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%I #14 Jan 24 2022 04:00:59
%S 1,15,136,969,5985,33649,177100,888030,4292145,20160075,92561040,
%T 417225900,1852482996,8122425444,35240152720,151532656696,
%U 646626422970,2741188875414,11554258485616,48459472266975
%N a(n) = binomial(2*n+1, n-6).
%H Chai Wah Wu, <a href="/A030056/b030056.txt">Table of n, a(n) for n = 6..500</a>
%H Milan Janjic, <a href="https://pmf.unibl.org/wp-content/uploads/2017/10/enumfor.pdf">Two Enumerative Functions</a>.
%F a(n+1) = a(n)*(2*n+2)*(2*n+3)/((n-5)*(n+8)). - _Chai Wah Wu_, Jan 26 2016
%F From _Amiram Eldar_, Jan 24 2022: (Start)
%F Sum_{n>=6} 1/a(n) = 40*Pi/(9*sqrt(3)) - 96827/13860.
%F Sum_{n>=6} (-1)^n/a(n) = 29248*log(phi)/(5*sqrt(5)) - 3486955/2772, where phi is the golden ratio (A001622). (End)
%t Table[Binomial[2*n + 1, n - 6], {n, 6, 25}] (* _Amiram Eldar_, Jan 24 2022 *)
%o (Python)
%o from __future__ import division
%o A030056_list, b = [], 1
%o for n in range(6,501):
%o A030056_list.append(b)
%o b = b*(2*n+2)*(2*n+3)//((n-5)*(n+8)) # _Chai Wah Wu_, Jan 26 2016
%Y Diagonal 14 of triangle A100257.
%Y Cf. A001622.
%K nonn
%O 6,2
%A _N. J. A. Sloane_.
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