OFFSET
0,3
COMMENTS
a(n) is always an integer since (2k-1)|C(2k,k) for any nonnegative integer k.
Conjecture: (i) The sequence a(n+1)/a(n) (n = 2,3,...) is strictly increasing to the limit 9, and the sequence a(n+1)^(1/(n+1))/a(n)^(1/n) (n = 1,2,3,...) is strictly decreasing to the limit 1.
(ii) sum_{k=0}^{n-1}a(k) == 0 (mod n^2) for all n > 0. Moreover, for any prime p we have sum_{k=0}^{p-1}a(k) == -p^2*(1+9*(p/3))/2 (mod p^3), where (p/3) is the Legendre symbol.
We are able to prove n | sum_{k=0}^{n-1}a(k). Note also that sum_{k=0}^{n-1}a(k)*9^(n-1-k) = -n^2*A086618(n-1) for all n > 0 since both sides satisfy the same recurrence via the Zeilberger algorithm.
The congruence (0 mod n^2) in (ii) is true, see the formula for A246138 in terms of A005802. - Mark van Hoeij, Nov 07 2023
LINKS
Zhi-Wei Sun, Table of n, a(n) for n = 0..150
Zhi-Wei Sun, A new kind of numbers and their arithmetic properties, preprint, arXiv:1408.5381 [math.NT], 2014-2018.
FORMULA
Recurrence (obtained via the Zeilberger algorithm):
9*(n+1)^2*a(n) -(19n^2+58n+63)*a(n+1) + (11n^2+46n+47)*a(n+2)-(n+3)^2*a(n+3) = 0.
a(n) ~ A086618(n)/2 ~ 3^(2*n + 5/2)/(16*Pi*n^2) as n tends to the infinity.
a(n) = -hypergeom([-1/2, -n, -n], [1, 1], 4). - Peter Luschny, Nov 07 2023
EXAMPLE
a(2) = 9 since Sum_{k=0,1,2}C(2,k)^2*C(2k,k)/(2k-1) = -1 + 8 + 6/3 = 9.
MAPLE
a := n -> -hypergeom([-1/2, -n, -n], [1, 1], 4):
seq(simplify(a(n)), n=0..23); # Peter Luschny, Nov 07 2023
ogf := -(1-9*x)^(1/4)*hypergeom([-1/4, 3/4], [1], 64*x^3/((1-9*x)*(x-1)^3))/(1-x)^(5/4);
series(ogf, x=0, 25); # Mark van Hoeij, Nov 12 2023
MATHEMATICA
a[n_]:=Sum[Binomial[n, k]^2*Binomial[2k, k]/(2k-1), {k, 0, n}]
Table[a[n], {n, 0, 20}]
CROSSREFS
KEYWORD
sign
AUTHOR
Zhi-Wei Sun, Aug 24 2014
STATUS
approved