%I #29 Jun 19 2019 17:57:13
%S 1,2,1,2,1,1,3,2,1,1,3,0,1,1,1,4,2,3,1,1,1,4,0,0,1,1,1,1,5,1,2,3,1,1,
%T 1,1,5,0,2,1,1,1,1,1,1,6,1,2,1,3,1,1,1,1,1,6,0,0,0,0,1,1,1,1,1,1,7,1,
%U 3,4,3,4,1,1,1,1,1,1,7,0,0,0,0,0,1,1,1,1,1,1,1
%N Triangle read by rows, n>=1, 1<=k<=n. T(n,n-k+1) = number of cells in the k-th row = number of cells in the k-th column of the diagram of the symmetric representation of sigma(n) in the first quadrant.
%C Since the diagram is symmetric the number of cells in the k-th row equals the number of cell in the k-th column, see example.
%C Row sums give A000203.
%C Column 1 gives A008619, n >= 1.
%C If n is an odd prime then row n lists (n+1)/2, ((n+1)/2 - 2) zeros, and (n+1)/2 ones.
%C Mirror of A240061.
%H <a href="/index/Si#SIGMAN">Index entries for sequences related to sigma(n)</a>
%e Triangle begins:
%e 1;
%e 2, 1;
%e 2, 1, 1;
%e 3, 2, 1, 1;
%e 3, 0, 1, 1, 1;
%e 4, 2, 3, 1, 1, 1;
%e 4, 0, 0, 1, 1, 1, 1;
%e 5, 1, 2, 3, 1, 1, 1, 1;
%e 5, 0, 2, 1, 1, 1, 1, 1, 1;
%e 6, 1, 2, 1, 3, 1, 1, 1, 1, 1;
%e 6, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1;
%e 7, 1, 3, 4, 3, 4, 1, 1, 1, 1, 1, 1;
%e 7, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1;
%e ...
%e For n = 9 the symmetric representation of sigma(9) = 13 in the first quadrant looks like this:
%e y
%e . Number of cells
%e ._ _ _ _ _
%e |_ _ _ _ _| 5
%e . |_ _ 0
%e . |_ | 2
%e . |_|_ _ 1
%e . | | 1
%e . | | 1
%e . | | 1
%e . | | 1
%e . . . . . . . . |_| . . x 1
%e .
%e So the 9th row of triangle is [5, 0, 2, 1, 1, 1, 1, 1, 1].
%e For n = 9 and k = 7 there are two cells in the 7th row of the diagram, also there are two cells in the 7th column of the diagram, so T(9,9-7+1) = T(9,3) = 2.
%Y Cf. A000203, A008619, A024916, A196020, A236104, A235791, A237270, A237271, A237591, A237593, A239660, A239931-A239934, A240061.
%K nonn,tabl
%O 1,2
%A _Omar E. Pol_, Apr 26 2014