OFFSET
0,5
COMMENTS
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.
LINKS
Jinyuan Wang, Rows n = 0..100 of triangle, flattened
FORMULA
T(n, k) = binomial(n, k)*Sum_{j=0..k} T(k, j).
EXAMPLE
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;
MAPLE
A202687 := proc(n, k)
if k = 0 or k = n then
1;
else
binomial(n, k)*add(procname(k, j), j=0..k) ;
end if;
end proc: # R. J. Mathar, Mar 15 2013
MATHEMATICA
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]
CROSSREFS
KEYWORD
AUTHOR
Alonso del Arte, Dec 22 2011
STATUS
approved