A295112 a(n) = Sum_{k=0..n} binomial(n,2*k)*binomial(2*k,k)/(2*k-1).

-1, -1, 1, 5, 13, 29, 63, 139, 317, 749, 1827, 4575, 11699, 30419, 80161, 213573, 574253, 1556077, 4244835, 11647151, 32122231, 88995879, 247573565, 691246369, 1936445619, 5441165699, 15331341373, 43308322049, 122624939677, 347957102909, 989335822559, 2818200111867

Offset: 0

Author

Zhi-Wei Sun, Nov 14 2017

Keywords

• sign
• easy
• changed

Comments

As binomial(2*k,k) = 2*(2*k-1)*A000108(k-1) for all k = 1,2,..., we see that a(n) is always an odd integer. Clearly, a(n) > 0 for all n > 1. a(n) can be viewed as an analog of Motzkin numbers, which should have some combinatorial interpretations.

Conjecture: The sequence a(n+1)/a(n) (n = 5,6,...) is strictly increasing with limit 3, and the sequence a(n+1)^(1/(n+1))/a(n)^(1/n) (n = 9,10,...) is strictly decreasing to the limit 1.

See also A295113 for a conjecture involving the current sequence.

Examples

			a(3) = 5 since binomial(3,2*0)*binomial(2*0,0)/(2*0-1) + binomial(3,2*1)*binomial(2*1,1)/(2*1-1) = -1 + 3*2 = 5.
		

Links

• Zhi-Wei Sun, Table of n, a(n) for n = 0..400
• Min Bian, Olivia X. M. Yao, Yan Zhang, and Alina F. Y. Zhao, Proof of a conjecture of Z. W. Sun, Contributions to Discrete Mathematics (2019) Vol. 14, No. 1, 214-221.
• Zhi-Wei Sun, Conjectures involving arithmetical sequences, arXiv:1208.2683 [math.CO], 2012-2013; in: Number Theory: Arithmetic in Shangri-La (eds., S. Kanemitsu, H. Li and J. Liu), Proc. 6th China-Japan Seminar (Shanghai, August 15-17, 2011), World Sci., Singapore, 2013, pp. 244-258.

Crossrefs

Cf. A000108, A001006, A167022, A295113.

Programs

Maple

a := n -> -hypergeom([-1/2, 1/2 - n/2, -n/2], [1/2, 1], 4):
seq(simplify(a(n)), n=0..31); # Peter Luschny, Nov 15 2017

Mathematica

W[n_]:=W[n]=Sum[Binomial[n,2k]Binomial[2k,k]/(2k-1),{k,0,n/2}]; Table[W[n],{n,0,35}]
a[n_] := -AppellF1[-n, -1/2, -1/2, 1, 2, -2]; Table[a[n], {n,0,31}] (* Peter Luschny, Nov 15 2017 *)

Formula

Via the Zeilberger algorithm we have the recurrence: (n+3)*a(n+3) = (3n+7)*a(n+2) + (n-5)*a(n+1) - 3*(n+1)*a(n) for any nonnegative integer n.

a(n) = -hypergeom([-1/2, 1/2 - n/2, -n/2], [1/2, 1], 4). - Peter Luschny, Nov 15 2017

a(n) ~ 3^(n + 3/2) / (4*sqrt(Pi)*n^(3/2)). - Vaclav Kotesovec, Nov 15 2017

From Mélika Tebni, Sep 03 2025: (Start)

G.f.: -sqrt((1 + x)*(1 - 3*x)) / (1 - x)^2.

a(n) = -Sum_{k=0..n} Sum_{j=0..k} A167022(j). (End)