column 1 equals the number of partitions of n: A000041(n) is the coefficient of q^n in the central q-binomial coefficient [2*n, n] for n > 0.

Let X = A003239, Y = A000013, Z = A000048.

Rotations producing the complementary and the same necklace: CR and SR

There are X(n) balanced bicolor necklaces (BBN) of length 2n. (Central numbers of A047996.)

Y(n) among them are self-complementary (SCBBN). (They can be rotated so that all beads change color.)

Z(n) among those are primitive (not periodic). Each has a unique CR and SR. (SR is trivial rotation.)

The other Y(n)-Z(n) = A115118(n) SCBBN have multiple CR and SR.

T(n,k) SCBBN have k different CR and SR.

Column 1 is Z. The other columns have the same positive entries, each preceded by k-1 zeros.

One could add a column 0 to this triangle, whose entries would be X(n)-Y(n) = 2*A386388(n).

Triangle A385666 does the same for SR of all BBN.

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)

a(n) = number of necklaces of n white beads and 2n black beads. - David Callan, Mar 28 2004

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

There are A003239(n) balanced binary necklaces of length 2n. (Central numbers of A047996.)

T(n,k) is the number of those that can be rotated into themselves in k different ways (at least 1 for the trivial rotation).

A022553(n) necklaces (corresponding to Lyndon words) have only the trivial rotation.

All columns have the same positive entries, each preceded by k-1 zeros.

Compare triangle A385665, which counts only self-complementary balanced binary necklaces.

Left half of even rows of triangle A047996 (with the leftmost edge discarded).

Left half of even rows of table A052307 with left column deleted.