OFFSET
0,2
COMMENTS
From Peter Bala, Mar 12 2023: (Start)
We conjecture that a(n+k) == a(n) (mod k) for all n and k. If true, then for each k, the sequence a(n) taken modulo k is a periodic sequence and the period divides k. For example, modulo 7 the sequence becomes [1, 2, 4, 1, 1, 4, 2, 1, 2, 4, 1, 1, 4, 2, ...], apparently a periodic sequence of period 7.
More generally, let F(x) and G(x) denote power series with integer coefficients with F(0) = G(0) = 1. Define b(n) = n! * [x^n] exp(x*G(x))*F(x)^n. Then we conjecture that b(n+k) == b(n) (mod k) for all n and k. The present sequence is the case F(x) = 1/(1 - x), G(x) = 1. Cf. A361281. (End)
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..370
FORMULA
a(-n) = a(n).
a(n) = n! [x^n] exp((1-h(x))/2)*(1+h(x))/(2*h(x)) with h(x) = sqrt(1-4*x).
a(n) = ((2*n-1)*a(n-2) + 4*(n*(2*n-4)+1)*a(n-1))/(2*n-3) for n>=2.
a(n) ~ 2^(2*n-1/2) * n^n / exp(n-1/2). - Vaclav Kotesovec, Nov 10 2016
a(n) = n!*Sum_{i=0..n}(binomial(2*n-i-1,n-i)/i!). - Vladimir Kruchinin, Nov 23 2016
a(n) = n! * [x^n] exp(x)/(1 - x)^n. - Ilya Gutkovskiy, Sep 21 2017
MAPLE
a := n -> hypergeom([n, -n], [], -1): seq(simplify(a(n)), n=0..18);
# Alternatively:
a := proc(n) option remember; `if`(n<2, n+1,
((2*n-1)*a(n-2) + 4*(n*(2*n-4)+1)*a(n-1))/(2*n-3)) end:
seq(a(n), n=0..18);
MATHEMATICA
Table[HypergeometricPFQ[{n, -n}, {}, -1], {n, 0, 20}] (* Vaclav Kotesovec, Nov 10 2016 *)
PROG
(Sage)
def a():
a, b, c, d, h, e = 1, 2, 1, 8, 4, 0
yield a
while True:
yield b
e = c; c += 2
a, b = b, (c*a + h*b)//e
d += 16; h += d
A278070 = a()
[next(A278070) for _ in range(19)]
(Maxima)
a(n):=n!*sum(binomial(2*n-i-1, n-i)/i!, i, 0, n); /* Vladimir Kruchinin, Nov 23 2016 */
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Peter Luschny, Nov 10 2016
STATUS
approved