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 A177728 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))) 0
 1, 45, 1085, 20925, 366141, 6120765, 100080445, 1618667325, 26038501181, 417737748285, 6692790374205, 107156587499325, 1715081133346621, 27445904805580605, 439171333486530365, 7027036201446788925 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,2 COMMENTS 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. REFERENCES 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 LINKS FORMULA 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] MATHEMATICA 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}] CROSSREFS Sequence in context: A162418 A010997 A163721 * A140346 A049447 A215769 Adjacent sequences:  A177725 A177726 A177727 * A177729 A177730 A177731 KEYWORD nonn,uned AUTHOR Roger L. Bagula, May 12 2010 STATUS approved

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