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%I #15 Mar 17 2023 21:34:07
%S 1,1,1,1,2,1,1,4,4,1,1,10,16,8,1,1,36,84,64,16,1,1,202,656,680,256,32,
%T 1,1,1828,8148,10816,5456,1024,64,1,1,27338,167568,274856,174336,
%U 43680,4096,128,1,1,692004,5866452,11622976,8909648,2794496,349504,16384,256,1
%N Infinite lower triangular matrix, M, that satisfies [M^2](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^(2k)](i,j) = [M^k](i+1,j+1) for all i,j,k>=0; thus [M^(2^n)](i,j) = M(i+n,j+n) for all n>=0.
%H Alois P. Heinz, <a href="/A078121/b078121.txt">Rows n = 0..80, flattened</a>
%F M(1,j) = A002577(j) (partitions of 2^j into powers of 2), M(j+1,j) = 2^j, M(j+2,j) = 4^j, M(j+3,j) = A016131(j).
%F M(n,k) = the coefficient of x^(2^n - 2^(n-k)) in the power series expansion of 1/Product_{j=0..n-k} (1-x^(2^j)) whenever 0<=k<n for all n>0 (conjecture).
%e The square of the matrix is the same matrix excluding the first row and column:
%e [1, 0, 0, 0, 0]^2 = [ 1, 0, 0, 0, 0]
%e [1, 1, 0, 0, 0] [ 2, 1, 0, 0, 0]
%e [1, 2, 1, 0, 0] [ 4, 4, 1, 0, 0]
%e [1, 4, 4, 1, 0] [10,16, 8, 1, 0]
%e [1,10,16, 8, 1] [36,84,64,16, 1]
%p M:= proc(i, j) option remember; `if`(j=0 or i=j, 1,
%p add(M(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 rows = 10; M[k_] := Table[ Which[j == 1, 1, i == j, 1, 1 < j < i, m[i, j], True, 0], {i, 1, k}, {j, 1, k}]; m2[i_, j_] := m[i+1, j+1]; M2[k_] := Table[ Which[j<i, m2[i, j], j == i, 1, True, 0], {i, 1, k}, {j, 1, k}]; sol[k_] := Thread[ Flatten[ M[k].M[k]] == Flatten[M2[k]]] // Solve; Table[M[rows][[i, j]], {i, 1, rows}, {j, 1, i}] /. sol[rows] // Flatten (* _Jean-François Alcover_, Feb 27 2015 *)
%t M[i_, j_] := M[i, j] = If[j == 0 || i == j, 1, Sum[M[i-1, k]*M[k, j-1], {k, 0, i-1}]]; Table[Table[M[n, k], {k, 0, n}], {n, 0, 10}] // Flatten (* _Jean-François Alcover_, Feb 27 2015, after _Alois P. Heinz_ *)
%Y Cf. A078122, A002577, A016131.
%K nonn,tabl
%O 0,5
%A _Paul D. Hanna_, Nov 18 2002
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