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A078536 Infinite lower triangular matrix, M, that satisfies [M^4](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.

%I A078536 #10 Oct 20 2019 11:18:34
%S A078536 1,1,1,1,4,1,1,28,16,1,1,524,496,64,1,1,29804,41136,8128,256,1,1,
%T A078536 5423660,10272816,2755264,130816,1024,1,1,3276048300,8220685104,
%U A078536 2804672704,178301696,2096128,4096,1,1,6744720496300,21934062166320,9139625620672,729250931456,11442760704,33550336,16384,1
%N A078536 Infinite lower triangular matrix, M, that satisfies [M^4](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 A078536 M also satisfies: [M^(4k)](i,j) = [M^k](i+1,j+1) for all i,j,k>=0; thus [M^(4^n)](i,j) = M(i+n,j+n) for all n>=0. Conjecture: sum of the n-th row equals the partitions of 4^n into powers of 4.
%F A078536 M(n, k) = the coefficient of x^(4^n - 4^(n-k)) in the power series expansion of 1/Product_{j=0..n-k}(1-x^(4^j)) whenever 0<=k<n for all n>0 (conjecture).
%e A078536 The 4th power of matrix is the same matrix excluding the first row and column:
%e A078536 [1,__0,__0,_0,0]^4=[____1,____0,___0,__0,0]
%e A078536 [1,__1,__0,_0,0]___[____4,____1,___0,__0,0]
%e A078536 [1,__4,__1,_0,0]___[___28,___16,___1,__0,0]
%e A078536 [1,_28,_16,_1,0]___[__524,__496,__64,__1,0]
%e A078536 [1,524,496,64,1]___[29804,41136,8128,256,1]
%t A078536 dim = 9;
%t A078536 a[_, 0] = 1; a[i_, i_] = 1; a[i_, j_] /; j > i = 0;
%t A078536 M = Table[a[i, j], {i, 0, dim-1}, {j, 0, dim-1}];
%t A078536 M4 = MatrixPower[M, 4];
%t A078536 sol = Table[M4[[i, j]] == M[[i+1, j+1]], {i, 1, dim-1}, {j, 1, dim-1}] // Flatten // Solve;
%t A078536 Table[a[i, j], {i, 0, dim-1}, {j, 0, i}] /. sol // Flatten (* _Jean-François Alcover_, Oct 20 2019 *)
%Y A078536 Cf. A078121, A078122, A078537.
%K A078536 nonn,tabl
%O A078536 0,5
%A A078536 _Paul D. Hanna_, Nov 29 2002
%E A078536 More terms from _Jean-François Alcover_, Oct 20 2019