A294786 Number of ways to choose a set partition of a factorization of n into distinct factors greater than one such that different blocks have different products.

1, 1, 1, 1, 1, 3, 1, 3, 1, 3, 1, 5, 1, 3, 3, 3, 1, 5, 1, 5, 3, 3, 1, 12, 1, 3, 3, 5, 1, 12, 1, 5, 3, 3, 3, 11, 1, 3, 3, 12, 1, 12, 1, 5, 5, 3, 1, 19, 1, 5, 3, 5, 1, 12, 3, 12, 3, 3, 1, 26, 1, 3, 5, 9, 3, 12, 1, 5, 3, 12, 1, 26, 1, 3, 5, 5, 3, 12, 1, 19, 3, 3

Offset: 1

Author

Gus Wiseman, Nov 08 2017

Keywords

• nonn

Examples

			The a(36)=11 ways are:
(2)*(3)*(6),
(2)*(3*6), (2*6)*(3), (2)*(18), (3)*(12), (4)*(9),
(2*3*6), (2*18), (3*12), (4*9), (36).
		

Links

• Michael De Vlieger, Table of n, a(n) for n = 1..16384

Crossrefs

Cf. A000258, A001055, A045778, A279375, A281113, A294787, A294788.

Programs

Mathematica

sfs[n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[sfs[n/d],Min@@#>d&]],{d,Rest[Divisors[n]]}]];
sps[{}]:={{}};sps[set:{i_,_}]:=Join@@Function[s,Prepend[#,s]&/@sps[Complement[set,s]]]/@Cases[Subsets[set],{i,_}];
Table[Length[Join@@Function[fac,Select[sps[fac],UnsameQ@@Times@@@#&]]/@sfs[n]],{n,100}]

Formula

a(product of n distinct primes) = A000258(n).

a(prime^n) = A279375(n).