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
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
Programs
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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 *)
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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.
Comments