A141828 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

Offset: 0

Author

Peter Bala, Jul 09 2008

Keywords

• easy
• nonn

Comments

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.

Links

• Harvey P. Dale, Table of n, a(n) for n = 0..181

Crossrefs

Cf. A001339, A007808, A141827.

Programs

Maple

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

Mathematica

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 *)

Formula

Sum_{n>=0} a(n)*x^n/n!^3 = (1/(1-x)^2)*Sum_{n>=0} (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!^3*k*(k-1))) for n > 0. [corrected by Jason Yuen, Jan 31 2025]

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.

Limit_{n->oo} a(n)/(n*n!^3) = Sum_{n>=0} (n^2+n+1)/n!^3 = 4.9367223378... .

Limit_{n->oo} a(n)/(n*n!^3) = 1 + Sum_{n>=0} 1/(Product_{k=0..n} A008620(k)).