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A020926
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Expansion of 1/(1-4*x)^(15/2).
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4
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1, 30, 510, 6460, 67830, 624036, 5200300, 40116600, 290845350, 2003601300, 13223768580, 84151254600, 518932736700, 3113596420200, 18236779032600, 104557533120240, 588136123801350, 3252046802195700, 17705588145287700, 95051052148386600, 503770576386448980
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OFFSET
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0,2
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LINKS
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FORMULA
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a(n) = ((2*n+13)*(2*n+11)*(2*n+9)*(2*n+7)*(2*n+5)*(2*n+3)*(2*n+1)/135135) * binomial(2*n, n). - Vincenzo Librandi, Jul 05 2013
Boas-Buck recurrence: a(n) = (30/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n+7, 7). See a comment there. - Wolfdieter Lang, Aug 10 2017
Sum_{n>=0} 1/a(n) = 2106*sqrt(3)*Pi - 13234624/1155.
Sum_{n>=0} (-1)^n/a(n) = 162500*sqrt(5)*log(phi) - 121172896/693, where phi is the golden ratio (A001622). (End)
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MATHEMATICA
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CoefficientList[Series[1/(1-4x)^(15/2), {x, 0, 20}], x] (* Vincenzo Librandi, Jul 05 2013 *)
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PROG
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(Magma) [&*[2*n+i: i in [1..13 by 2]]*Binomial(2*n, n)/135135: n in [0..20]]; // Vincenzo Librandi, Jul 05 2013
(PARI) vector(20, n, n--; m=n+7; binomial(2*m, m)*binomial(m, 7)/3432) \\ G. C. Greubel, Jul 21 2019
(Sage) [binomial(2*(n+7), n+7)*binomial(n+7, 7)/3432 for n in (0..20)] # G. C. Greubel, Jul 21 2019
(GAP) List([0..20], n-> Binomial(2*(n+7), n+7)*Binomial(n+7, 7)/3432); # G. C. Greubel, Jul 21 2019
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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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