%I #26 Feb 05 2014 10:58:52
%S 1,1,1,1,1,2,1,1,2,4,1,1,2,4,8,1,1,2,4,8,16,1,1,2,4,8,16,32,1,1,2,4,8,
%T 16,32,64,1,1,2,4,8,16,32,64,128,1,1,2,4,8,16,32,64,128,256,1,1,2,4,8,
%U 16,32,64,128,256,512
%N Table read by rows, which consist of 1 followed by 2^k, 0 <= k < n ; n = 0,1,2,3,...
%C The n-th row consists of the n+1 terms A011782(k), k=0,...,n. Thus the rows converge to A011782, which is also equal to the diagonal = last element of each row.
%C This (read as a "linear" sequence) is also the limit of the rows of A232088; more precisely, for n>0, each row of A232088 consists of the first n(n+1)/2 elements of this sequence, followed by 2^(n-1). See the LINK there for one motivation for this sequence.
%F T(n,k) = max(1,2^(k-1)) = A011782(k); 0 <= k <= n.
%e The table reads:
%e 1,
%e 1, 1,
%e 1, 1, 2,
%e 1, 1, 2, 4,
%e 1, 1, 2, 4, 8,
%e 1, 1, 2, 4, 8, 16,
%e 1, 1, 2, 4, 8, 16, 32,
%e 1, 1, 2, 4, 8, 16, 32, 64, etc.
%o (PARI) for(n=0,10,print1("1,");for(k=0,n-1,print1(2^k,",")))
%K nonn,tabl,easy
%O 0,6
%A _M. F. Hasler_, Jan 20 2014