| Sum {n = 0..inf} a(n)*x^n/n!^2 = 1/(1-x)^2*sum {n = 0..inf} (n+1)*x^n/n!^2.
a(n) = n!^2*sum {k = 0..n} (n-k+1)(k+1)/k!^2.
a(n) := n* n!^2*(4 - sum{k = 2..n} 1/(k!^2*k*(k-1)).
Congruence property: a(n) == (1+n+n^2) (mod n^3).
The recurrence a(n) = (n^2+n+2)*a(n-1) - (n-1)^2*a(n-2), n >= 2, shows that a(n) is always a positive integer. The sequence b(n) := n*n!^2 also satisfies the same recurrence with b(0) = 0, b(1) = 1. Hence we obtain the finite continued fraction expansion a(n)/(n*n!^2) = 4 - 1^2/(8 - 2^2/(14 - 3^2/(22 -...-(n-1)^2/(n^2+n+2)))), for n > 1. a(n)*b(n+1) - b(n)*a(n+1) = n!^2.
Lim n -> infinity a(n)/(n*n!^2) = sum {n = 0..inf} (n+1)/n!^2 = BesselI(0,2) + BesselI(1,2) = 3.87022 21569 ..., using the values of the modified Bessel function, BesselI(0, 2) = 2.27958 53023 ... and BesselI(1, 2) = 1.59063 68546 ... (see A070910 and A096789; Cf. A130820). This yields the continued fraction expansion BesselI(0,2) + BesselI(1,2) = 4 - 1^2/(8 - 2^2/(14 - 3^2/(22 -...-(n-1)^2/(n^2+n+2 - ... )))).
Lim n -> infinity a(n)/(n*n!^2) = sum {n = 1..inf} (n+n^2)/n!^2 = sum {n = 1..inf} n^3/n!^2 = 1/2 * sum {n = 1..inf} n^4/n!^2.
Lim n -> infinity a(n)/(n*n!^2) = sum {n = 0..inf} A001405 (n)/n!.
Lim n -> infinity a(n)/(n*n!^2) = 1 + sum {n = 0..inf} 1/(prod{k = 0..n} (A008619(k)).
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