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Harvey P. Dale, <a href="/A141828/b141828.txt">Table of n, a(n) for n = 0..181</a>

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nxt[{n_, a_}]:={n+1, ((n+1)^4*a-1)/n}; Join[{1}, NestList[nxt, {1, 5}, 15][[All, 2]]] (* Harvey P. Dale, Mar 12 2017 *)

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_Peter Bala (pbala(AT)toucansurf.com), _, Jul 09 2008

a(n) = (n^4*a(n-1)-1)/(n-1) for n >= 2, with a(0) = 1, a(1) = 5.

1, 5, 79, 3199, 272981, 42653281, 11055730435, 4424134795739, 2588750874763849, 2123099311165701661, 2358999234628557401111, 3453810779419670890966615, 6510747302004208690462157149, 15496121141045183700690805861049

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 A141827 (k = 3). 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!^3 = 1/(1-x)^2*sum {n = 0..inf} (n^2+n+1)*x^n/n!^3.

a(n) = n!^3*sum {k = 0..n} (n-k+1)(k^2+k+1)/k!^3.

a(n) := n* n!^3*(5 - sum{k = 2..n} 1/(k!^2*k*(k-1)).

Congruence property: a(n) == (1+n+n^2+n^3) (mod n^4).

The recurrence a(n) = (n^3+n^2+n+2)*a(n-1) - (n-1)^3*a(n-2), n >= 2, shows that a(n) is always a positive integer. The sequence b(n) := n*n!^3 also satisfies the same recurrence with b(0) = 0, b(1) = 1. Hence we obtain the finite continued fraction expansion a(n)/(n*n!^3) = 5 - 1^3/(16 - 2^3/(41 - 3^3/(86 -...-(n-1)^3/(n^3+n^2+n+2)))), for n >= 1. a(n)*b(n+1) - b(n)*a(n+1) = n!^3.

Lim n -> infinity a(n)/(n*n!^3) = sum {n = 0..inf} (n^2+n+1)/n!^3 = 4.93672 23378... .

Lim n -> infinity a(n)/(n*n!^3) = 1 + sum {n = 0..inf} 1/(prod{k = 0..n}(A008620 (k)).

a := n -> n!^3*add((n-k+1)*(k^2+k+1)/k!^3, k = 0..n): seq(a(n), n = 0..16);

Cf. A001339, A007808, A141827.

easy,nonn

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

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