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A118969
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a(n) = 2*binomial(5*n+1,n)/(4*n+2).
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8
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1, 2, 11, 80, 665, 5980, 56637, 556512, 5620485, 57985070, 608462470, 6474009360, 69682358811, 757366074080, 8300675584120, 91634565938880, 1018002755977245, 11372548404732930, 127677890035721025, 1439777493407492640
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OFFSET
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0,2
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COMMENTS
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If y = x + 2*x^3 + x^5, the series reversion is x = y - 2*y^3 + 11*y^5 - 80*y^7 + 665*y^9 - ... - R. J. Mathar, Sep 29 2012
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LINKS
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FORMULA
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a(n) is sum of top row terms in M^n, where M is an infinite square production matrix with the tetrahedral series in each column (A000292), as follows:
1, 1, 0, 0, 0, 0, ...
4, 1, 1, 0, 0, 0, ...
10, 10, 4, 1, 0, 0, ...
20, 20, 10, 4, 1, 0, ...
35, 35, 20, 10, 4, 1, ...
... (End)
G.f.: hypergeom([1/5, 2/5, 3/5, 4/5],[1/2, 3/4, 5/4], 3125*x/256)^2. - Mark van Hoeij, Apr 19 2013
a(n) = 2*binomial(5n+1,n-1)/n for n>0, a(0)=1. - Bruno Berselli, Jan 19 2014
D-finite with recurrence 8*n*(4*n+1)*(2*n+1)*(4*n-1)*a(n) - 5*(5*n+1)*(5*n-3)*(5*n-2)*(5*n-1)*a(n-1) = 0. - R. J. Mathar, Oct 10 2014
G.f. A(x) satisfies: A(x) = 1 / (1 - x * A(x)^2)^2. - Ilya Gutkovskiy, Nov 13 2021
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EXAMPLE
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a(3) = 80 = sum of top row terms in M^n = (35 + 35 + 9 + 1).
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MATHEMATICA
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Table[2*Binomial[5n+1, n]/(4n+2), {n, 0, 20}] (* Harvey P. Dale, Aug 21 2011 *)
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PROG
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(Magma) [2*Binomial(5*n+1, n)/(4*n+2): n in [0..20]]; // Vincenzo Librandi, Aug 12 2011
(PARI) a(n)=2*binomial(5*n+1, n)/(4*n+2); \\ Joerg Arndt, Apr 20 2013
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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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