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

%I #10 Jun 25 2022 08:25:28

%S 1,2,9,72,900,16200,396900,12700800,514382400,25719120000,

%T 1556006760000,112032486720000,9466745127840000,927741022528320000,

%U 104370865034436000000,13359470724407808000000,1930443519676928256000000,312731850187662377472000000

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

%C Self-convolution of A184359.

%F From _Amiram Eldar_, Jun 25 2022: (Start)

%F Sum_{n>=0} 1/a(n) = (BesselI(0, 2*sqrt(2)) - 1)/2.

%F Sum_{n>=0} (-1)^n/a(n) = (1 - BesselJ(0, 2*sqrt(2)))/2. (End)

%e G.f.: A(x) = 1 + 2*x + 9*x^2 + 72*x^3 + 900*x^4 + 16200*x^5 +...

%e A(x)^(1/2) = 1 + x + 4*x^2 + 32*x^3 + 410*x^4 + 7562*x^5 + 188736*x^6 +...+ A184359(n)*x^n +...

%t a[n_] := (n + 1)!^2/2^n; Array[a, 20, 0] (* _Amiram Eldar_, Jun 25 2022 *)

%o (PARI) {a(n)=(n+1)!^2/2^n}

%Y Cf. A184359, A184360, A184361, A006472.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Jan 16 2011