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%I #62 Apr 04 2024 10:12:57
%S 1,1,3,7,14,26,42,66,99,143,201,273,364,476,612,776,969,1197,1463,
%T 1771,2126,2530,2990,3510,4095,4751,5481,6293,7192,8184,9276,10472,
%U 11781,13209,14763,16451,18278,20254,22386,24682,27151,29799,32637,35673
%N Molien series for cyclic group of order 5.
%C a(n) is the number of necklaces with 5 black beads and n white beads.
%C 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
%D B. Sturmfels, Algorithms in Invariant Theory, Springer, '93, p. 65.
%H Vincenzo Librandi, <a href="/A008646/b008646.txt">Table of n, a(n) for n = 0..1000</a>
%H Mónica A. Reyes, Cristina Dalfó, Miguel Àngel Fiol, and Arnau Messegué, <a href="https://arxiv.org/abs/2403.20148">A general method to find the spectrum and eigenspaces of the k-token of a cycle, and 2-token through continuous fractions</a>, arXiv:2403.20148 [math.CO], 2024. See p. 6.
%H <a href="/index/Gre#groups">Index entries for sequences related to groups</a>
%H <a href="/index/Mo#Molien">Index entries for Molien series</a>
%H <a href="/index/Ne#necklaces">Index entries for sequences related to necklaces</a>
%H <a href="/index/Rec#order_09">Index entries for linear recurrences with constant coefficients</a>, signature (4,-6,4,-1,1,-4,6,-4,1).
%F 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)).
%F a(-5-n) = a(n) for all integers.
%F a(n) = ceiling( binomial(n+5, 5) / (n+5) ).
%F G.f.: (1 -3*x +5*x^2 -3*x^3 +x^4)/((1-x)^4*(1-x^5)). - _Michael Somos_, Dec 04, 2001
%F 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
%F G.f.: (4/(1-x^5) + 1/(1-x)^5)/5. - _Herbert Kociemba_, Oct 15 2016
%p 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
%p seq(ceil(binomial(n,4)/5), n=4..41); # _Zerinvary Lajos_, Jan 12 2009
%t k = 5; Table[Apply[Plus, Map[EulerPhi[ # ]Binomial[n/#, k/# ] &, Divisors[GCD[n, k]]]]/n, {n, k, 50}] (* _Robert A. Russell_, Sep 27 2004 *)
%t 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 *)
%t 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 *)
%o (PARI) a(n)=ceil((n+4)*(n+3)*(n+2)*(n+1)/120)
%o (Magma) [Ceiling((n+4)*(n+3)*(n+2)*(n+1)/120): n in [0..50]]; // _Vincenzo Librandi_, Jun 11 2013
%o (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
%o (Sage) [ceil(binomial(n+5,5)/(n+5)) for n in (0..50)] # _G. C. Greubel_, Sep 06 2019
%Y Cf. A000031, A047996.
%K nonn,easy
%O 0,3
%A _N. J. A. Sloane_
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