%I
%S 1,1,1,1,2,1,1,3,1,1,1,4,1,4,2,1,5,1,9,4,1,1,6,1,16,6,1,1,1,7,1,25,8,
%T 1,6,3,1,8,1,36,10,1,17,8,1,1,9,1,49,12,1,36,15,2,1,1,10,1,64,14,1,65,
%U 24,3,1,1,1,11,1,81,16,1,106,35,4,1,12,4,1,12,1,100,18,1,161,48,5,1,61,26
%N Triangular matrix T, read by rows, where diagonal n equals T^(n+1) when flattened (as read by rows), for n>=0.
%C The ratio of consecutive row sums seems to converge to a value near 2, but it is not yet certain that the limit exists.
%e Triangle begins:
%e 1;
%e 1,1;
%e 1,2,1;
%e 1,3,1,1;
%e 1,4,1,4,2;
%e 1,5,1,9,4,1;
%e 1,6,1,16,6,1,1;
%e 1,7,1,25,8,1,6,3;
%e 1,8,1,36,10,1,17,8,1;
%e 1,9,1,49,12,1,36,15,2,1;
%e 1,10,1,64,14,1,65,24,3,1,1; ...
%e which when flattened (read by rows) forms diagonal 0:
%e 1,1,1,1,2,1,1,3,1,1,1,4,1,4,2,1,5,1,9,4,1,...
%e The matrix square, T^2 = A105537, begins:
%e 1;
%e 2,1;
%e 4,4,1;
%e 6,8,2,1;
%e 12,26,7,12,4; ...
%e which when flattened (read by rows) forms diagonal 1:
%e 1,2,1,4,4,1,6,8,2,1,12,26,7,12,4,...
%e The matrix cube, T^3 = A105539, begins:
%e 1;
%e 3,1;
%e 9,6,1;
%e 17,15,3,1;
%e 61,92,23,28,8; ...
%e which when flattened (read by rows) forms diagonal 2:
%e 1,3,1,9,6,1,17,15,3,1,61,92,23,28,8,...
%o (PARI) {T(n,k)=local(A=matrix(n+1,n+1,r,c,if(r>=c,1)),B, TK=floor((sqrt(8*k+1)1)/2),TC);B=A; for(r=0,n, for(c=0,r,TC=floor((sqrt(8*c+1)1)/2); B[r+1,c+1]=(A^(rc+1))[TC+1,cTC*(TC+1)/2+1]);A=B); return((A^(nk+1))[TK+1,kTK*(TK+1)/2+1])}
%Y Cf. A105536 (row sums), A105537 (matrix square), A105539 (matrix cube).
%K nonn,tabl
%O 0,5
%A _Paul D. Hanna_, Apr 12 2005
