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A240060
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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.
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1
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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, 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, 3, 4, 3, 4, 1, 1, 1, 1, 1, 1, 7, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1
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OFFSET
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1,2
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COMMENTS
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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.
If n is an odd prime then row n lists (n+1)/2, ((n+1)/2 - 2) zeros, and (n+1)/2 ones.
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LINKS
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EXAMPLE
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Triangle begins:
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, 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, 3, 4, 3, 4, 1, 1, 1, 1, 1, 1;
7, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1;
...
For n = 9 the symmetric representation of sigma(9) = 13 in the first quadrant looks like this:
y
. Number of cells
._ _ _ _ _
|_ _ _ _ _| 5
. |_ _ 0
. |_ | 2
. |_|_ _ 1
. | | 1
. | | 1
. | | 1
. | | 1
. . . . . . . . |_| . . x 1
.
So the 9th row of triangle is [5, 0, 2, 1, 1, 1, 1, 1, 1].
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.
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CROSSREFS
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Cf. A000203, A008619, A024916, A196020, A236104, A235791, A237270, A237271, A237591, A237593, A239660, A239931-A239934, A240061.
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KEYWORD
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AUTHOR
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STATUS
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approved
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