cp's OEIS Frontend

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)

Also number of bracelets (or necklaces) with n red or blue beads such that the beads switch colors when bracelet is turned over.

a(n) is also the number of frieze patterns generated by filling a 1 X n block with n copies of an asymmetric motif (where the copies are chosen from original motif or a 180-degree rotated copy) and then repeating the block by translation to produce an infinite frieze pattern. (Pisanski et al.)

a(n) is also the number of minimal fibrations of a bidirectional n-cycle over the 2-bouquet up to precompositions with automorphisms of the n-cycle. (Boldi et al.) - Sebastiano Vigna, Jan 08 2018

Also, number of "twills" (Grünbaum and Shephard). - N. J. A. Sloane, Oct 21 2015

a(n) is the coefficient of c_1^n*c_2^n in the cycle index polynomial for the dihedral group D_{2*n} evaluated with the figure counting polynomial c = c_1 + c_2, n>=1, abbreviated as Z(D_{2*n},c). See, e.g., the Harary-Palmer reference (given under A212355), p. 42, Theorem (PET), and the example for all 6 two-colored 4-bracelets (called there necklaces) on p. 44, Figure 2.4.2. - Wolfdieter Lang, Jun 05 2012

For k>=1 is column k asymptotic to (k+1)^((k+1)*n-1/2) / (sqrt(2*Pi) * k^(k*n+1/2) * n^(3/2)). - Vaclav Kotesovec, Aug 22 2015

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

Number of chiral pairs of set partitions of a cycle of n elements using exactly two different elements. - Robert A. Russell, Oct 02 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)

For n > 2, a(n) is also number of asymmetric bracelets with n beads of two colors. - Herbert Kociemba, Nov 29 2016