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%I #4 Mar 30 2012 18:36:45
%S 1,1,1,1,2,1,1,1,3,1,2,4,1,4,1,1,4,9,1,5,1,1,1,6,16,1,6,1,1,6,1,8,25,
%T 1,7,1,3,8,21,1,10,36,1,8,1,1,6,21,52,1,12,49,1,9,1,2,1,9,40,105,1,14,
%U 64,1,10,1,4,13,1,12,65,186,1,16,81,1,11,1,1,14,51,1,15,96,301,1,18,100,1
%N Triangular matrix T, read by rows, where column 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 sqrt(2).
%e Triangle T begins:
%e 1;
%e 1,1;
%e 1,2,1;
%e 1,1,3,1;
%e 2,4,1,4,1;
%e 1,4,9,1,5,1;
%e 1,1,6,16,1,6,1;
%e 1,6,1,8,25,1,7,1;
%e 3,8,21,1,10,36,1,8,1;
%e 1,6,21,52,1,12,49,1,9,1;
%e 2,1,9,40,105,1,14,64,1,10,1;
%e 4,13,1,12,65,186,1,16,81,1,11,1;
%e 1,14,51,1,15,96,301,1,18,100,1,12,1; ...
%e which when flattened (read by rows) forms column 0:
%e 1,1,1,1,2,1,1,1,3,1,2,4,1,4,1,1,4,9,1,5,1,...
%e The matrix square, T^2 = A105542, begins:
%e 1;
%e 2,1;
%e 4,4,1;
%e 6,8,6,1;
%e 13,14,14,8,1;
%e 26,47,26,22,10,1; ...
%e which when flattened (read by rows) forms column 1:
%e 1,2,1,4,4,1,6,8,6,1,13,14,14,8,1,...
%e The matrix cube, T^3 = A105545, begins:
%e 1;
%e 3,1;
%e 9,6,1;
%e 21,21,9,1;
%e 51,54,39,12,1;
%e 142,165,111,63,15,1; ...
%e which when flattened (read by rows) forms column 2:
%e 1,3,1,9,6,1,21,21,9,1,51,54,39,12,1,...
%o (PARI) {T(n,k)=local(A=matrix(n+1,n+1,r,c,if(r>=c,1)),B, TK=floor((sqrt(8*(n-k)+1)-1)/2),TC);B=A; for(r=0,n, for(c=0,r,TC=floor((sqrt(8*(r-c)+1)-1)/2); B[r+1,c+1]=(A^(c+1))[TC+1,r-c-TC*(TC+1)/2+1]);A=B); return((A^(k+1))[TK+1,n-k-TK*(TK+1)/2+1])}
%Y Cf. A105541 (row sums), A105542 (matrix square), A105545 (matrix cube).
%K nonn,tabl
%O 0,5
%A _Paul D. Hanna_, Apr 13 2005
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