A120914 Cascadence of (1+2x)^2; a triangle, read by rows of 2n+1 terms, that retains its original form upon convolving each row with [1,4,4] 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.
1, 4, 4, 4, 20, 36, 32, 16, 20, 116, 256, 288, 212, 144, 80, 116, 720, 1776, 2388, 2144, 1504, 1012, 784, 464, 720, 4656, 12372, 18800, 19632, 15604, 10848, 7648, 5712, 4736, 2880, 4656, 30996, 86912, 144320, 169332, 151792, 113456, 79696, 58176
Offset: 0
Examples
Triangle begins: 1; 4, 4, 4; 20, 36, 32, 16, 20; 116, 256, 288, 212, 144, 80, 116; 720, 1776, 2388, 2144, 1504, 1012, 784, 464, 720; 4656, 12372, 18800, 19632, 15604, 10848, 7648, 5712, 4736, 2880, 4656; Convolution of [1,4,4] with each row produces: [1,4,4]*[1] = [1,4,4]; [1,4,4]*[4,4,4] = [4,20,36,32,16]; [1,4,4]*[20,36,32,16,20] = [20,116,256,288,212,144,80]; [1,4,4]*[116,256,288,212,144,80,116] = [116,720,1776,2388,2144,1504,1012,784,464]; These convoluted rows, when concatenated, yield the sequence: 1,4,4, 4,20,36,32,16, 20,116,256,288,212,144,80, 116,720,1776,2388,... which equals the concatenated rows of this original triangle: 1, 4,4,4, 20,36,32,16,20, 116,256,288,212,144,80,116, 720,1776,2388,...
Crossrefs
Programs
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PARI
/* Generate Triangle by the Recurrence: */ {T(n,k)=if(2*n -
PARI
/* Generate Triangle by the G.F.: */ {T(n,k)=local(A,F=1+4*x+4*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,n,print1(T(n,k),", "));print(""))
Formula
G.f.: A(x,y) = ( x*H(x) - y*H(x*y^2) )/( x*(1+2y)^2 - y ), where H(x) satisfies: H(x) = G*H(x*G^2)/x and G satisfies: G = x*(1 + 2G)^2 ; also, H(x) is the g.f. of column 0.
Comments