%I #25 Mar 30 2023 15:31:49

%S 1,18,540,22680,1224720,80831520,6304858560,567437270400,

%T 57878601580800,6598160580211200,831368233106611200,

%U 114728816168712345600,17209322425306851840000,2787910232899709998080000,485096380524549539665920000,90227926777566214377861120000

%N a(n) = 3^n*(2*n + 1)!/n!.

%F E.g.f.: 1/(1 - 12*x)^(3/2) = 1 + 18*x + 540*x^2/2! + 22680*x^3/3! + ....

%F Recurrence equation: a(n) = 6*(2*n + 1)*a(n-1) with a(0) = 1.

%F 2nd order recurrence equation: a(n) = 8*(n + 1)*a(n-1) + 12*(2*n - 1)^2*a(n-2) with a(0) = 1, a(1) = 18.

%F Define a sequence b(n) := a(n)*sum {k = 0..n} (-1)^k/((2*k + 1)*3^k) beginning [1, 16, 492, 20544, 1111056, 73299456, 5718022848, ...]. It is not difficult to check that b(n) also satisfies the previous 2nd order recurrence equation (and so is an integer sequence). Using this observation we obtain the continued fraction expansion Pi/(2*sqrt(3)) = Sum {k >= 0} (-1)^k/( (2*k + 1)*3^k ) = 1 - 2/(18 + 12*3^2/(24 + 12*5^2/(32 + ... + 12*(2*n - 1)^2/((8*n + 8) + ... )))). Cf. A254619 and A254620.

%p seq(3^n*(2*n + 1)!/n!, n = 0..13);

%t Table[3^n(2n + 1)!/n!, {n, 0, 19}] (* _Alonso del Arte_, Feb 04 2015 *)

%Y Cf. A002391, A034910, A073000, A093766, A254619, A254620.

%K nonn,easy

%O 0,2

%A _Peter Bala_, Feb 04 2015