This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A295112 #52 Sep 03 2025 10:37:25
%S A295112 -1,-1,1,5,13,29,63,139,317,749,1827,4575,11699,30419,80161,213573,
%T A295112 574253,1556077,4244835,11647151,32122231,88995879,247573565,
%U A295112 691246369,1936445619,5441165699,15331341373,43308322049,122624939677,347957102909,989335822559,2818200111867
%N A295112 a(n) = Sum_{k=0..n} binomial(n,2*k)*binomial(2*k,k)/(2*k-1).
%C A295112 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.
%C A295112 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.
%C A295112 See also A295113 for a conjecture involving the current sequence.
%H A295112 Zhi-Wei Sun, <a href="/A295112/b295112.txt">Table of n, a(n) for n = 0..400</a>
%H A295112 Min Bian, Olivia X. M. Yao, Yan Zhang, and Alina F. Y. Zhao, <a href="https://doi.org/10.11575/cdm.v14i1.62700">Proof of a conjecture of Z. W. Sun</a>, Contributions to Discrete Mathematics (2019) Vol. 14, No. 1, 214-221.
%H A295112 Zhi-Wei Sun, <a href="http://arxiv.org/abs/1208.2683">Conjectures involving arithmetical sequences</a>, 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.
%F A295112 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.
%F A295112 a(n) = -hypergeom([-1/2, 1/2 - n/2, -n/2], [1/2, 1], 4). - _Peter Luschny_, Nov 15 2017
%F A295112 a(n) ~ 3^(n + 3/2) / (4*sqrt(Pi)*n^(3/2)). - _Vaclav Kotesovec_, Nov 15 2017
%F A295112 From _Mélika Tebni_, Sep 03 2025: (Start)
%F A295112 G.f.: -sqrt((1 + x)*(1 - 3*x)) / (1 - x)^2.
%F A295112 a(n) = -Sum_{k=0..n} Sum_{j=0..k} A167022(j). (End)
%e A295112 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.
%p A295112 a := n -> -hypergeom([-1/2, 1/2 - n/2, -n/2], [1/2, 1], 4):
%p A295112 seq(simplify(a(n)), n=0..31); # _Peter Luschny_, Nov 15 2017
%t A295112 W[n_]:=W[n]=Sum[Binomial[n,2k]Binomial[2k,k]/(2k-1),{k,0,n/2}]; Table[W[n],{n,0,35}]
%t A295112 a[n_] := -AppellF1[-n, -1/2, -1/2, 1, 2, -2]; Table[a[n], {n,0,31}] (* _Peter Luschny_, Nov 15 2017 *)
%Y A295112 Cf. A000108, A001006, A167022, A295113.
%K A295112 sign,easy,changed
%O A295112 0,4
%A A295112 _Zhi-Wei Sun_, Nov 14 2017