OFFSET
2,7
COMMENTS
An orderless tree-factorization (see A292504 for definition) is complete if all leaves are prime numbers. This sequence first differs from A281119 at a(64)=33.
a(n) depends only on the prime signature of n. - Andrew Howroyd, Nov 18 2018
LINKS
Andrew Howroyd, Table of n, a(n) for n = 2..10000
FORMULA
a(p^n) = A000669(n) for prime p. - Andrew Howroyd, Nov 18 2018
EXAMPLE
The a(60)=17 complete orderless tree-factorizations are: (2(2(35))), (2(3(25))), (2(5(23))), (2(235)), (3(2(25))), (3(5(22))), (3(225)), (5(2(23))), (5(3(22))), (5(223)), ((22)(35)), ((23)(25)), (22(35)), (23(25)), (25(23)), (35(22)), (2235).
MATHEMATICA
postfacs[n_]:=If[n<=1, {{}}, Join@@Table[Map[Prepend[#, d]&, Select[postfacs[n/d], Min@@#>=d&]], {d, Rest[Divisors[n]]}]];
oltfacs[n_]:=If[n<=1, {{}}, Prepend[Union@@Function[q, Sort/@Tuples[oltfacs/@q]]/@DeleteCases[postfacs[n], {n}], n]];
Table[Length[Select[oltfacs[n], FreeQ[#, _Integer?(!PrimeQ[#]&)]&]], {n, 2, 100}]
PROG
(PARI) seq(n)={my(v=vector(n), w=vector(n)); v[1]=1; for(k=2, n, w[k]=v[k]+isprime(k); forstep(j=n\k*k, k, -k, my(i=j, e=0); while(i%k==0, i/=k; e++; v[j]+=binomial(e+w[k]-1, e)*v[i]))); w[2..n]} \\ Andrew Howroyd, Nov 18 2018
CROSSREFS
KEYWORD
nonn
AUTHOR
Gus Wiseman, Sep 17 2017
STATUS
approved