A118650 Center column a(2n,n) of A106597.
1, 3, 14, 72, 388, 2150, 12134, 69370, 400398, 2328162, 13616892, 80022612, 472133638, 2794893246, 16592160944, 98743919468, 588913687966, 3518968100982, 21062444053040, 126256612255260, 757853797478188
Offset: 0
Keywords
Crossrefs
Cf. A106597.
Programs
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Mathematica
Table[Sum[Binomial[2k,k](-1)^(n-k)Sum[Binomial[k,j]Binomial[k+1,n-k-j]2^(k-j),{j,0,k}],{k,0,n}],{n,0,40}] (* Emanuele Munarini, Feb 06 2017 *) -
Maxima
makelist(sum(binomial(2*k,k)*(-1)^(n-k)*sum(binomial(k,j)*binomial(k+1,n-k-j)*2^(k-j),j,0,k),k,0,n),n,0,12); /* Emanuele Munarini, Feb 06 2017 */
Formula
G.f.: (1-x)/sqrt(1-8*x+12*x^2-4*x^3). - Mark van Hoeij, Apr 16 2013
Conjecture: n*a(n) +3*(-3*n+2)*a(n-1) +4*(5*n-8)*a(n-2) +2*(-8*n+21)*a(n-3) +2*(2*n-7)*a(n-4)=0. - R. J. Mathar, Nov 10 2013
From Emanuele Munarini, Feb 06 2017: (Start)
a(n) = Sum_{k=0..n} binomial(2*k,k)*(-1)^(n-k) * Sum_{j=0..k} binomial(k,j)*binomial(k+1,n-k-j)*2^(k-j).
Proof of Mathar's recurrence.
Let A(t) be the g.f. of the coefficients a(n). Then we have the identity (1 - 9*t + 20*t^2 - 16*t^3 + 4*t^4)*A'(t) = (3 - 8*t + 6*t^2 - 2*t^3)*A(t).
Let R be the incremental ratio, i.e. the operator defined by RA(t) = (A(t)-A(0))/t, giving the g.f. of the shifted sequence a(n+1), then we have
R^3A'(t) - 9*R^2A'(t) + 20*RA'(t) - 16*A'(t) + 4*t*A'(t) = 3*R^3A(t) - 8*R^2A(t) + 6*RA(t) - 2*A(t), from which we obtain the recurrence:
(n+4)*a(n+4)-3*(3*n+10)*a(n+3)+4*(5*n+12)*a(n+2)-2*(8*n+11)*a(n+1)+2*(2*n+1)*a(n)=0.
This proves the above conjecture.
(End)
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