A020924 Expansion of 1/(1-4*x)^(13/2).
1, 26, 390, 4420, 41990, 352716, 2704156, 19315400, 130378950, 840219900, 5209363380, 31256180280, 182327718300, 1037865473400, 5782393351800, 31610416989840, 169905991320390, 899502306990300, 4697400936504900, 24228699567235800, 123566367792902580, 623715951716555880
Offset: 0
Links
- Vincenzo Librandi, Table of n, a(n) for n = 0..200
Programs
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GAP
List([0..30], n-> Binomial(n+6,n)*Binomial(2*n+12,n+6)/924); # G. C. Greubel, Jul 20 2019
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Magma
[&*[2*n+i: i in [1..11 by 2]]*Binomial(2*n, n)/10395: n in [0..20]]; // Vincenzo Librandi, Jul 05 2013
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Magma
[Binomial(n+6,n)*Binomial(2*n+12,n+6)/924: n in [0..30]]; // G. C. Greubel, Jul 20 2019
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Mathematica
CoefficientList[Series[1/(1-4x)^(13/2), {x,0,30}], x] (* Vincenzo Librandi, Jul 05 2013 *) -
PARI
vector(30, n, n--; m=n+6; binomial(m,n)*binomial(2*m,m)/924)
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Sage
[binomial(n+6,n)*binomial(2*n+12,n+6)/924 for n in (0..30)] # G. C. Greubel, Jul 20 2019
Formula
a(n) = binomial(n+6, 6)*A000984(n+6)/A000984(6), where A000984 are the central binomial coefficients. - Wolfdieter Lang
a(n) = (2*n+11)*(2*n+9)*(2*n+7)*(2*n+5)*(2*n+3)*(2*n+1)*Binomial(2*n, n)/10395. - Vincenzo Librandi, Jul 05 2013
n*a(n) - 2*(2*n+11)*a(n-1) = 0. - Bruno Berselli, Jul 05 2013
Boas-Buck recurrence: a(n) = (26/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n+6, 6). See a comment there. - Wolfdieter Lang, Aug 10 2017
a(n) = 45*binomial(n+6,n)*binomial(2*n+12,n+6)/(4*binomial(2*n,n)). - G. C. Greubel, Jul 20 2019
From Amiram Eldar, Mar 25 2022: (Start)
Sum_{n>=0} 1/a(n) = 1018468/315 - 594*sqrt(3)*Pi.
Sum_{n>=0} (-1)^n/a(n) = 27500*sqrt(5)*log(phi) - 1864148/63, where phi is the golden ratio (A001622). (End)