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A316790
Number of orderless same-tree-factorizations of n.
2
1, 1, 1, 2, 1, 1, 1, 2, 2, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 9, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1
OFFSET
1,4
COMMENTS
A constant factorization of n is a finite nonempty constant multiset of positive integers greater than 1 with product n. Constant factorizations correspond to perfect divisors (A089723). An orderless same-tree-factorization of n is either (case 1) the number n itself or (case 2) a finite multiset of two or more orderless same-tree-factorizations, one of each factor in a constant factorization of n.
a(n) depends only on the prime signature of n. - Andrew Howroyd, Nov 18 2018
LINKS
FORMULA
a(n) = 1 + Sum_{n = x^y, y > 1} binomial(a(x) + y - 1, y).
a(2^n) = A289078(n).
EXAMPLE
The a(64) = 9 orderless same-tree-factorizations:
64
(8*8)
(4*4*4)
(4*4*(2*2))
(8*(2*2*2))
(2*2*2*2*2*2)
(4*(2*2)*(2*2))
((2*2*2)*(2*2*2))
((2*2)*(2*2)*(2*2))
MATHEMATICA
a[n_]:=1+Sum[Binomial[a[n^(1/d)]+d-1, d], {d, Rest[Divisors[GCD@@FactorInteger[n][[All, 2]]]]}]
Array[a, 100]
PROG
(PARI) a(n)={my(z, e=ispower(n, , &z)); 1 + if(e, sumdiv(e, d, if(d>1, binomial(a(z^(e/d)) + d - 1, d))))} \\ Andrew Howroyd, Nov 18 2018
KEYWORD
nonn
AUTHOR
Gus Wiseman, Jul 14 2018
STATUS
approved