A371734 Maximal length of a factorization of n into factors > 1 all having different sums of prime indices.
0, 1, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 2, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 2, 3, 1, 2, 2, 3, 1, 3, 1, 2, 2, 2, 1, 3, 1, 2, 2, 2, 1, 3, 2, 3, 2, 2, 1, 3, 1, 2, 2, 3, 2, 3, 1, 2, 2, 3, 1, 3, 1, 2, 2, 2, 2, 3, 1, 3, 2, 2, 1, 3, 2, 2, 2, 3, 1, 3, 2, 2, 2, 2, 2, 3, 1, 2, 2, 3, 1, 3, 1, 3, 3
Offset: 1
Keywords
Examples
The factorizations of 90 of this type are (2*3*15), (2*5*9), (2*45), (3*30), (5*18), (6*15), (90), so a(90) = 3.
Links
Crossrefs
Programs
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Mathematica
facs[n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&, Select[facs[n/d],Min@@#>=d&]],{d,Rest[Divisors[n]]}]]; hwt[n_]:=Total[Cases[FactorInteger[n],{p_,k_}:>PrimePi[p]*k]]; Table[Max[Length/@Select[facs[n],UnsameQ@@hwt/@#&]],{n,100}]
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PARI
A056239(n) = if(1==n, 0, my(f=factor(n)); sum(i=1, #f~, f[i, 2] * primepi(f[i, 1]))); all_have_different_sum_of_pis(facs) = if(!#facs, 1, (#Set(apply(A056239,facs)) == #facs)); A371734(n, m=n, facs=List([])) = if(1==n, if(all_have_different_sum_of_pis(facs),#facs,0), my(s=0, newfacs); fordiv(n, d, if((d>1)&&(d<=m), newfacs = List(facs); listput(newfacs,d); s = max(s,A371734(n/d, d, newfacs)))); (s)); \\ Antti Karttunen, Jan 20 2025
Extensions
Data section extended to a(105) by Antti Karttunen, Jan 20 2025
Comments