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A047750 If n mod 2 = 0 then m := n/2 and a(n) = (3*m)!*(5*m+1)/((m+1)!*(2*m+1)!); otherwise m := (n-1)/2, a(n) = 6*(3*m+2)!/(m!*(2*m+3)!). 3
1, 2, 3, 6, 11, 24, 48, 110, 231, 546, 1183, 2856, 6324, 15504, 34884, 86526, 197087, 493350, 1134705, 2861430, 6633315, 16829280, 39268320, 100134216, 234930276, 601661144, 1418201268, 3645533040, 8627761528, 22249511328 (list; graph; refs; listen; history; text; internal format)
OFFSET

0,2

LINKS

Table of n, a(n) for n=0..29.

L. W. Beineke and R. E. Pippert, Enumerating dissectable polyhedra by their automorphism groups, Canad. J. Math., 26 (1974), 50-67.

FORMULA

From Gary W. Adamson, Jul 14 2011: (Start)

a(n) = sum of top row terms in M^n, M = the infinite square production matrix:

1, 1, 0, 0, 0, 0,...

0, 0, 1, 0, 0, 0,...

1, 1, 0, 1, 0, 0,...

0, 0, 1, 0, 1, 0,...

1, 1, 0, 1, 0, 1,...

... (End)

8*n*(n+2)*a(n) +4*(7*n^2-7*n-17)*a(n-1) +6*(-9*n^2+9*n-17)*a(n-2) -21*(3*n-5)*(3*n-7)*a(n-3)=0. - R. J. Mathar, Jul 10 2013

MAPLE

series(RootOf(x*A^3-2*A^2+3*A-1, A)^2, x=0, 30);  # Mark van Hoeij, May 16 2013

MATHEMATICA

a[0] = 1; a[1] = 2; a[n_] := a[n] = 3(2n+3)(3n-4)(3n-2)a[n-2]/(4n(n+2)(2n+1)) + (3(18n+16)a[n-1])/(4n(n+2)(2n+1)); Table[a[n], {n, 0, 29}] (* Jean-Fran├žois Alcover, Dec 02 2016 *)

CROSSREFS

Sequence in context: A176425 A000992 A036648 * A072187 A072374 A122852

Adjacent sequences:  A047747 A047748 A047749 * A047751 A047752 A047753

KEYWORD

nonn

AUTHOR

N. J. A. Sloane.

STATUS

approved

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Last modified February 23 21:20 EST 2020. Contains 332195 sequences. (Running on oeis4.)