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A008646
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Molien series for cyclic group of order 5.
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17
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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
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OFFSET
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0,3
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COMMENTS
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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
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REFERENCES
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B. Sturmfels, Algorithms in Invariant Theory, Springer, '93, p. 65.
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LINKS
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FORMULA
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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
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MAPLE
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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
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MATHEMATICA
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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 *)
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PROG
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(PARI) a(n)=ceil((n+4)*(n+3)*(n+2)*(n+1)/120)
(Magma) [Ceiling((n+4)*(n+3)*(n+2)*(n+1)/120): n in [0..50]]; // Vincenzo Librandi, Jun 11 2013
(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
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CROSSREFS
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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