A193092 Augmentation of the triangular array P given by p(n,k)=k! for 0<=k<=n. See Comments.
1, 1, 1, 1, 2, 3, 1, 3, 7, 13, 1, 4, 12, 32, 69, 1, 5, 18, 58, 173, 421, 1, 6, 25, 92, 321, 1058, 2867, 1, 7, 33, 135, 523, 1977, 7159, 21477, 1, 8, 42, 188, 790, 3256, 13344, 53008, 175769, 1, 9, 52, 252, 1134, 4986, 21996, 97956, 427401, 1567273
Offset: 0
Examples
First 7 rows: 1 1...1 1...2....3 1...3....7.....13 1...4....12....32....69 1...5....18....58....173...421 1...6....25....92....321...1058...2867 The matrix method described at A193091 shows that row 3 arises from row 2 as the matrix product ............. (1...1...2...4) (1...2...3) * (0...1...1...2) = (1...3...7...13) ............. (0...0...1...1). The equivalent polynomial substitution method: x^2+2x+3 -> (x^3+x^2+2x+4)+2(x^2+x+2)+3(x+1)= x^3+3x^2+7x+13.
Programs
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Mathematica
p[n_, k_] := k! Table[p[n, k], {n, 0, 5}, {k, 0, n}] m[n_] := Table[If[i <= j, p[n + 1 - i, j - i], 0], {i, n}, {j, n + 1}] TableForm[m[4]] w[0, 0] = 1; w[1, 0] = p[1, 0]; w[1, 1] = p[1, 1]; v[0] = w[0, 0]; v[1] = {w[1, 0], w[1, 1]}; v[n_] := v[n - 1].m[n] TableForm[Table[v[n], {n, 0, 6}]] (* A193092 *) Flatten[Table[v[n], {n, 0, 8}]]
Comments