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a(n) = Sum_{k=0..floor(n/2)} (-1)^k * binomial(2*n-2*k-1,n-2*k).
5

%I #22 Dec 28 2024 11:44:25

%S 1,1,2,7,26,96,356,1331,5014,19006,72412,277058,1063856,4097510,

%T 15823432,61245987,237536326,922906150,3591500972,13996328322,

%U 54614894396,213360770840,834409399672,3266370155262,12797894251276,50184309630196,196936674150296

%N a(n) = Sum_{k=0..floor(n/2)} (-1)^k * binomial(2*n-2*k-1,n-2*k).

%H Harvey P. Dale, <a href="/A371798/b371798.txt">Table of n, a(n) for n = 0..1000</a>

%H Sergi Elizalde, Nadia Lafrenière, Joel Brewster Lewis, Erin McNicholas, Jessica Striker, and Amanda Welch, <a href="https://arxiv.org/abs/2412.16368">Enumeration of interval-closed sets via Motzkin paths and quarter-plane walks</a>, arXiv:2412.16368 [math.CO], 2024. See p. 13.

%F a(n) = [x^n] 1/((1+x^2) * (1-x)^n).

%F a(n) = binomial(2*n-1, n)*hypergeom([1, (1-n)/2, -n/2], [1/2-n, 1-n], -1). - _Stefano Spezia_, Apr 06 2024

%F a(n) ~ 2^(2*n+1) / (5*sqrt(Pi*n)). - _Vaclav Kotesovec_, Apr 07 2024

%F Conjectured g.f.: 1 + x*(4 - 10*x + 8*x^2)/(2 - 11*x + 14*x^2 - 8*x^3 + (2 - 3*x)*sqrt(1 - 4*x)) (see Elizalde et al. at p. 13). - _Stefano Spezia_, Dec 27 2024

%t Table[Sum[(-1)^k Binomial[2n-2k-1,n-2k],{k,0,Floor[n/2]}],{n,0,30}] (* _Harvey P. Dale_, Oct 31 2024 *)

%o (PARI) a(n) = sum(k=0, n\2, (-1)^k*binomial(2*n-2*k-1, n-2*k));

%Y Cf. A026641, A120305.

%K nonn

%O 0,3

%A _Seiichi Manyama_, Apr 06 2024