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A202687
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Triangle arising in the computation of hypersigma, definition 2 (A191161).
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1
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1, 1, 1, 1, 4, 1, 1, 6, 18, 1, 1, 8, 36, 104, 1, 1, 10, 60, 260, 750, 1, 1, 12, 90, 520, 2250, 6492, 1, 1, 14, 126, 910, 5250, 22722, 65562, 1, 1, 16, 168, 1456, 10500, 60592, 262248, 756688, 1, 1, 18, 216, 2184, 18900, 136332, 786744, 3405096, 9825030, 1
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OFFSET
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0,5
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COMMENTS
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Given a squarefree number with n distinct prime factors put through the hypersigma function (A191161), each smaller divisor contributes a given number of "loose" 1s. Row n of this triangle tells how many loose 1s are contributed by prime divisors (column 1), by semiprime divisors (column 2), sphenic divisors (column 3) and so on up to column n - 1.
It may appear somewhat artificial that the leftmost and rightmost columns are all filled with 1s since under the hypersigma function, the smallest divisor of a number, 1, may be said to contribute two loose 1s (itself and the 1 in the recursion level immediately below). However, the reason for attributing one of these 1s to the number itself and the 1 in the recursion level below to that 1 in the recursion level above is to more clearly show how this triangle is tied to Pascal's triangle.
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LINKS
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FORMULA
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T(n, k) = binomial(n, k)*Sum_{j=0..k} T(k, j).
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EXAMPLE
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Triangle starts:
1;
1, 1;
1, 4, 1;
1, 6, 18, 1;
1, 8, 36, 104, 1;
1, 10, 60, 260, 750, 1;
1, 12, 90, 520, 2250, 6492, 1;
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MAPLE
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if k = 0 or k = n then
1;
else
binomial(n, k)*add(procname(k, j), j=0..k) ;
end if;
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MATHEMATICA
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a[0, k_] := 1; a[n_, n_] := 1; a[n_, k_] := a[n, k] = Binomial[n, k] Sum[a[k, j], {j, 0, k}]; ColumnForm[Table[a[n, k], {n, 0, 9}, {k, 0, n}], Center]
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CROSSREFS
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KEYWORD
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AUTHOR
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STATUS
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approved
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