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A120898
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Cascadence of 1+2x+x^2; a triangle, read by rows of 2n+1 terms, that retains its original form upon convolving each row with [1,2,1] and then letting excess terms spill over from each row into the initial positions of the next row such that only 2n+1 terms remain in row n for n>=0.
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10
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1, 2, 1, 2, 5, 6, 5, 2, 5, 16, 22, 18, 14, 12, 5, 16, 54, 78, 72, 58, 43, 38, 37, 16, 54, 186, 282, 280, 231, 182, 156, 128, 123, 124, 54, 186, 654, 1030, 1073, 924, 751, 622, 535, 498, 425, 418, 426, 186, 654, 2338, 3787, 4100, 3672, 3048, 2530, 2190, 1956, 1766
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OFFSET
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0,2
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COMMENTS
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In this case, the g.f. of column 0, H(x), satisfies: H(x) = H(x*G^2)*G/x where G satisfies: G = x*(1+2G+G^2), so that 1+G = g.f. of Catalan numbers (A000108). More generally, the cascadence of polynomial F(x) of degree d, F(0)=1, is a triangle with d*n+1 terms in row n where the g.f. H(x) of column 0 satisfies: H(x) = H(x*G^d)*G/x where G = x*F(G); thus G = series_reversion(x/F(x)), or, equivalently, [x^n] G = [x^n] x*F(x)^n/n for n>=1.
Further, the g.f. of the cascadence triangle for polynomial F(x) of degree d is given by: A(x,y) = ( x*H(x) - y*H(x*y^d) )/( x*F(y) - y ), where H(x) = G*H(x*G^d)/x and G = x*F(G). - Paul D. Hanna, Jul 17 2006
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LINKS
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FORMULA
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G.f.: A(x,y) = ( x*H(x) - y*H(x*y^2) )/( x*F(y) - y ), where H(x) = G*H(x*G^2)/x, G = x*F(G), F(x)=1+2x+x^2. - Paul D. Hanna, Jul 17 2006
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EXAMPLE
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Triangle begins:
1;
2, 1, 2;
5, 6, 5, 2, 5;
16, 22, 18, 14, 12, 5, 16;
54, 78, 72, 58, 43, 38, 37, 16, 54;
186, 282, 280, 231, 182, 156, 128, 123, 124, 54, 186;
654, 1030, 1073, 924, 751, 622, 535, 498, 425, 418, 426, 186, 654;
2338, 3787, 4100, 3672, 3048, 2530, 2190, 1956, 1766, 1687, 1456, 1452, 1494, 654, 2338; ...
Convolution of [1,2,1] with each row produces:
[1,2,1]*[1] = [1,2,1];
[1,2,1]*[2,1,2] = [2,5,6,5,2];
[1,2,1]*[5,6,5,2,5] = [5,16,22,18,14,12,5];
[1,2,1]*[16,22,18,14,12,5,16] = [16,54,78,72,58,43,38,37,16];
These convoluted rows, when concatenated, yield the sequence:
1,2,1, 2,5,6,5,2, 5,16,22,18,14,12,5, 16,54,78,72,58,43,38,37,16, ...
which equals the concatenated rows of this original triangle:
1, 2,1,2, 5,6,5,2,5, 16,22,18,14,12,5,16, 54,78,72,58,43,38,37,16,54,
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PROG
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(PARI) T(n, k)=if(2*n<k || k<0, 0, if(n==0 && k==0, 1, if(k==0, T(n-1, 1)+2*T(n-1, 0), if(k==2*n, T(n, 0), T(n-1, k-1)+2*T(n-1, k)+T(n-1, k+1)))))
for(n=0, 10, for(k=0, 2*n, print1(T(n, k), ", ")); print(""))
(PARI) /* Generated by the G.F.: */
{T(n, k)=local(A, F=1+2*x+x^2, d=2, G=x, H=1+x, S=ceil(log(n+1)/log(d+1))); for(i=0, n, G=x*subst(F, x, G+x*O(x^n))); for(i=0, S, H=subst(H, x, x*G^d+x*O(x^n))*G/x); A=(x*H-y*subst(H, x, x*y^d +x*O(x^n)))/(x*subst(F, x, y)-y); polcoeff(polcoeff(A, n, x), k, y)}
for(n=0, 10, for(k=0, 2*n, print1(T(n, k), ", ")); print()) \\ Paul D. Hanna, Jul 17 2006
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CROSSREFS
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KEYWORD
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nonn,tabf
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AUTHOR
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STATUS
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approved
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