A132615 Triangle T, read by rows, where row n+1 of T = row n of T^(2n-1) with appended '1' for n>=0 with T(0,0)=1.

1, 1, 1, 1, 1, 1, 6, 3, 1, 1, 80, 25, 5, 1, 1, 1666, 378, 56, 7, 1, 1, 47232, 8460, 1020, 99, 9, 1, 1, 1694704, 252087, 26015, 2134, 154, 11, 1, 1, 73552752, 9392890, 855478, 61919, 3848, 221, 13, 1, 1, 3744491970, 420142350, 34461260, 2257413, 125760, 6290, 300, 15, 1, 1

Offset: 0

Author

Paul D. Hanna, Aug 24 2007

Keywords

• nonn
• tabl

Examples

			Triangle begins:
  1;
  1, 1;
  1, 1, 1;
  6, 3, 1, 1;
  80, 25, 5, 1, 1;
  1666, 378, 56, 7, 1, 1;
  47232, 8460, 1020, 99, 9, 1, 1;
  1694704, 252087, 26015, 2134, 154, 11, 1, 1;
  73552752, 9392890, 855478, 61919, 3848, 221, 13, 1, 1; ...
GENERATE T FROM ODD MATRIX POWERS OF T.
Matrix cube, T^3, begins:
  1;
  3, 1;
  6, 3, 1; <-- row 3 of T
  31, 12, 3, 1;
  357, 100, 18, 3, 1;
  6786, 1455, 205, 24, 3, 1; ...
where row 3 of T = row 2 of T^3 with appended '1'.
Matrix fifth power, T^5, begins:
  1;
  5, 1;
  15, 5, 1;
  80, 25, 5, 1; <-- row 4 of T
  855, 215, 35, 5, 1;
  15171, 3065, 410, 45, 5, 1; ...
where row 4 of T = row 3 of T^5 with appended '1'.
Matrix seventh power, T^7, begins:
  1;
  7, 1;
  28, 7, 1;
  161, 42, 7, 1;
  1666, 378, 56, 7, 1; <-- row 5 of T
  28119, 5348, 679, 70, 7, 1; ...
where row 5 of T = row 4 of T^7 with appended '1'.
ALTERNATE GENERATING METHOD.
Row 4: start with a '1' followed by 4 zeros;
take partial sums and append 2 zeros; then
take partial sums thrice more:
  (1), 0, 0, 0, 0;
  1, 1, 1, 1, (1), 0, 0;
  1, 2, 3, 4, 5, 5, (5);
  1, 3, 6, 10, 15, 20, (25);
  1, 4, 10, 20, 35, 55, (80);
the final nonzero terms form row 4: [80, 25, 5, 1, 1].
Row 5: start with a '1' followed by 6 zeros;
take partial sums and append 4 zeros;
take partial sums and append 2 zeros; then
take partial sums thrice more:
  (1), 0, 0, 0, 0, 0, 0;
  1, 1, 1, 1, 1, 1, (1), 0, 0, 0, 0;
  1, 2, 3, 4, 5, 6, 7, 7, 7, 7, (7), 0, 0;
  1, 3, 6, 10, 15, 21, 28, 35, 42, 49, 56, 56, (56);
  1, 4, 10, 20, 35, 56, 84, 119, 161, 210, 266, 322, (378);
  1, 5, 15, 35, 70, 126, 210, 329, 490, 700, 966, 1288, (1666);
the final nonzero terms form row 5: [1666, 378, 56, 7, 1, 1].
Continuing in this way produces all the rows of this triangle.
		

Links

• Alois P. Heinz, Rows n = 0..140, flattened

Crossrefs

Cf. columns: A132616, A132617, A132618; A132619; variants: A132610, A101479.

Cf. A304190, A304191, A304193.

Programs

Maple

b:= proc(n) option remember;
      Matrix(n, (i,j)-> T(i-1,j-1))^(2*n-3)
    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}], 2n-3];
T[n_, k_] := T[n, k] = If[n == k, 1, If[k > n, 0, b[n][[n, k + 1]]]];
Table[T[n, k], {n, 0, 10}, {k, 0, n}] // Flatten (* Jean-François Alcover, Apr 27 2020, after Alois P. Heinz *)

PARI

{T(n, k)=local(A=vector(n+1), p); A[1]=1; for(j=1, n-k-1, p=(n-1)*(n-2)-(n-j-1)*(n-j-2); A=Vec((Polrev(A)+x*O(x^p))/(1-x))); A=Vec((Polrev(A)+x*O(x^p))/(1-x)); A[p+1]}

Formula

T(n+1,1) is divisible by 2n-1 for n>=1.