OFFSET
0,2
COMMENTS
From Peter Bala, Mar 20 2022: (Start)
The congruence a(n+k) == a(n) (mod k) holds for all n and k.
It follows that the sequence obtained by taking a(n) modulo a fixed positive integer k is periodic with exact period dividing k. For example, taken modulo 5 the sequence becomes [2, 3, 4, 0, 1, 2, 3, 4, 0, 1, ...], a purely periodic sequence with period 5.
More generally, the same property holds for any sequence with an e.g.f. of the form F(x)*exp(x*G(x)), where F(x) and G(x) are power series with integer coefficients and G(0) = 1 (see Bala, Theorem 1). (End)
LINKS
FORMULA
In Maple notation: a(n) = hypergeom([1, 1/3, 2/3, -n], [], -27), n = 0, 1, ....
a(n) = Integral_{x = 0..infinity} (1+x^3)^n*exp(-x) dx. - Gerald McGarvey, Oct 12 2007
From Vaclav Kotesovec, Oct 30 2017: (Start)
a(n) = (27*n^3 - 27*n^2 + 6*n + 1)*a(n-1) - 3*(n-1)*(27*n^2 - 45*n + 20)*a(n-2) + 27*(n-2)*(n-1)*(3*n - 4)*a(n-3) - 27*(n-3)*(n-2)*(n-1)*a(n-4).
a(n) ~ sqrt(2*Pi) * (3*n)^(3*n + 1/2) / exp(3*n). (End)
From Peter Bala, Mar 20 2022: (Start)
a(n) = Sum_{k = 0..n} binomial(n,k)*(3*k)!.
a(n) = 1 + 6*n*(1 + 60*(n-1)*(1 + 168*(n-2)*(1 + 330*(n-3)*(1 + 546*(n-4)*(1 + ...*(1 + (3*(3*m+1)*(3*m+2))*(n-m)*(1 + ...))))))).
a(n) == 1 (mod 6).
O.g.f: Sum_{k >= 0} (3*k)!*x^k/(1 - x)^(k+1).
E.g.f.: exp(x)*Sum_{k >= 0} (3*k)!/k!*x^k. (End)
MATHEMATICA
Table[Sum[Binomial[n, k] * (3*k)!, {k, 0, n}], {n, 0, 12}] (* Vaclav Kotesovec, Oct 30 2017 *)
PROG
(PARI) for(n=0, 12, print1(round(intnum(x=0, 999, exp(-x)*(1+x^3)^n)), ", ")) - Gerald McGarvey, Oct 12 2007
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Karol A. Penson, Sep 20 2001
EXTENSIONS
Corrected and extended by N. J. A. Sloane, Oct 29 2006
STATUS
approved