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A226731 a(n) = (2n - 1)!/(2n).

%I A226731 #39 Mar 10 2022 03:39:10
%S A226731 20,630,36288,3326400,444787200,81729648000,19760412672000,
%T A226731 6082255020441600,2322315553259520000,1077167364120207360000,
%U A226731 596585001666576384000000,388888194657798291456000000
%N A226731 a(n) = (2n - 1)!/(2n).
%C A226731 For n < 3, the formula does not produce an integer.
%C A226731 The ratio of the product of the partition parts of 2n into exactly two parts to the sum of the partition parts of 2n into exactly two parts. For example, a(3) = 20, and 2*3 = 6 has 3 partitions into exactly two parts: (5,1), (4,2), (3,3). Forming the ratio of product to sum (of parts), we have (5*1*4*2*3*3)/(5+1+4+2+3+3) = 360/18 = 20. - _Wesley Ivan Hurt_, Jun 24 2013
%H A226731 Seiichi Manyama, <a href="/A226731/b226731.txt">Table of n, a(n) for n = 3..225</a>
%H A226731 <a href="/index/Par#part">Index entries for sequences related to partitions</a>.
%F A226731 a(n) = A009445(n-1)/A005843(n) = A002674(n)/A001105(n). - _Wesley Ivan Hurt_, Jun 24 2013
%F A226731 a(n) ~ sqrt(Pi)*2^(2*n-1)*n^(2*n-3/2)/exp(2*n). - _Ilya Gutkovskiy_, Nov 01 2016
%F A226731 From _Amiram Eldar_, Mar 10 2022: (Start)
%F A226731 Sum_{n>=3} 1/a(n) = e - 8/3.
%F A226731 Sum_{n>=3} (-1)^(n+1)/a(n) = cos(1) + sin(1) - 4/3. (End)
%e A226731 a(3) = (2*3 - 1)!/(2*3) = 5!/6 = 120/6 = 20.
%p A226731 seq((2*k-1)!/(2*k),k=1..20); # _Wesley Ivan Hurt_, Jun 24 2013
%t A226731 Table[(2n-1)!/(2n),{n,3,20}] (* _Harvey P. Dale_, Jun 19 2013 *)
%o A226731 (PARI) a(n) = (2*n-1)!/(2*n); \\ _Michel Marcus_, Nov 01 2016
%Y A226731 Cf. A001105, A002674, A005843, A009445, A093353, A143623, A211374.
%K A226731 nonn,easy
%O A226731 3,1
%A A226731 _Wesley Ivan Hurt_, Jun 15 2013