0,3

a(n) is the number of necklaces with 4 black beads and n white beads.

Also nonnegative integer 2 X 2 matrices with sum of elements equal to n, up to rotational symmetry.

The g.f. is Z(C_4,x), the 4-variate cycle index polynomial for the cyclic group C_4, with substitution x[i]->1/(1-x^i), i=1,...,4. Therefore by Polya enumeration a(n) is the number of cyclically inequivalent 4-necklaces whose 4 beads are labeled with nonnegative integers such that the sum of labels is n, for n=0,1,2,... See A102190 for Z(C_4,x). - Wolfdieter Lang, Feb 15 2005

D. J. Benson, Polynomial Invariants of Finite Groups, Cambridge, 1993, p. 104.

E. V. McLaughlin, Numbers of factorizations in non-unique factorial domains, Senior Thesis, Allegeny College, Meadville, PA, April 2004.

G. C. Greubel, 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 linear recurrences with constant coefficients, signature (2,0,-2,2,-2,0,2,-1).

G.f.: (1+2*x^3+x^4)/((1-x)*(1-x^2)^2*(1-x^4)) = (1-x+x^2+x^3)/((1-x)^2*(1-x^2)*(1-x^4)).

a(n) = (1/48)*(2*n^3 + 3*(-1)^n*(n + 4) + 12*n^2 + 25*n + 24 + 12*cos(n*Pi/2)). - Ralf Stephan, Apr 29 2014

G.f.: (1/4)*(1/(1-x)^4 + 1/(1-x^2)^2 + 2/(1-x^4)). - Herbert Kociemba, Oct 22 2016

a(n) = -A032801(-n), per formulae of Colin Barker (A032801) and R. Stephan (above). Also, a(n) - A032801(n+4) = (1+(-1)^signum(n mod 4))/2, i.e., (1,0,0,0,1,0,0,0,...) repeating, (offset 0). - Gregory Gerard Wojnar, Jul 09 2022

There are 10 inequivalent nonnegative integer 2 X 2 matrices with sum of elements equal to 4, up to rotational symmetry:

[0 0] [0 0] [0 0] [0 0] [0 1] [0 1] [0 1] [0 2] [0 2] [1 1]

[0 4] [1 3] [2 2] [3 1] [1 2] [2 1] [3 0] [1 1] [2 0] [1 1].

1/(1-x)/(1-x^2)^2/(1-x^4)*(1+2*x^3+x^4);

seq(coeff(series(%, x, n+1), x, n), n=0..40);

k = 4; Table[Apply[Plus, Map[EulerPhi[ # ]Binomial[n/#, k/# ] &, Divisors[GCD[n, k]]]]/n, {n, k, 30}] (* Robert A. Russell, Sep 27 2004 *)

LinearRecurrence[{2, 0, -2, 2, -2, 0, 2, -1}, {1, 1, 3, 5, 10, 14, 22, 30}, 50] (* G. C. Greubel, Jan 31 2020 *)

(PARI) a(n)=if(n, ([0, 1, 0, 0, 0, 0, 0, 0; 0, 0, 1, 0, 0, 0, 0, 0; 0, 0, 0, 1, 0, 0, 0, 0; 0, 0, 0, 0, 1, 0, 0, 0; 0, 0, 0, 0, 0, 1, 0, 0; 0, 0, 0, 0, 0, 0, 1, 0; 0, 0, 0, 0, 0, 0, 0, 1; -1, 2, 0, -2, 2, -2, 0, 2]^n*[1; 1; 3; 5; 10; 14; 22; 30])[1, 1], 1) \\ Charles R Greathouse IV, Oct 22 2015

(PARI) my(x='x+O('x^50)); Vec((1+2*x^3+x^4)/((1-x)*(1-x^2)^2*(1-x^4))) \\ G. C. Greubel, Jan 31 2020

(Magma) R<x>:=PowerSeriesRing(Integers(), 50); Coefficients(R!( (1+2*x^3+x^4)/((1-x)*(1-x^2)^2*(1-x^4)) )); // G. C. Greubel, Jan 31 2020

(SageMath)

def A008610_list(prec):

P.<x> = PowerSeriesRing(ZZ, prec)

return P( (1+2*x^3+x^4)/((1-x)*(1-x^2)^2*(1-x^4)) ).list()

A008610_list(50) # G. C. Greubel, Jan 31 2020

(GAP) a:=[1, 1, 3, 5, 10, 14, 22, 30];; for n in [9..50] do a[n]:=2*a[n-1]-2*a[n-3] +2*a[n-4]-2*a[n-5]+2*a[n-7]-a[n-1]; od; a; # G. C. Greubel, Jan 31 2020

Row n=2 of A343874.

Column k=4 of A037306 and A047996.

Cf. A000031, A005232, A008804, A047996, A032801.

Sequence in context: A266793 A176222 A365763 * A394589 A281688 A078411

Adjacent sequences: A008607 A008608 A008609 * A008611 A008612 A008613

nonn,easy

Comment and example from Vladeta Jovovic, May 18 2000

approved