A121412 Triangular matrix T, read by rows, where row n of T equals row (n-1) of T^(n+1) with an appended '1'.

1, 1, 1, 3, 1, 1, 18, 4, 1, 1, 170, 30, 5, 1, 1, 2220, 335, 45, 6, 1, 1, 37149, 4984, 581, 63, 7, 1, 1, 758814, 92652, 9730, 924, 84, 8, 1, 1, 18301950, 2065146, 199692, 17226, 1380, 108, 9, 1, 1, 508907970, 53636520, 4843125, 387567, 28365, 1965, 135, 10, 1, 1

Offset: 0

Author

Paul D. Hanna, Jul 30 2006

Keywords

• nonn
• tabl

Comments

Related to the number of subpartitions of a partition as defined in A115728; for examples involving column k of successive matrix powers, see A121430, A121431, A121432 and A121433. Essentially the same as triangle A101479, but this form best illustrates the nice properties of this triangle.

Examples

			Triangle T begins:
1;
1, 1;
3, 1, 1;
18, 4, 1, 1;
170, 30, 5, 1, 1;
2220, 335, 45, 6, 1, 1;
37149, 4984, 581, 63, 7, 1, 1;
758814, 92652, 9730, 924, 84, 8, 1, 1;
18301950, 2065146, 199692, 17226, 1380, 108, 9, 1, 1;
508907970, 53636520, 4843125, 387567, 28365, 1965, 135, 10, 1, 1;
To get row 4 of T, append '1' to row 3 of matrix power T^5:
1;
5, 1;
25, 5, 1;
170, 30, 5, 1; ...
To get row 5 of T, append '1' to row 4 of matrix power T^6:
1;
6, 1;
33, 6, 1;
233, 39, 6, 1;
2220, 335, 45, 6, 1; ...
Likewise, get row n of T by appending '1' to row (n-1) of T^(n+1).
		

Links

• Alois P. Heinz, Rows n = 0..45, flattened

Crossrefs

Cf. A121416 (T^2), A121420 (T^3), columns: A121413, A121414, A121415; related tables: A121424, A121426, A121428; related subpartitions: A121430, A121431, A121432, A121433.

Programs

Mathematica

T[n_, k_] := Module[{A = {{1}}, B}, Do[B = Array[0&, {m, m}]; Do[Do[B[[i, j]] = If[j == i, 1, MatrixPower[A, i][[i-1, j]]], {j, 1, i}], {i, 1, m}]; A = B, {m, 1, n+1}]; A[[n+1, k+1]]];
Table[T[n, k], {n, 0, 9}, {k, 0, n}] // Flatten (* Jean-François Alcover, Oct 03 2019 *)

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)[i-1, j]); )); A=B); return((A^1)[n+1, k+1])}
for(n=0,12, for(k=0,n, print1( T(n,k),", "));print(""))

Formula

G.f.: Column k of successive powers of T satisfy the amazing relation given by: 1 = Sum_{n>=0} (1-x)^(n+1) * x^(n(n+1)/2 + k*n) * Sum_{j=0..n+k} [T^(j+1)](n+k,k) * x^j.