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A008610
Molien series of 4-dimensional representation of cyclic group of order 4 over GF(2) (not Cohen-Macaulay).
17
1, 1, 3, 5, 10, 14, 22, 30, 43, 55, 73, 91, 116, 140, 172, 204, 245, 285, 335, 385, 446, 506, 578, 650, 735, 819, 917, 1015, 1128, 1240, 1368, 1496, 1641, 1785, 1947, 2109, 2290, 2470, 2670, 2870, 3091, 3311, 3553, 3795, 4060, 4324, 4612, 4900, 5213, 5525, 5863
OFFSET
0,3
COMMENTS
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
REFERENCES
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.
FORMULA
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
EXAMPLE
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].
MAPLE
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);
MATHEMATICA
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 *)
PROG
(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
(Sage)
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
CROSSREFS
Row n=2 of A343874.
Column k=4 of A037306 and A047996.
Sequence in context: A266793 A176222 A365763 * A281688 A078411 A137630
KEYWORD
nonn,easy
EXTENSIONS
Comment and example from Vladeta Jovovic, May 18 2000
STATUS
approved