%I #31 Mar 25 2022 09:13:42

%S 1,22,286,2860,24310,184756,1293292,8498776,53117350,318704100,

%T 1848483780,10418726760,57302997180,308554600200,1630931458200,

%U 8480843582640,43464323361030,219878341708740,1099391708543700,5439095821216200,26651569523959380,129450480544945560

%N Expansion of 1/(1-4*x)^(11/2).

%C Also convolution of A000984 with A040075, also convolution of A000302 with A020920, also convolution of A002457 with A038846, also convolution of A002697 with A020918, also convolution of A002802 with A038845. - _Rui Duarte_, Oct 08 2011

%H Vincenzo Librandi, <a href="/A020922/b020922.txt">Table of n, a(n) for n = 0..200</a>

%F a(n) = binomial(n+5, 5)*A000984(n+5)/A000984(5), where A000984 are central binomial coefficients. - _Wolfdieter Lang_

%F From _Rui Duarte_, Oct 08 2011: (Start)

%F a(n) = ((2n+9)(2n+7)(2n+5)(2n+3)(2n+1)/(9*7*5*3*1)) * binomial(2n, n).

%F a(n) = binomial(2n+10, 10) * binomial(2n, n) / binomial(n+5, 5).

%F a(n) = binomial(n+5, 5) * binomial(2n+10, n+5) / binomial(10, 5).

%F a(n) = Sum_{ i_1+i_2+i_3+i_4+i_5+i_6+i_7+i_8+i_9+i_10+i_11 = n } f(i_1)* f(i_2)*f(i_3)*f(i_4)*f(i_5)*f(i_6)*f(i_7)*f(i_8)*f(i_9)*f(i_10)*f(i_11) with f(k)=A000984(k). (End)

%F Boas-Buck recurrence: a(n) = (22/n)*Sum_{k=0..n-1} 4^(n-k-1)*a(k), n >= 1, a(0) = 1. Proof from a(n) = A046521(n+5, 5). See a comment there. - _Wolfdieter Lang_, Aug 10 2017

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

%F Sum_{n>=0} 1/a(n) = 162*sqrt(3)*Pi - 30816/35.

%F Sum_{n>=0} (-1)^n/a(n) = 4500*sqrt(5)*log(phi) - 33888/7, where phi is the golden ratio (A001622). (End)

%t CoefficientList[Series[1/(1-4x)^(11/2), {x,0,30}], x] (* _Vincenzo Librandi_, Jul 05 2013 *)

%o (Magma) [(2*n+9)*(2*n+7)*(2*n+5)*(2*n+3)*(2*n+1)*Binomial(2*n, n)/945: n in [0..30]] // _Vincenzo Librandi_, Jul 05 2013

%o (PARI) vector(30, n, n--; m=n+5; binomial(m, 5)*binomial(2*m, m)/252) \\ _G. C. Greubel_, Jul 20 2019

%o (Sage) [binomial(n+5, 5)*binomial(2*n+10, n+5)/252 for n in (0..30)] # _G. C. Greubel_, Jul 20 2019

%o (GAP) List([0..30], n-> Binomial(n+5, 5)*Binomial(2*n+10, n+5)/252); # _G. C. Greubel_, Jul 20 2019

%Y Cf. A000302, A000984, A001622, A002457, A002697, A002802, A020918, A020920, A038845, A038846, A040075, A046521 (sixth column).

%K nonn,easy

%O 0,2

%A _N. J. A. Sloane_