%I #5 Mar 30 2012 18:36:43
%S 1,1,1,1,2,1,2,5,4,1,1,5,8,5,1,3,13,22,18,7,1,1,9,26,35,24,8,1,4,26,
%T 70,101,84,40,10,1,1,14,61,131,160,116,49,11,1,5,45,171,363,476,400,
%U 215,71,13,1,1,20,120,363,654,752,565,275,83,14,1,6,71,356,1017,1856,2282,1932
%N Triangle, read by rows, such that row n equals the inverse binomial transform of column n of the triangle of trinomial coefficients (A027907), omitting leading zeros.
%C Row sums form A099603, where A099603(n) = fibonacci(n+1)*2^[(n+1)/2]. Central coefficients of even-indexed rows form A082759, where A082759(n) = Sum_{k=0..n} binomial(n,k)*Trinomial(n,k). Antidiagonal sums form A099604.
%C Matrix inverse equals triangle A104495, which is generated from self-convolutions of the Catalan sequence (A000108).
%F G.f.: (1 + (y+1)*x - (y+1)*x^2)/(1 - (y+1)*(y+2)*x^2 + (y+1)^2*x^4).
%e Rows begin:
%e [1],
%e [1,1],
%e [1,2,1],
%e [2,5,4,1],
%e [1,5,8,5,1],
%e [3,13,22,18,7,1],
%e [1,9,26,35,24,8,1],
%e [4,26,70,101,84,40,10,1],
%e [1,14,61,131,160,116,49,11,1],
%e [5,45,171,363,476,400,215,71,13,1],
%e [1,20,120,363,654,752,565,275,83,14,1],...
%e The binomial transform of row 2 equals column 2 of A027907:
%e BINOMIAL[1,2,1] = [1,3,6,10,15,21,28,36,45,55,...].
%e The binomial transform of row 3 equals column 3 of A027907:
%e BINOMIAL[2,5,4,1] = [2,7,16,30,50,77,112,156,210,...].
%e The binomial transform of row 4 equals column 4 of A027907:
%e BINOMIAL[1,5,8,5,1] = [1,6,19,45,90,161,266,414,615,...].
%e The binomial transform of row 5 equals column 5 of A027907:
%e BINOMIAL[3,13,22,18,7,1] = [3,16,51,126,266,504,882,1452,...].
%o (PARI) {T(n,k)=polcoeff(polcoeff((1+(y+1)*x-(y+1)*x^2)/(1-(y+1)*(y+2)*x^2+(y+1)^2*x^4)+x*O(x^n),n,x)+y*O(y^k),k,y)}
%o (PARI) {T(n,k)=(matrix(n+1,n+1,i,j,if(i>=j,polcoeff(polcoeff( (1+x*y/(1+x))/(1+x-y^2*(1-(1+4*x+O(x^i))^(1/2))^2/4+O(y^j)),i-1,x),j-1,y)))^-1)[n+1,k+1]}
%Y Cf. A027907, A082759, A099603, A099604.
%Y Cf. A104495.
%K nonn,tabl
%O 0,5
%A _Paul D. Hanna_, Oct 25 2004
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