%I #39 Aug 11 2024 22:06:06
%S 0,1,7,42,242,1375,7773,43876,247684,1399293,7913955,44812878,
%T 254064726,1442131899,8195232633,46621424520,265490365448,
%U 1513290869881,8633347134975,49293941140402,281670686307130
%N a(n) = T(n,n+3), array T as in A050143.
%H Vincenzo Librandi, <a href="/A050152/b050152.txt">Table of n, a(n) for n = 0..200</a>
%H Milan Janjic, <a href="http://www.pmfbl.org/janjic/">Two Enumerative Functions</a>
%F G.f.: (x*sqrt(x^2-6*x+1)-x^2+3*x)/(-x^4+sqrt(x^2-6*x+1)*(x^3-3*x^2-3*x+1)+6*x^3-2*x^2+6*x-1)+1/(4*x)-1/(4*x^2). - _Vladimir Kruchinin_, May 24 2014
%F a(n) = Sum_{i=0..n+2} 2^(i-2)*(-1)^(n-i)*binomial(n,n-i+2)*binomial(n+i-1,n-1). - _Vladimir Kruchinin_, May 24 2014
%F a(n) ~ sqrt(48+34*sqrt(2)) * (3+2*sqrt(2))^n / (8*sqrt(Pi*n)). - _Vaclav Kotesovec_, May 24 2014
%F a(n) = (-1)^n*(n+1)*(n/2)*hypergeom([-n, n+2], [3], 2). - _Peter Luschny_, May 24 2014
%F n^2*(n+1)*a(n-1) = Sum_{k=0..n-1} (2*k^3+k^2+k)*binomial(n-1,k)*binomial(n+k,k) for all n > 0. This follows from the Zeilberger algorithm. - _Zhi-Wei Sun_, Aug 30 2014
%F a(n) = Sum_{k=0..n} (binomial(n,k)*binomial(2*n-k+1,n-k-1)). - _Vladimir Kruchinin_, Oct 26 2016
%p a := n -> (-1)^n*(n+1)*(n/2)*hypergeom([-n,n+2], [3], 2);
%p seq(round(evalf(a(n),32)), n=0..20); # _Peter Luschny_, May 24 2014
%t Table[JacobiP[n-1, 1, 2, 3], {n, 0, 20}] (* _Vladimir Joseph Stephan Orlovsky_, Sep 12 2008 *)
%o (Maxima)
%o a(n):=sum(2^(i-2)*(-1)^(n-i)*binomial(n,n-i+2)*binomial(n+i-1,n-1),i,0,n+2); /* _Vladimir Kruchinin_, May 24 2014 */
%Y Cf. A050143.
%K nonn
%O 0,3
%A _Clark Kimberling_, Dec 11 1999
%E Typo in Mathematica code fixed by _Vincenzo Librandi_, May 26 2013