A020930 Expansion of 1/(1-4*x)^(19/2).
1, 38, 798, 12236, 152950, 1651860, 15967980, 141430680, 1166803110, 9075135300, 67156001220, 476197099560, 3254013513660, 21526550936520, 138384970306200, 867212480585520, 5311676443586310, 31870058661517860, 187679234340049620, 1086563988284497800
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
Programs
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GAP
List([0..20], n-> Binomial(2*(n+9), n+9)*Binomial(n+9, 9)/Binomial(18, 9)); # G. C. Greubel, Jul 21 2019
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Magma
[&*[2*n+i: i in [1..17 by 2]]*Binomial(2*n, n)/34459425: n in [0..20]]; // Vincenzo Librandi, Jul 05 2013
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Mathematica
CoefficientList[Series[1/(1-4x)^(19/2), {x,0,20}], x] (* Vincenzo Librandi, Jul 05 2013 *) -
PARI
vector(20, n, n--; m=n+9; binomial(2*m,m)*binomial(m, 9)/binomial(18,9) ) \\ G. C. Greubel, Jul 21 2019
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Sage
[binomial(2*(n+9),n+9)*binomial(n+9, 9)/binomial(18,9) for n in (0..20)] # G. C. Greubel, Jul 21 2019
Formula
a(n) = binomial(n+9, 9)*A000984(n+9)/A000984(9), where A000984 are the central binomial coefficients. - Wolfdieter Lang
a(n) = ((2*n+17)*(2*n+15)*(2*n+13)*(2*n+11)*(2*n+9)*(2*n+7)*(2*n+5)*(2*n+3)*(2*n+1)/34459425)*binomial(2*n, n). - Vincenzo Librandi, Jul 05 2013
Boas-Buck recurrence: a(n) = (38/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n+9, 9). See a comment there. - Wolfdieter Lang, Aug 10 2017
a(n) = binomial(2*(n+9),n+9)*binomial(n+9, 9)/binomial(18,9). - G. C. Greubel, Jul 21 2019
From Amiram Eldar, Mar 27 2022: (Start)
Sum_{n>=0} 1/a(n) = 24786*sqrt(3)*Pi - 2025065024/15015.
Sum_{n>=0} (-1)^n/a(n) = 5312500*sqrt(5)*log(phi) - 257493786304/45045, where phi is the golden ratio (A001622). (End)