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%I #26 May 11 2020 05:43:34
%S 1,3,14,72,388,2150,12134,69370,400398,2328162,13616892,80022612,
%T 472133638,2794893246,16592160944,98743919468,588913687966,
%U 3518968100982,21062444053040,126256612255260,757853797478188
%N Center column a(2n,n) of A106597.
%C Number of lattice paths from (0,0) to (n,n) using steps (1,0), (0,1), and (s,s) for s>=1. [_Joerg Arndt_, Jul 01 2011]
%F G.f.: (1-x)/sqrt(1-8*x+12*x^2-4*x^3). - _Mark van Hoeij_, Apr 16 2013
%F 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
%F From _Emanuele Munarini_, Feb 06 2017: (Start)
%F 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).
%F Proof of Mathar's recurrence.
%F 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).
%F 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
%F 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:
%F (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.
%F This proves the above conjecture.
%F (End)
%t 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 *)
%o (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 */
%Y Cf. A106597.
%K nonn
%O 0,2
%A _Joshua Zucker_, May 10 2006
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