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%I #42 Dec 10 2016 16:54:13
%S 1,0,1,1,1,1,0,0,1,1,1,1,1,1,1,0,1,0,1,1,1,1,0,1,1,1,1,1,0,1,1,0,1,1,
%T 1,1,1,1,1,1,1,1,1,1,1,0,0,0,1,0,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,1,0,1,
%U 1,1,1,0,1,1,1,1,1,1,1,0,1,0,1,1,1,1,1,1,1,1,1,0,1,0,1,1,1,0,1,1,1,1,1,1,1
%N Square array read by downwards antidiagonals: T(n,k) = 1 if k mod (n+1) > 0, T(n,k) = 0 if k mod (n+1) = 0.
%C Sum of antidiagonals generates sequence A049820 beginning at n=3.
%C When written as a triangular array, TR(n,j), by shifting row n of square array to the right by n columns (i.e., TR(n,j) = T(n,k-n) see second part of Example below), the sum of antidiagonals then becomes the column sums. TR(n,j) is then the trivial indicator of all the proper factors of the integers (N) starting at N = 3, where a value of zero in the array indicates the presence of a proper factor. [More details on that if you like: the row index n+1 of a zero in any column of TR(n,j) is factor, N = j + 2 (column index + 2) is the integer of interest.]
%C Row n is characteristic function of numbers that are not multiples of n+1. Rows 1..9 are A059841, A011655, A166486, A011558, A097325, A109720, A168181, A168182, A168184. - _Boris Putievskiy_, May 08 2013
%F T(n,k) = 1 if k mod (n+1)> 0; T(n,k) = 0 if k mod (n+1) = 0.
%F Or, in simple words, each row is a repeating pattern that starts with n instances of 1 followed by one instance of 0.
%F From _Boris Putievskiy_, May 08 2013: (Start)
%F As table T(n,k)= 1-floor(k/(n+1))+floor((k-1)/(n+1)).
%F As linear sequence a(n) = 1-floor(A004736(n)/(A002260(n)+1) + floor((A004736(n)-1)/(A002260(n)+1));
%F a(n)=1-floor(j/(i+1))+floor((j-1)/(i+1)), where i=n-t*(t+1)/2, j=(t*t+3*t+4)/2-n, t=floor((-1+sqrt(8*n-7))/2). (End)
%e As the Square Array T(n,k)
%e 1,0,1,0,1,0,1,0,1,0,1,0 ...
%e 1,1,0,1,1,0,1,1,0,1,1,0 ...
%e 1,1,1,0,1,1,1,0,1,1,1,0 ...
%e 1,1,1,1,0,1,1,1,1,0,1,1 ...
%e 1,1,1,1,1,0,1,1,1,1,1,0 ...
%e 1,1,1,1,1,1,0,1,1,1,1,1 ...
%e Now, as a Triangular Array TR(n,j):
%e 1,0,1,0,1,0,1,0,1,0,1,0 ...
%e 0,1,1,0,1,1,0,1,1,0,1,1,0 ...
%e 0,0,1,1,1,0,1,1,1,0,1,1,1,0 ...
%e 0,0,0,1,1,1,1,0,1,1,1,1,0,1,1 ...
%e 0,0,0,0,1,1,1,1,1,0,1,1,1,1,1,0 ...
%e 0,0,0,0,0,1,1,1,1,1,1,0,1,1,1,1,1 ...
%t max = 15; row[n_] := Table[{Table[1, {n}], 0}, {max/(n+1)}] // Flatten; t = Table[row[n], {n, max - 1}]; Table[t[[n-k+1, k]], {n, 1, max-1}, {k, n, 1, -1}] // Flatten (* _Jean-François Alcover_, May 02 2013 *)
%Y Cf. A002260, A004736, A059841, A011655, A011558, A097325, A109720, A166486, A168181, A168182, A168184.
%K nonn,tabl
%O 1
%A _Richard R. Forberg_, May 01 2013
%E More terms from _Jean-François Alcover_, May 02 2013
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