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%I #16 Jun 14 2020 00:49:37
%S 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,1,0,0,0,0,
%T 0,1,0,0,0,1,0,1,0,0,0,0,0,3,0,0,0,0,0,1,0,1,0,0,0,4,0,0,0,1,0,1,0,0,
%U 0,1,0,5,0,0,0,0,0,1,0,3,0,0,0,4,0,0,0,1,0,4,0,0,0,0,0,8,0,0,0,1
%N Number of unordered factorizations of n with 3 different parts > 1.
%C a(n) depends only on the prime signature of n. E.g. a(12)=a(75), since 12=2^2*3 and 75=5^2*3 share the same prime signature (2,1).
%H Jacob Sprittulla, <a href="/A334740/b334740.txt">Table of n, a(n) for n = 1..1000</a>
%e a(48) = 3 = #{ (6,4,2), (8,3,2), (4,3,2,2) }.
%o (R)
%o maxe <- function(n, d) { i=0; while( n%%(d^(i+1))==0 ) { i=i+1 }; i }
%o uhRec <- function(n, l=1) {
%o uh = 0
%o if( n<=0 ) {
%o return(0)
%o } else if(n==1) {
%o return(ifelse(l==0, 1, 0))
%o } else if(l<=0) {
%o return(0)
%o } else if( (n>=2) && (l>=1) ) {
%o for(d in 2:n) {
%o m = maxe(n, d)
%o if(m>=1) for(i in 1:m) for(j in 1:min(i, l)) {
%o uhj = uhRec( n/d^i, l-j )
%o uh = uh + log(d)/log(n) * (-1)^(j+1) * choose(i, j) * uhj
%o }
%o }
%o return(round(uh, 3))
%o }
%o }
%o n=100; l=2; sapply(1:n, uhRec, l) # A334739
%o n=100; l=3; sapply(1:n, uhRec, l) # A334740
%Y Cf. A334739 (2 different parts), A072670 (2 parts), A122179 (3 parts), A211159 (2 distinct parts), A122180 (3 distinct parts), A001055, A045778
%K nonn
%O 1,48
%A _Jacob Sprittulla_, May 09 2020