cp's OEIS Frontend

From Vladimir Shevelev, Apr 23 2011: (Start)

Also number of non-equivalent necklaces of 11 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=11 (see our comment to A032279).

(End)

This version ignores the offset of A_n and just counts from the beginning of the terms shown in the OEIS entry.

Thus a(8) = 6 because A_8 begins 1,1,2,2,3,4,5,6,... [even though A_8(8) is really 7].

The value a(n) = -1 could arise in two different ways, but it will be easy to decide which. - N. J. A. Sloane, Nov 27 2016

From M. F. Hasler, Sep 22 2013: (Start)

The value of a(91967) can be chosen at will.

Note that this sequence may change if the initial terms in A_n are altered, which does happen from time to time, usually because of the addition of an initial term.

After a(47), currently unknown, the sequence continues with a(48) = A48(47) = 1497207322929, a(49) = A49(48) = unknown, a(50) = A50(49) = unknown, a(51) = A51(50) = 1125899906842625, a(52)=97, a(53) = -1 (since A000053 has only 29 terms). (End)

a(58) = A000058(57) = 138752...985443 (29334988649136302 digits) is too large to include in the b-file. - Pontus von Brömssen, May 21 2022

Two color patterns are equivalent if the colors are permuted. An unoriented cycle counts each chiral pair as one, i.e., they are equivalent.

Adnk[d,n,k] in Mathematica program is coefficient of x^k in A(d,n)(x) in Gilbert and Riordan reference.

T(n,k)=Pi_k(C_n) which is the number of non-equivalent partitions of the cycle on n vertices, with at most k parts. Two partitions P1 and P2 of a graph G are said to be equivalent if there is a nontrivial automorphism of G which maps P1 onto P2. - Bahman Ahmadi, Aug 21 2019

In other words, the number of n-bead bracelet structures using a maximum of k different colored beads. - Andrew Howroyd, Oct 30 2019

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)

For n>2 also number of asymmetric bracelets with n beads of four colors. - Herbert Kociemba, Nov 29 2016

Turning over will not create a new bracelet. Permuting the colors of the beads will not change the structure.

A Lyndon word is the aperiodic necklace representative which is lexicographically earliest among its cyclic shifts. Thus we can apply the fixed density formulas: L_k(n,d)=sum L(n-d, n_1,..., n_(k-1)); n_1+...+n_(k-1)=d where L(n_0, n_1,...,n_(k-1))=(1/n)sum mu(j)*[(n/j)!/((n_0/j)!(n_1/j)!...(n_(k-1)/j)!)]; j|gcd(n_0, n_1,...,n_(k-1)). For this sequence, sum over n_0=even. Alternatively, a(n)=(sum mu(d)*3^(n/d)/n; d|n) - (sum mu(d)*(3^(n/d)-1)/(2n); d|n, d odd).

From Vladimir Shevelev, Apr 23 2011: (Start)

Also number of non-equivalent (turnover) necklaces of 10 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=10 (see our comment to A032279). (End)

From Petros Hadjicostas, Aug 26 2018: (Start)

Assume n >= k >= 2. If a_k(n) is the number of aperiodic necklaces of n beads of 2 colors such that k of them are black and n-k of them are white, then a_k(n) = (1/k)*Sum_{d|gcd(n,k)} mu(d)*binomial(n/d - 1, k/d - 1) = (1/n)*Sum_{d|gcd(n,k)} mu(d)*binomial(n/d, k/d). This follows from Herbert Kociemba's general formula for the g.f. of (a_k(n): n>=1) that can be found in the comments for sequence A032168.

For k prime, we get a_k(n) = floor(binomial(n-1, k-1)/k). In such a case, the sequence becomes a column for triangle A011847. (This is not true when k is composite >= 4.)

(End)