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%I #10 Mar 17 2023 22:10:36
%S 1,1,1,1,3,1,1,12,9,1,1,93,117,27,1,1,1632,3033,1080,81,1,1,68457,
%T 177507,86373,9801,243,1,1,7112055,24975171,15562314,2371761,88452,
%U 729,1,1,1879090014,8786827629,6734916423,1291958181,64392813,796797,2187,1
%N 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.
%C 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.
%C Conjecture: the sum of the n-th row equals the number of partitions of 3^n into powers of 3 (A078125).
%H Alois P. Heinz, <a href="/A078122/b078122.txt">Rows n = 0..60, flattened</a>
%F M(1, j) = A078124(j), M(j+1, j)=3^j, M(j+2, j) = A016142(j).
%F 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).
%e The cube of the matrix is the same matrix excluding the first row and column:
%e [1, 0, 0, 0]^3 = [ 1, 0, 0, 0]
%e [1, 1, 0, 0] [ 3, 1, 0, 0]
%e [1, 3, 1, 0] [12, 9, 1, 0]
%e [1,12, 9, 1] [93,117,27, 1]
%p S:= proc(i, j) option remember;
%p add(M(i, k)*M(k, j), k=0..i)
%p end:
%p M:= proc(i, j) option remember; `if`(j=0 or i=j, 1,
%p add(S(i-1, k)*M(k, j-1), k=0..i-1))
%p end:
%p seq(seq(M(n,k), k=0..n), n=0..10); # _Alois P. Heinz_, Feb 27 2015
%t 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}]]
%Y Cf. A078121, A078123, A078124, A078125, A016142.
%K nonn,tabl
%O 0,5
%A _Paul D. Hanna_, Nov 18 2002
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