login
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.
11

%I #10 Nov 24 2017 11:47:12

%S 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,

%T 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,

%U 3,12,1,5,3,12,1,26,1,3,5,5,3,12,1,19,3,3

%N 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.

%H Michael De Vlieger, <a href="/A294786/b294786.txt">Table of n, a(n) for n = 1..16384</a>

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

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

%e The a(36)=11 ways are:

%e (2)*(3)*(6),

%e (2)*(3*6), (2*6)*(3), (2)*(18), (3)*(12), (4)*(9),

%e (2*3*6), (2*18), (3*12), (4*9), (36).

%t sfs[n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[sfs[n/d],Min@@#>d&]],{d,Rest[Divisors[n]]}]];

%t sps[{}]:={{}};sps[set:{i_,___}]:=Join@@Function[s,Prepend[#,s]&/@sps[Complement[set,s]]]/@Cases[Subsets[set],{i,___}];

%t Table[Length[Join@@Function[fac,Select[sps[fac],UnsameQ@@Times@@@#&]]/@sfs[n]],{n,100}]

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

%K nonn

%O 1,6

%A _Gus Wiseman_, Nov 08 2017