A003435 Number of directed Hamiltonian circuits on n-octahedron with a marked starting node.
8, 192, 11904, 1125120, 153262080, 28507207680, 6951513784320, 2153151603671040, 826060810479206400, 384600188992919961600, 213656089636192754073600, 139620366072628402087526400, 106033731334825319789808844800
Offset: 2
Examples
n=2: label vertices of a square 1,2,3,4. Then the 8 Hamiltonian circuits are 1234, 1432, 2341, 2143, 3412, 3214, 4123, 4321; so a(2) = 8.
References
- N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
Links
- Vincenzo Librandi, Table of n, a(n) for n = 2..100
- Kenneth P. Bogart and Peter G. Doyle, Nonsexist solution of the menage problem, Amer. Math. Monthly 93 (1986), no. 7, 514-519.
- D. Singmaster, Enumerating unlabeled Hamiltonian circuts, Preprint (1974).
- D. Singmaster, Hamiltonian circuits on the n-dimensional octahedron, J. Combinatorial Theory Ser. B 19 (1975), no. 1, 1-4.
- D. Singmaster, Letter to N. J. A. Sloane, May 1975
- Eric Weisstein's World of Mathematics, Cocktail Party Graph
Programs
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Magma
[(&+[ (-1)^k*2^(k+1)*n*Binomial(n, k)*Factorial(2*n-k-1): k in [0..n]]) : n in [2..20]]; // G. C. Greubel, Nov 17 2022
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Maple
A003435 := n->add((-1)^k*binomial(n,k)*((2*n)/(2*n-k))*2^k*(2*n-k)!,k=0..n);
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Mathematica
a[n_] := 2^n*n!*(2n-1)!!*Hypergeometric1F1[-n, 1-2n, -2]; Table[ a[n], {n, 2, 14}] (* Jean-François Alcover, Nov 04 2011 *) -
PARI
a(n)=sum(k=0,n,(-1)^k*binomial(n,k)*((2*n)/(2*n-k))*2^k*(2*n-k)!) \\ Charles R Greathouse IV, Nov 04 2011
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SageMath
[sum( (-1)^k*2^(k+1)*n*binomial(n, k)*factorial(2*n-k-1) for k in (0..n)) for n in (2..20)] # G. C. Greubel, Nov 17 2022
Formula
For n >= 2, a(n) = Sum_{k=0..n}(-1)^k*binomial(n, k)*((2*n)/(2*n-k))*2^k*(2*n-k)!.
Conjecture: a(n) -(4*n^2 - 2*n + 5)*a(n-1) + 2*(n-1)*(4*n-17)*a(n-2) + 12*(n-1)*(n-2)*a(n-3) = 0. - R. J. Mathar, Oct 02 2013
Recurrence: (2*n-3)*a(n) = 2*n*(4*n^2 - 8*n + 5)*a(n-1) + 4*(n-1)*n*(2*n-1)*a(n-2). - Vaclav Kotesovec, Feb 12 2014
a(n) ~ sqrt(Pi) * 2^(2*n+1) * n^(2*n+1/2) / exp(2*n+1). - Vaclav Kotesovec, Feb 12 2014
a(n) = -(-2)^(n+1)*n!*hypergeom([n, -n], [], 1/2). - Peter Luschny, Nov 10 2016
Extensions
Name made more precise by Andrew Howroyd, May 14 2017
Comments