A134090 Triangle, read by rows, where T(n,k) = [(I + D*C)^n](n,k); that is, row n of T = row n of (I + D*C)^n for n>=0 where C denotes Pascal's triangle, I the identity matrix and D a matrix where D(n+1,n)=1 and zeros elsewhere.
1, 1, 1, 3, 2, 1, 13, 9, 3, 1, 71, 46, 18, 4, 1, 456, 285, 110, 30, 5, 1, 3337, 2021, 780, 215, 45, 6, 1, 27203, 16023, 6167, 1729, 371, 63, 7, 1, 243203, 139812, 53494, 15176, 3346, 588, 84, 8, 1, 2357356, 1326111, 504030, 143814, 32376, 5886, 876, 108, 9, 1
Offset: 0
Examples
Triangle T begins: 1; 1, 1; 3, 2, 1; 13, 9, 3, 1; 71, 46, 18, 4, 1; 456, 285, 110, 30, 5, 1; 3337, 2021, 780, 215, 45, 6, 1; 27203, 16023, 6167, 1729, 371, 63, 7, 1; 243203, 139812, 53494, 15176, 3346, 588, 84, 8, 1; 2357356, 1326111, 504030, 143814, 32376, 5886, 876, 108, 9, 1; ... Let P denote the matrix equal to Pascal's triangle shift down 1 row: P(n,k) = C(n+1,k) for n>k>=0, with P(n,n)=1 for n>=0. Illustrate row n of T = row n of P^n as follows. Matrix P = I + D*C begins: 1; 1, 1; 1, 1, 1; 1, 2, 1, 1; 1, 3, 3, 1, 1; 1, 4, 6, 4, 1, 1; ... Matrix cube P^3 begins: 1; 3, 1; 6, 3, 1; 13, 9, 3, 1; <== row 3 of P^3 = row 3 of T 30, 25, 12, 3, 1; 73, 72, 40, 15, 3, 1; ... Matrix 4th power P^4 begins: 1; 4, 1; 10, 4, 1; 26, 14, 4, 1; 71, 46, 18, 4, 1; <== row 4 of P^4 = row 4 of T 204, 155, 70, 22, 4, 1; ... Matrix 5th power P^5 begins: 1; 5, 1; 15, 5, 1; 45, 20, 5, 1; 140, 75, 25, 5, 1; 456, 285, 110, 30, 5, 1; <== row 5 of P^5 = row 5 of T.
Programs
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PARI
\\ As generated by the g.f. {T(n,k)=polcoeff(sum(j=0,n,binomial(n,j)*x^j/(1-j*x)^k/prod(i=0,j,1-i*x+x*O(x^(n-k)))),n-k)} -
PARI
\\ As generated by matrix power: row n of T equals row n of P^n {T(n,k)=local(P=matrix(n+1,n+1,r,c,if(r==c,1,if(r>c,binomial(r-2,c-1)))));(P^n)[n+1,k+1]}
Formula
T(n,k) = [x^(n-k)] Sum_{j=0..n} C(n,j)*x^j/(1-j*x)^k /[Product_{i=0..j}(1-i*x)].
Comments