A101479 Triangular matrix T, read by rows, where row n equals row (n-1) of T^(n-1) after appending '1' for the main diagonal.

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, 46737, 11961, 2220, 335, 45, 6, 1, 1, 1003150, 224504, 37149, 4984, 581, 63, 7, 1, 1, 25330125, 5051866, 758814, 92652, 9730, 924, 84, 8, 1, 1, 735180292, 132523155, 18301950, 2065146, 199692, 17226, 1380, 108, 9, 1, 1

Offset: 0

Author

Paul D. Hanna, Jan 21 2005, Jul 26 2006, May 27 2007

Keywords

• nonn
• tabl

Comments

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.

Examples

			Triangle begins:
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;
46737, 11961, 2220, 335, 45, 6, 1, 1;
1003150, 224504, 37149, 4984, 581, 63, 7, 1, 1;
25330125, 5051866, 758814, 92652, 9730, 924, 84, 8, 1, 1;
735180292, 132523155, 18301950, 2065146, 199692, 17226, 1380, 108, 9, 1, 1; ...
Row 4 starts with row 3 of T^3 which begins:
1;
3, 1;
6, 3, 1;
19, 9, 3, 1; ...
row 5 starts with row 4 of T^4 which begins:
1;
4, 1;
10, 4, 1;
34, 14, 4, 1;
191, 70, 18, 4, 1; ...
An ALTERNATE GENERATING METHOD is illustrated as follows.
For row 4:
Start with a '1' and append 2 zeros,
take partial sums and append 1 zero,
take partial sums thrice more, resulting in:
1, 0, 0;
1, 1, 1, 0;
1, 2, 3, 3;
1, 3, 6, 9;
1, 4,10,19.
Final nonzero terms form row 4: [19,9,3,1,1].
For row 5:
Start with a '1' and append 3 zeros,
take partial sums and append 2 zeros,
take partial sums and append 1 zero,
take partial sums thrice more, resulting in:
1, 0, 0, 0;
1, 1, 1, 1, 0,  0;
1, 2, 3, 4, 4,  4,  0;
1, 3, 6,10,14, 18, 18;
1, 4,10,20,34, 52, 70;
1, 5,15,35,69,121,191;
where the final nonzero terms form row 5: [191,70,18,4,1,1].
Likewise, for row 6:
1, 0, 0, 0,  0;
1, 1, 1, 1,  1,  0,  0,  0;
1, 2, 3, 4,  5,  5,  5,  5,   0,   0;
1, 3, 6,10, 15, 20, 25, 30,  30,  30,   0;
1, 4,10,20, 35, 55, 80,110, 140, 170, 170;
1, 5,15,35, 70,125,205,315, 455, 625, 795;
1, 6,21,56,126,251,456,771,1226,1851,2646;
where the final nonzero terms form row 6: [2646,795,170,30,5,1,1].
Continuing in this way generates all rows of this triangle.
		

Links

• Alois P. Heinz, Rows n = 0..140, flattened (first 31 rows from Paul D. Hanna)

Crossrefs

Columns are A101481, A101482, A101483, row sums form A101484.

Cf. A107876 (dual triangle).

Cf. A304184, A304185, A304186, A304187.

Programs

Maple

b:= proc(n) option remember;
      Matrix(n, (i,j)-> T(i-1,j-1))^(n-1)
    end:
T:= proc(n,k) option remember;
     `if`(n=k, 1, `if`(k>n, 0, b(n)[n,k+1]))
    end:
seq(seq(T(n,k), k=0..n), n=0..10);  # Alois P. Heinz, Apr 13 2020

Mathematica

b[n_] := b[n] = MatrixPower[Table[T[i-1, j-1], {i, n}, {j, n}], n-1];
T[n_, k_] := T[n, k] = If[n == k, 1, If[k > n, 0, b[n][[n, k+1]]]];
Table[Table[T[n, k], {k, 0, n}], {n, 0, 10}] // Flatten (* Jean-François Alcover, Apr 25 2020, after Alois P. Heinz *)

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])}
for(n=0,10, for(k=0,n, print1(T(n,k),", ")); print(""))

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]}
for(n=0,10, for(k=0,n, print1(T(n,k),", ")); print(""))