OFFSET
0,2
COMMENTS
Conjecture: a(p*n) == a(n) (mod p^2) for prime p == 1 (mod 9) and all positive integers n except those n of the form n = m*p + k for 0 <= m <= (p-1)/9 and 1 <= k <= (p-1)/9. Cf. A034171, A004981 and A004982. - Peter Bala, Dec 23 2019
LINKS
Seiichi Manyama, Table of n, a(n) for n = 0..700
FORMULA
a(n) = 3^n/n! * Product_{k=0..n-1} (9*k + 1) for n > 0.
a(n) ~ 3^(3*n) / (Gamma(1/9) * n^(8/9)). - Vaclav Kotesovec, Jun 23 2018
From Peter Luschny, Dec 26 2019: (Start)
a(n) = (-27)^n*binomial(-1/9, n).
a(n) = n! * [x^n] hypergeom([1/9], [1], 27*x). (End)
D-finite with recurrence: n*a(n) +3*(-9*n+8)*a(n-1)=0. - R. J. Mathar, Jan 20 2020
MAPLE
seq(coeff(series((1-27*x)^(-1/9), x, n+1), x, n), n=0..20); # Muniru A Asiru, Jun 23 2018
# Alternative:
A298799 := n -> (-27)^n*binomial(-1/9, n):
seq(A298799(n), n=0..17); # Peter Luschny, Dec 26 2019
PROG
(PARI) N=20; x='x+O('x^N); Vec((1-27*x)^(-1/9))
(GAP) List([0..20], n->(3^n/Factorial(n))*Product([0..n-1], k->9*k+1)); # Muniru A Asiru, Jun 23 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
Seiichi Manyama, Jun 22 2018
STATUS
approved