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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.

%I A134090 #12 Mar 16 2025 23:59:28
%S A134090 1,1,1,3,2,1,13,9,3,1,71,46,18,4,1,456,285,110,30,5,1,3337,2021,780,
%T A134090 215,45,6,1,27203,16023,6167,1729,371,63,7,1,243203,139812,53494,
%U A134090 15176,3346,588,84,8,1,2357356,1326111,504030,143814,32376,5886,876,108,9,1
%N 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.
%C A134090 Column 0 equals A122455 if we define A122455(0)=1.
%F A134090 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)].
%e A134090 Triangle T begins:
%e A134090 1;
%e A134090 1, 1;
%e A134090 3, 2, 1;
%e A134090 13, 9, 3, 1;
%e A134090 71, 46, 18, 4, 1;
%e A134090 456, 285, 110, 30, 5, 1;
%e A134090 3337, 2021, 780, 215, 45, 6, 1;
%e A134090 27203, 16023, 6167, 1729, 371, 63, 7, 1;
%e A134090 243203, 139812, 53494, 15176, 3346, 588, 84, 8, 1;
%e A134090 2357356, 1326111, 504030, 143814, 32376, 5886, 876, 108, 9, 1; ...
%e A134090 Let P denote the matrix equal to Pascal's triangle shift down 1 row:
%e A134090 P(n,k) = C(n+1,k) for n>k>=0, with P(n,n)=1 for n>=0.
%e A134090 Illustrate row n of T = row n of P^n as follows.
%e A134090 Matrix P = I + D*C begins:
%e A134090 1;
%e A134090 1, 1;
%e A134090 1, 1, 1;
%e A134090 1, 2, 1, 1;
%e A134090 1, 3, 3, 1, 1;
%e A134090 1, 4, 6, 4, 1, 1; ...
%e A134090 Matrix cube P^3 begins:
%e A134090 1;
%e A134090 3, 1;
%e A134090 6, 3, 1;
%e A134090 13, 9, 3, 1; <== row 3 of P^3 = row 3 of T
%e A134090 30, 25, 12, 3, 1;
%e A134090 73, 72, 40, 15, 3, 1; ...
%e A134090 Matrix 4th power P^4 begins:
%e A134090 1;
%e A134090 4, 1;
%e A134090 10, 4, 1;
%e A134090 26, 14, 4, 1;
%e A134090 71, 46, 18, 4, 1; <== row 4 of P^4 = row 4 of T
%e A134090 204, 155, 70, 22, 4, 1; ...
%e A134090 Matrix 5th power P^5 begins:
%e A134090 1;
%e A134090 5, 1;
%e A134090 15, 5, 1;
%e A134090 45, 20, 5, 1;
%e A134090 140, 75, 25, 5, 1;
%e A134090 456, 285, 110, 30, 5, 1; <== row 5 of P^5 = row 5 of T.
%o A134090 (PARI) \\ As generated by the g.f.
%o A134090 {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)}
%o A134090 (PARI) \\ As generated by matrix power: row n of T equals row n of P^n
%o A134090 {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]}
%Y A134090 Cf. columns: A134091, A134092, A134093; A134094 (row sums).
%K A134090 nonn,tabl
%O A134090 0,4
%A A134090 _Paul D. Hanna_, Oct 07 2007