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Column 0 of triangle A116077.
2

%I #31 Jul 25 2019 11:12:04

%S 1,2,7,28,117,496,2110,8968,38017,160648,676626,2840872,11892562,

%T 49650368,206773372,859186768,3562780057,14746127608,60929182282,

%U 251358948328,1035479267542,4260071237728,17505144024292

%N Column 0 of triangle A116077.

%C a(n) equals the number of sequences (x(1),...,x(n)) of n numbers from {0,1,...,n} such that x(i+1) <= x(i)+1 for i=1,...,n-1 and x(1) <= x(n)+1. This is proved in a linked pdf, as well as another interpretation as the downward closed subsets of a certain poset. - _Clayton Thomas_, Jul 16 2019

%H Vincenzo Librandi, <a href="/A116078/b116078.txt">Table of n, a(n) for n = 0..300</a>

%H Clayton Thomas, <a href="/A116078/a116078.pdf">A note on the cylinder poset</a>

%F G.f.: A(x) = 1/sqrt(1-4*x) + 4*x^2/(1+sqrt(1-4*x))^2/(1-4*x)^(3/2).

%F a(n) = (n+3)*C(2*n-1,n) - 2^(2*n-1), a(n) ~ 2^(2*n - 1) * sqrt(n) / sqrt(Pi). - _Vaclav Kotesovec_, Oct 28 2012

%F Conjecture: a(n) = 2^(2*n)*(Sum_{j=1..n+2-floor((n+3)/2)} (cos(j*Pi/(n+3)))^(2*n)). - _L. Edson Jeffery_, Nov 23 2013

%t Flatten[{1,Table[(n+3)*Binomial[2*n-1,n]-2^(2*n-1),{n,1,20}]}] (* _Vaclav Kotesovec_, Oct 28 2012 *)

%o (PARI) {a(n)=local(X=x+x*O(x^n)); polcoeff(1/sqrt(1-4*X)+4*X^2/(1+sqrt(1-4*X))^2/(1-4*X)^(3/2),n,x)}

%Y Cf. A116077.

%K nonn

%O 0,2

%A _Paul D. Hanna_, Feb 04 2006