A008646 Molien series for cyclic group of order 5.

1, 1, 3, 7, 14, 26, 42, 66, 99, 143, 201, 273, 364, 476, 612, 776, 969, 1197, 1463, 1771, 2126, 2530, 2990, 3510, 4095, 4751, 5481, 6293, 7192, 8184, 9276, 10472, 11781, 13209, 14763, 16451, 18278, 20254, 22386, 24682, 27151, 29799, 32637, 35673

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

Comments

a(n) is the number of necklaces with 5 black beads and n white beads.

The g.f. is Z(C_5,x), the 5-variate cycle index polynomial for the cyclic group C_5, with substitution x[i]->1/(1-x^i), i=1,...,5. Therefore by Polya enumeration a(n) is the number of cyclically inequivalent 5-necklaces whose 5 beads are labeled with nonnegative integers such that the sum of labels is n, for n=0,1,2,... See A102190 for Z(C_5,x). - Wolfdieter Lang, Feb 15 2005

References

• B. Sturmfels, Algorithms in Invariant Theory, Springer, '93, p. 65.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• David Broadhurst and Xavier Roulleau, Number of partitions of modular integers, arXiv:2502.19523 [math.NT], 2025. See p. 19.
• Mónica A. Reyes, Cristina Dalfó, Miguel Àngel Fiol, and Arnau Messegué, A general method to find the spectrum and eigenspaces of the k-token of a cycle, and 2-token through continuous fractions, arXiv:2403.20148 [math.CO], 2024. See p. 6.
• Index entries for sequences related to groups
• Index entries for Molien series
• Index entries for sequences related to necklaces
• Index entries for linear recurrences with constant coefficients, signature (4,-6,4,-1,1,-4,6,-4,1).

Crossrefs

Cf. A000031, A047996.

Programs

Magma

[Ceiling((n+4)*(n+3)*(n+2)*(n+1)/120): n in [0..50]]; // Vincenzo Librandi, Jun 11 2013

Maple

seq(coeff(series((1+x^2+3*x^3+4*x^4+6*x^5+4*x^6+3*x^7+x^8+x^10)/((1-x)* (1-x^2)*(1-x^3)*(1- x^4)*(1-x^5)), x, n+1), x, n), n = 0..50); # corrected by G. C. Greubel, Sep 06 2019
seq(ceil(binomial(n,4)/5), n=4..41); # Zerinvary Lajos, Jan 12 2009

Mathematica

k = 5; Table[Apply[Plus, Map[EulerPhi[ # ]Binomial[n/#, k/# ] &, Divisors[GCD[n, k]]]]/n, {n, k, 50}] (* Robert A. Russell, Sep 27 2004 *)
CoefficientList[Series[(1 +x^2 +3*x^3 +4*x^4 +6*x^5 +4*x^6 +3*x^7 +x^8 +x^10)/((1-x)*(1-x^2)*(1-x^3)*(1- x^4)*(1-x^5)), {x,0,50}], x] (* Vincenzo Librandi, Jun 11 2013 *)
LinearRecurrence[{4,-6,4,-1,1,-4,6,-4,1}, {1,1,3,7,14,26,42,66,99}, 50] (* Harvey P. Dale, Jan 11 2017 *)

PARI

a(n)=ceil((n+4)*(n+3)*(n+2)*(n+1)/120)

PARI

Vec((1-3*x+5*x^2-3*x^3+x^4)/((1-x)^4*(1-x^5)) + O(x^50)) \\ Altug Alkan, Oct 31 2015

Sage

[ceil(binomial(n+5,5)/(n+5)) for n in (0..50)] # G. C. Greubel, Sep 06 2019

Formula

G.f.: (1 +x^2 +3*x^3 +4*x^4 +6*x^5 +4*x^6 +3*x^7 +x^8 +x^10)/((1-x)*(1-x^2)*(1-x^3)*(1- x^4)*(1-x^5)).

a(-5-n) = a(n) for all integers.

a(n) = ceiling( binomial(n+5, 5) / (n+5) ).

G.f.: (1 -3*x +5*x^2 -3*x^3 +x^4)/((1-x)^4*(1-x^5)). - Michael Somos, Dec 04 2001

a(n) = (n^4 +10*n^3 +35*n^2 +50*n +24*(3 -2*(-1)^(2^(n-5*floor(n/5)) )))/120. - Luce ETIENNE, Oct 31 2015

G.f.: (4/(1-x^5) + 1/(1-x)^5)/5. - Herbert Kociemba, Oct 15 2016