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Expansion of 1/(1-4*x)^(15/2).
4

%I #27 Mar 25 2022 09:14:07

%S 1,30,510,6460,67830,624036,5200300,40116600,290845350,2003601300,

%T 13223768580,84151254600,518932736700,3113596420200,18236779032600,

%U 104557533120240,588136123801350,3252046802195700,17705588145287700,95051052148386600,503770576386448980

%N Expansion of 1/(1-4*x)^(15/2).

%H Vincenzo Librandi, <a href="/A020926/b020926.txt">Table of n, a(n) for n = 0..200</a>

%F a(n) = binomial(n+7, 7)*A000984(n+7)/A000984(7), A000984: central binomial coefficients. - _Wolfdieter Lang_

%F 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

%F 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

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

%F Sum_{n>=0} 1/a(n) = 2106*sqrt(3)*Pi - 13234624/1155.

%F Sum_{n>=0} (-1)^n/a(n) = 162500*sqrt(5)*log(phi) - 121172896/693, where phi is the golden ratio (A001622). (End)

%t CoefficientList[Series[1/(1-4x)^(15/2), {x,0,20}], x] (* _Vincenzo Librandi_, Jul 05 2013 *)

%o (Magma) [&*[2*n+i: i in [1..13 by 2]]*Binomial(2*n, n)/135135: n in [0..20]]; // _Vincenzo Librandi_, Jul 05 2013

%o (PARI) vector(20, n, n--; m=n+7; binomial(2*m,m)*binomial(m,7)/3432) \\ _G. C. Greubel_, Jul 21 2019

%o (Sage) [binomial(2*(n+7), n+7)*binomial(n+7,7)/3432 for n in (0..20)] # _G. C. Greubel_, Jul 21 2019

%o (GAP) List([0..20], n-> Binomial(2*(n+7), n+7)*Binomial(n+7,7)/3432); # _G. C. Greubel_, Jul 21 2019

%Y Cf. A000984, A001622, A020924, A046521 (eighth column).

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_