A020924 - OEIS

%I #23 Mar 25 2022 09:13:55

%S 1,26,390,4420,41990,352716,2704156,19315400,130378950,840219900,

%T 5209363380,31256180280,182327718300,1037865473400,5782393351800,

%U 31610416989840,169905991320390,899502306990300,4697400936504900,24228699567235800,123566367792902580,623715951716555880

%N Expansion of 1/(1-4*x)^(13/2).

%H Vincenzo Librandi, <a href="/A020924/b020924.txt">Table of n, a(n) for n = 0..200</a>

%F a(n) = binomial(n+6, 6)*A000984(n+6)/A000984(6), where A000984 are the central binomial coefficients. - _Wolfdieter Lang_

%F 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

%F n*a(n) - 2*(2*n+11)*a(n-1) = 0. - _Bruno Berselli_, Jul 05 2013

%F 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

%F a(n) = 45*binomial(n+6,n)*binomial(2*n+12,n+6)/(4*binomial(2*n,n)). - _G. C. Greubel_, Jul 20 2019

%F From _Amiram Eldar_, Mar 25 2022: (Start)

%F Sum_{n>=0} 1/a(n) = 1018468/315 - 594*sqrt(3)*Pi.

%F Sum_{n>=0} (-1)^n/a(n) = 27500*sqrt(5)*log(phi) - 1864148/63, where phi is the golden ratio (A001622). (End)

%t CoefficientList[Series[1/(1-4x)^(13/2), {x,0,30}], x] (* _Vincenzo Librandi_, Jul 05 2013 *)

%o (Magma) [&*[2*n+i: i in [1..11 by 2]]*Binomial(2*n, n)/10395: n in [0..20]]; // _Vincenzo Librandi_, Jul 05 2013

%o (Magma) [Binomial(n+6,n)*Binomial(2*n+12,n+6)/924: n in [0..30]]; // _G. C. Greubel_, Jul 20 2019

%o (PARI) vector(30, n, n--; m=n+6; binomial(m,n)*binomial(2*m,m)/924)

%o (Sage) [binomial(n+6,n)*binomial(2*n+12,n+6)/924 for n in (0..30)] # _G. C. Greubel_, Jul 20 2019

%o (GAP) List([0..30], n-> Binomial(n+6,n)*Binomial(2*n+12,n+6)/924); # _G. C. Greubel_, Jul 20 2019

%Y Cf. A000984, A001622, A020922, A046521 (seventh column).

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_