A254381 a(n) = 3^n*(2*n + 1)!/n!.

1, 18, 540, 22680, 1224720, 80831520, 6304858560, 567437270400, 57878601580800, 6598160580211200, 831368233106611200, 114728816168712345600, 17209322425306851840000, 2787910232899709998080000, 485096380524549539665920000, 90227926777566214377861120000

Offset

0,2

Links

Table of n, a(n) for n=0..15.

Formula

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

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

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.

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.

Maple

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

Mathematica

Table[3^n(2n + 1)!/n!, {n, 0, 19}] (* Alonso del Arte, Feb 04 2015 *)

Crossrefs

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

Sequence in context: A255859 A334179 A183498 * A177098 A133401 A211708

Adjacent sequences: A254378 A254379 A254380 * A254382 A254383 A254384

Keyword

nonn,easy

Author

Peter Bala, Feb 04 2015

Status

approved