A326515 Number of factorizations of n into factors > 1 where every factor has the same average of prime indices.
1, 1, 1, 2, 1, 1, 1, 3, 2, 1, 1, 1, 1, 1, 1, 5, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 3, 1, 1, 2, 1, 7, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 5, 1, 1, 2, 1, 1, 1, 1, 1, 4, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 2
Offset: 1
Keywords
Examples
The a(900) = 9 factorizations: (3*3*10*10), (3*3*100), (3*10*30), (9*10*10), (3*300), (9*100), (10*90), (30*30), (900).
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Crossrefs
Programs
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Mathematica
primeMS[n_]:=If[n==1,{},Flatten[Cases[FactorInteger[n],{p_,k_}:>Table[PrimePi[p],{k}]]]]; facs[n_]:=If[n<=1,{{}},Join@@Table[Map[Prepend[#,d]&,Select[facs[n/d],Min@@#>=d&]],{d,Rest[Divisors[n]]}]]; Table[Length[Select[facs[n],SameQ@@Mean/@primeMS/@#&]],{n,100}]
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PARI
avgpis(n) = { my(f=factor(n)); f[,1] = apply(primepi,f[,1]); (1/bigomega(n))*sum(i=1,#f~,f[i,2]*f[i,1]); }; has_same_average_of_pis(facs) = if(!#facs, 1, my(avg=0); for(i=1,#facs,if(!avg, avg=avgpis(facs[i]), if(avg!=avgpis(facs[i]), return(0)))); (1)); A326515(n, m=n, facs=List([])) = if(1==n, has_same_average_of_pis(facs), my(s=0, newfacs); fordiv(n, d, if((d>1)&&(d<=m), newfacs = List(facs); listput(newfacs,d); s += A326515(n/d, d, newfacs))); (s)); \\ Antti Karttunen, Jan 20 2025
Extensions
Data section extended to a(105) by Antti Karttunen, Jan 20 2025
Comments