A078122 Infinite lower triangular matrix, M, that satisfies [M^3](i,j) = M(i+1,j+1) for all i,j>=0 where [M^n](i,j) denotes the element at row i, column j, of the n-th power of matrix M, with M(0,k)=1 and M(k,k)=1 for all k>=0.

1, 1, 1, 1, 3, 1, 1, 12, 9, 1, 1, 93, 117, 27, 1, 1, 1632, 3033, 1080, 81, 1, 1, 68457, 177507, 86373, 9801, 243, 1, 1, 7112055, 24975171, 15562314, 2371761, 88452, 729, 1, 1, 1879090014, 8786827629, 6734916423, 1291958181, 64392813, 796797, 2187, 1

Offset

0,5

Comments

M also satisfies: [M^(3k)](i,j) = [M^k](i+1,j+1) for all i,j,k >=0; thus [M^(3^n)](i,j) = M(i+n,j+n) for all n >= 0.

Conjecture: the sum of the n-th row equals the number of partitions of 3^n into powers of 3 (A078125).

Links

Alois P. Heinz, Rows n = 0..60, flattened

Formula

M(1, j) = A078124(j), M(j+1, j)=3^j, M(j+2, j) = A016142(j).

M(n, k) = the coefficient of x^(3^n - 3^(n-k)) in the power series expansion of 1/Product_{j=0..n-k}(1-x^(3^j)) whenever 0<=k<n for all n>0 (conjecture).

Example

The cube of the matrix is the same matrix excluding the first row and column:
  [1, 0, 0, 0]^3 = [ 1,  0, 0, 0]
  [1, 1, 0, 0]     [ 3,  1, 0, 0]
  [1, 3, 1, 0]     [12,  9, 1, 0]
  [1,12, 9, 1]     [93,117,27, 1]

Maple

S:= proc(i, j) option remember;
       add(M(i, k)*M(k, j), k=0..i)
    end:
M:= proc(i, j) option remember; `if`(j=0 or i=j, 1,
       add(S(i-1, k)*M(k, j-1), k=0..i-1))
    end:
seq(seq(M(n, k), k=0..n), n=0..10);  # Alois P. Heinz, Feb 27 2015

Mathematica

m[i_, j_] := m[i, j]=If[j==0||i==j, 1, m3[i-1, j-1]]; m2[i_, j_] := m2[i, j]=Sum[m[i, k]m[k, j], {k, j, i}]; m3[i_, j_] := m3[i, j]=Sum[m[i, k]m2[k, j], {k, j, i}]; Flatten[Table[m[i, j], {i, 0, 8}, {j, 0, i}]]

Crossrefs

Cf. A078121, A078123, A078124, A078125, A016142.

Sequence in context: A113340 A134523 A098778 * A128592 A156584 A209424

Adjacent sequences: A078119 A078120 A078121 * A078123 A078124 A078125

Keyword

nonn,tabl

Author

Paul D. Hanna, Nov 18 2002

Status

approved