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a(n) = 2^n * n! * [x^n](exp(2*x) * BesselI(1, x)).
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%I #12 Mar 06 2022 08:42:03

%S 0,1,8,51,304,1770,10224,58947,340064,1964862,11374000,65966318,

%T 383289504,2230877428,13005037920,75923905635,443837331648,

%U 2597761611894,15221636471088,89283411393018,524194439193120,3080311943546124,18115458433730592,106618075368243534

%N a(n) = 2^n * n! * [x^n](exp(2*x) * BesselI(1, x)).

%F a(n) = [x^n] (1/(2*x))*(1 - (4*x - 1)/(sqrt((6*x - 1)*(2*x - 1)))).

%F D-finite with recurrence a(n) = 4*(3*(n^2 - n)*a(n - 2) - (2*n^2 - n)*a(n - 1))/(1 - n^2) for n >= 2.

%F The INVERT transform of A052177.

%F a(n) ~ 2^(n - 1/2) * 3^(n + 1/2) / sqrt(Pi*n). - _Vaclav Kotesovec_, May 12 2021

%p gf := exp(2*x)*BesselI(1, x):

%p ser := series(gf, x, 32): seq(2^n*n!*coeff(ser, x, n), n = 0..23);

%p # Or:

%p gf := (1/(2*x))*(1 - (4*x - 1)/(sqrt((6*x - 1)*(2*x - 1)))):

%p ser := series(gf, x, 32): seq(coeff(ser, x, n), n = 0..23);

%t RecurrenceTable[{(1 - n^2) a[n] == 4 (3 (n^2 - n) a[n - 2] - (2 n^2 - n) a[n - 1]), a[0] == 0, a[1] == 1}, a, {n, 0, 23}]

%Y Cf. A052177.

%K nonn,easy

%O 0,3

%A _Peter Luschny_, May 12 2021