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A177728
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New form hypergeometric q-form product term recursion:a0 = 4; b0 = 2; c0 = 1;q=2; a(n)=a(n-1)*(1 - q^(a0 + n - 1))*(1 - q^(b0 + n - 1))/((1 - q^n)*(1 - q^(c0 + n - 1)))
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0
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1, 45, 1085, 20925, 366141, 6120765, 100080445, 1618667325, 26038501181, 417737748285, 6692790374205, 107156587499325, 1715081133346621, 27445904805580605, 439171333486530365, 7027036201446788925
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OFFSET
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0,2
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COMMENTS
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The hypergeometric product term recursion is:
a(n) = ((a0 + n - 1)*(b0 + n - 1)/(n*(c0 + n - 1)))*a(n - 1)
I've used a q-form in each term's place to get this recursion.
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REFERENCES
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Harry Hochstadt, The Functions of Mathematical Physics, Dover, New York, 1986 ,page 93
Steve Roman, The Umbral Calculus, Dover Publications, New York (1984), pp.174-182
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LINKS
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Table of n, a(n) for n=0..15.
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FORMULA
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a0 = 4; b0 = 2; c0 = 1;q=2;
a(n)=a(n-1)*(1 - q^(a0 + n - 1))*(1 - q^(b0 + n - 1))/((1 - q^n)*(1 - q^(c0 + n - 1)))
Empirical G.f.: -(14*x+1)/((x-1)*(2*x-1)*(4*x-1)*(8*x-1)*(16*x-1)). [Colin Barker, Nov 27 2012]
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MATHEMATICA
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a0 = 4; b0 = 2; c0 = 1;
a[0, q_] = 1;
a[n_, q_] := a[n, q] = (1 - q^(a0 + n - 1))*(1 - q^(b0 + n - 1))/((1 - q^n)*(1 - q^(c0 + n - 1)))*a[n - 1, q];
Table[Table[a[n, q], {n, 0, 20}], {q, 2, 10}]
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CROSSREFS
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Sequence in context: A162418 A010997 A163721 * A140346 A049447 A215769
Adjacent sequences: A177725 A177726 A177727 * A177729 A177730 A177731
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KEYWORD
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nonn,uned
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AUTHOR
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Roger L. Bagula, May 12 2010
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STATUS
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approved
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