%I #26 Apr 25 2020 08:44:08
%S 1,1,1,1,1,1,3,2,1,1,19,9,3,1,1,191,70,18,4,1,1,2646,795,170,30,5,1,1,
%T 46737,11961,2220,335,45,6,1,1,1003150,224504,37149,4984,581,63,7,1,1,
%U 25330125,5051866,758814,92652,9730,924,84,8,1,1,735180292,132523155,18301950,2065146,199692,17226,1380,108,9,1,1
%N Triangular matrix T, read by rows, where row n equals row (n-1) of T^(n-1) after appending '1' for the main diagonal.
%C Remarkably, T equals the product of these triangular matrices: T = A107867*A107862^-1 = A107870*A107867^-1 = A107873*A107870^-1; reversing the order of these products yields triangle A107876.
%H Alois P. Heinz, <a href="/A101479/b101479.txt">Rows n = 0..140, flattened</a> (first 31 rows from Paul D. Hanna)
%e Triangle begins:
%e 1;
%e 1, 1;
%e 1, 1, 1;
%e 3, 2, 1, 1;
%e 19, 9, 3, 1, 1;
%e 191, 70, 18, 4, 1, 1;
%e 2646, 795, 170, 30, 5, 1, 1;
%e 46737, 11961, 2220, 335, 45, 6, 1, 1;
%e 1003150, 224504, 37149, 4984, 581, 63, 7, 1, 1;
%e 25330125, 5051866, 758814, 92652, 9730, 924, 84, 8, 1, 1;
%e 735180292, 132523155, 18301950, 2065146, 199692, 17226, 1380, 108, 9, 1, 1; ...
%e Row 4 starts with row 3 of T^3 which begins:
%e 1;
%e 3, 1;
%e 6, 3, 1;
%e 19, 9, 3, 1; ...
%e row 5 starts with row 4 of T^4 which begins:
%e 1;
%e 4, 1;
%e 10, 4, 1;
%e 34, 14, 4, 1;
%e 191, 70, 18, 4, 1; ...
%e An ALTERNATE GENERATING METHOD is illustrated as follows.
%e For row 4:
%e Start with a '1' and append 2 zeros,
%e take partial sums and append 1 zero,
%e take partial sums thrice more, resulting in:
%e 1, 0, 0;
%e 1, 1, 1, 0;
%e 1, 2, 3, 3;
%e 1, 3, 6, 9;
%e 1, 4,10,19.
%e Final nonzero terms form row 4: [19,9,3,1,1].
%e For row 5:
%e Start with a '1' and append 3 zeros,
%e take partial sums and append 2 zeros,
%e take partial sums and append 1 zero,
%e take partial sums thrice more, resulting in:
%e 1, 0, 0, 0;
%e 1, 1, 1, 1, 0, 0;
%e 1, 2, 3, 4, 4, 4, 0;
%e 1, 3, 6,10,14, 18, 18;
%e 1, 4,10,20,34, 52, 70;
%e 1, 5,15,35,69,121,191;
%e where the final nonzero terms form row 5: [191,70,18,4,1,1].
%e Likewise, for row 6:
%e 1, 0, 0, 0, 0;
%e 1, 1, 1, 1, 1, 0, 0, 0;
%e 1, 2, 3, 4, 5, 5, 5, 5, 0, 0;
%e 1, 3, 6,10, 15, 20, 25, 30, 30, 30, 0;
%e 1, 4,10,20, 35, 55, 80,110, 140, 170, 170;
%e 1, 5,15,35, 70,125,205,315, 455, 625, 795;
%e 1, 6,21,56,126,251,456,771,1226,1851,2646;
%e where the final nonzero terms form row 6: [2646,795,170,30,5,1,1].
%e Continuing in this way generates all rows of this triangle.
%p b:= proc(n) option remember;
%p Matrix(n, (i,j)-> T(i-1,j-1))^(n-1)
%p end:
%p T:= proc(n,k) option remember;
%p `if`(n=k, 1, `if`(k>n, 0, b(n)[n,k+1]))
%p end:
%p seq(seq(T(n,k), k=0..n), n=0..10); # _Alois P. Heinz_, Apr 13 2020
%t b[n_] := b[n] = MatrixPower[Table[T[i-1, j-1], {i, n}, {j, n}], n-1];
%t T[n_, k_] := T[n, k] = If[n == k, 1, If[k > n, 0, b[n][[n, k+1]]]];
%t Table[Table[T[n, k], {k, 0, n}], {n, 0, 10}] // Flatten (* _Jean-François Alcover_, Apr 25 2020, after _Alois P. Heinz_ *)
%o (PARI) {T(n,k) = my(A=Mat(1),B); for(m=1,n+1, B=matrix(m,m); for(i=1,m, for(j=1,i, if(j==i,B[i,j]=1, B[i,j] = (A^(i-2))[i-1,j]);)); A=B); return(A[n+1,k+1])}
%o for(n=0,10, for(k=0,n, print1(T(n,k),", ")); print(""))
%o (PARI) {T(n,k) = my(A=vector(n+1),p); A[1]=1; for(j=1,n-k-1, p=(n-1)*(n-2)/2-(n-j-1)*(n-j-2)/2; A = Vec((Polrev(A)+x*O(x^p))/(1-x))); A = Vec((Polrev(A) +x*O(x^p)) / (1-x) ); A[p+1]}
%o for(n=0,10, for(k=0,n, print1(T(n,k),", ")); print(""))
%Y Columns are A101481, A101482, A101483, row sums form A101484.
%Y Cf. A107876 (dual triangle).
%Y Cf. A304184, A304185, A304186, A304187.
%K nonn,tabl
%O 0,7
%A _Paul D. Hanna_, Jan 21 2005, Jul 26 2006, May 27 2007