A096918 Intermediate prime factor of n-th product of 3 distinct primes.

3, 3, 3, 5, 3, 3, 5, 5, 3, 5, 3, 7, 5, 5, 3, 7, 3, 5, 5, 3, 5, 7, 7, 3, 5, 3, 7, 7, 3, 5, 11, 5, 5, 3, 7, 5, 3, 7, 3, 5, 11, 7, 7, 3, 7, 5, 11, 3, 11, 5, 7, 5, 3, 13, 7, 5, 5, 3, 7, 13, 3, 11, 7, 5, 3, 5, 11, 7, 3, 5, 7, 13, 7, 3, 7, 5, 5, 3, 11, 11, 3, 5, 17, 7, 3, 7, 13, 7, 5, 3, 11, 5, 5, 11, 5

Offset: 1

Author

Reinhard Zumkeller, Jul 15 2004

Keywords

• nonn

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A007304, A096917, A096919.

Programs

Mathematica

f[n_]:=Last/@FactorInteger[n]=={1,1,1};f1[n_]:=Min[First/@FactorInteger[n]];f2[n_]:=Max[First/@FactorInteger[n]];f3[n_]:=First/@FactorInteger[n][[2,1]];lst={};Do[If[f[n],AppendTo[lst,f3[n]]],{n,0,7!}];lst (* Vladimir Joseph Stephan Orlovsky, Apr 10 2010 *)

Python

from math import isqrt
from sympy import primepi, primerange, integer_nthroot, primefactors
def A096918(n):
    def f(x): return int(n+x-sum(primepi(x//(k*m))-b for a,k in enumerate(primerange(integer_nthroot(x,3)[0]+1),1) for b,m in enumerate(primerange(k+1,isqrt(x//k)+1),a+1)))
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    return sorted(primefactors(bisection(f)))[1] # Chai Wah Wu, Aug 30 2024

Formula

A096917(n)*a(n)*A096919(n) = A007304(n).

A096917(n) < a(n) < A096919(n).