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A120894
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Cascadence of 1+x+x^2; a triangle, read by rows of 2n+1 terms, that retains its original form upon convolving each row with [1,1,1] 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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9
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1, 1, 1, 1, 2, 3, 2, 1, 2, 5, 7, 6, 5, 3, 2, 5, 12, 18, 18, 14, 10, 10, 7, 5, 12, 30, 48, 50, 42, 34, 27, 22, 24, 17, 12, 30, 78, 128, 140, 126, 103, 83, 73, 63, 53, 59, 42, 30, 78, 206, 346, 394, 369, 312, 259, 219, 189, 175, 154, 131, 150, 108, 78, 206, 552, 946, 1109
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OFFSET
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0,5
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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+G+G^2), so that G/x = g.f. of Motzkin numbers (A001006). 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+x+x^2. - Paul D. Hanna, Jul 17 2006
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EXAMPLE
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Triangle begins:
1;
1, 1, 1;
2, 3, 2, 1, 2;
5, 7, 6, 5, 3, 2, 5;
12, 18, 18, 14, 10, 10, 7, 5, 12;
30, 48, 50, 42, 34, 27, 22, 24, 17, 12, 30;
78, 128, 140, 126, 103, 83, 73, 63, 53, 59, 42, 30, 78;
206, 346, 394, 369, 312, 259, 219, 189, 175, 154, 131, 150, 108, 78, 206;
552, 946, 1109, 1075, 940, 790, 667, 583, 518, 460, 435, 389, 336, 392, 284, 206, 552;
1498, 2607, 3130, 3124, 2805, 2397, 2040, 1768, 1561, 1413, 1284, 1160, 1117, 1012, 882, 1042, 758, 552, 1498; ...
Convolution of [1,1,1] with each row produces:
[1,1,1]*[1] = [1,1,1];
[1,1,1]*[1,1,1] = [1,2,3,2,1];
[1,1,1]*[2,3,2,1,2] = [2,5,7,6,5,3,2];
[1,1,1]*[5,7,6,5,3,2,5] = [5,12,18,18,14,10,10,7,5];
[1,1,1]*[12,18,18,14,10,10,7,5,12] = [12,30,48,50,42,34,27,22,24,17,12]; ...
These convoluted rows, when concatenated, yield the sequence:
1,1,1, 1,2,3,2,1, 2,5,7,6,5,3,2, 5,12,18,18,14,10,10,7,5, ...
which equals the concatenated rows of this original triangle:
1, 1,1,1, 2,3,2,1,2, 5,7,6,5,3,2,5, 12,18,18,14,10,10,7,5,12, ...
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PROG
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(PARI) T(n, k)=if(2*n<k || k<0, 0, if(n<=1, 1, if(k==0, T(n-1, 0)+T(n-1, 1), if(k==2*n, T(n, 0), T(n-1, k-1)+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+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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