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%I #14 Aug 15 2017 03:15:46
%S 1,1,2,2,1,4,9,10,5,1,6,21,44,57,42,14,1,8,38,116,240,336,308,168,42,
%T 1,10,60,240,680,1392,2060,2160,1530,660,132,1,12,87,430,1545,4152,
%U 8449,13014,14985,12540,7227,2574,429,1,14,119,700,3045,10122,26173,53048
%N Triangle read by rows in which row n >= 1 gives coefficients in expansion of the polynomial Sum_{k=1..n} (1/n)*binomial(n,k)*binomial(n,k-1)*x^(2k)*(1+x)^(2n-2k) / x^2 in powers of x.
%C Row n has 2n-1 terms.
%H C. Coker, <a href="http://dx.doi.org/10.1016/S0012-365X(03)00037-2">Enumerating a class of lattice paths</a>, Discrete Math., 271 (2003), 13-28.
%e For n=3 the polynomial is 1 + 4x + 9x^2 + 10x^3 + 5x^4.
%e 1;
%e 1, 2, 2;
%e 1, 4, 9, 10, 5;
%e 1, 6, 21, 44, 57, 42, 14;
%e 1, 8, 38, 116, 240, 336, 308, 168, 42;
%e 1, 10, 60, 240, 680, 1392, 2060, 2160, 1530, 660, 132;
%e 1, 12, 87, 430, 1545, 4152, 8449, 13014, 14985, 12540, 7227, 2574, 429;
%p j := 0:f := n->sum(binomial(n,k)*binomial(n,k-1)/n*x^(2*k)*(1+x)^(2*n-2*k),k=1..n): for n from 1 to 15 do p := expand(f(n)/x^2):for l from 0 to 2*n-2 do j := j+1: a[j] := coeff(p,x,l):od:od:seq(a[l],l=1..j); # _Sascha Kurz_
%o (PARI) for(n=1,8,p=sum(k=1,n,(1/n)*binomial(n,k)*binomial(n,k-1)*x^(2*k)*(1+x)^(2*n-2*k))/x^2; for(i=1,2*n-1,print1(polcoeff(p,i-1) ","); ); print; ); \\ _Ray Chandler_, Sep 17 2003
%Y A059231 gives row sums.
%K nonn,easy,tabf
%O 1,3
%A _N. J. A. Sloane_, Sep 16 2003
%E More terms from _Vladeta Jovovic_ and _Ray Chandler_, Sep 17 2003
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