login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A141828
a(n) = (n^4*a(n-1)-1)/(n-1) for n >= 2, with a(0) = 1, a(1) = 5.
2
1, 5, 79, 3199, 272981, 42653281, 11055730435, 4424134795739, 2588750874763849, 2123099311165701661, 2358999234628557401111, 3453810779419670890966615, 6510747302004208690462157149, 15496121141045183700690805861049
OFFSET
0,2
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
FORMULA
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)).
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 *)
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Peter Bala, Jul 09 2008
STATUS
approved