0,2

For related recurrences of the form a(n) = (n^k*a(n-1)-1)/(n-1) see A001339, A007808 (both k = 2) and A141828 (k = 4). a(n) is a difference divisibility sequence, that is, the difference a(n) - a(m) is divisible by n - m for all n and m (provided n is not equal to m). See A000522 for further properties of difference divisibility sequences.

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)).

a(n) := n -> n!^2*sum((n-k+1)*(k+1)/k!^2, k = 0..n): seq(a(n), n = 0..16);

Cf. A001339, A007808, A070910, A082425, A096789, A130820, A141828.

Sequence in context: A138860 A145087 A005046 this_sequence A143077 A005841 A005828

Adjacent sequences: A141824 A141825 A141826 this_sequence A141828 A141829 A141830

easy,nonn

Peter Bala (pbala(AT)toucansurf.com), Jul 09 2008, Oct 06 2008