OFFSET
0,2
COMMENTS
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..200
FORMULA
a(n) = binomial(n+5, 5)*A000984(n+5)/A000984(5), where A000984 are central binomial coefficients. - Wolfdieter Lang
From Rui Duarte, Oct 08 2011: (Start)
a(n) = ((2n+9)(2n+7)(2n+5)(2n+3)(2n+1)/(9*7*5*3*1)) * binomial(2n, n).
a(n) = binomial(2n+10, 10) * binomial(2n, n) / binomial(n+5, 5).
a(n) = binomial(n+5, 5) * binomial(2n+10, n+5) / binomial(10, 5).
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)
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
From Amiram Eldar, Mar 25 2022: (Start)
Sum_{n>=0} 1/a(n) = 162*sqrt(3)*Pi - 30816/35.
Sum_{n>=0} (-1)^n/a(n) = 4500*sqrt(5)*log(phi) - 33888/7, where phi is the golden ratio (A001622). (End)
MATHEMATICA
CoefficientList[Series[1/(1-4x)^(11/2), {x, 0, 30}], x] (* Vincenzo Librandi, Jul 05 2013 *)
PROG
(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
(PARI) vector(30, n, n--; m=n+5; binomial(m, 5)*binomial(2*m, m)/252) \\ G. C. Greubel, Jul 20 2019
(Sage) [binomial(n+5, 5)*binomial(2*n+10, n+5)/252 for n in (0..30)] # G. C. Greubel, Jul 20 2019
(GAP) List([0..30], n-> Binomial(n+5, 5)*Binomial(2*n+10, n+5)/252); # G. C. Greubel, Jul 20 2019
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved