A096917 Smallest prime factor of the n-th product of 3 distinct primes.

2, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 3, 2, 2, 2, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 2, 3, 2, 3, 2, 2, 2, 2, 2, 3, 2, 3, 2, 2, 2, 5, 3, 2, 2, 2, 2, 2, 3, 2, 2, 3, 2, 2, 5, 3, 2, 2, 3, 2, 2, 2, 2, 2, 2, 3, 3, 2, 2, 2, 5, 2, 2, 2, 3, 2, 3, 2, 3, 2, 2, 3, 2, 3, 2, 2, 3, 5, 2, 2, 2, 3, 2, 5, 2, 3, 2, 2, 3, 2, 3, 2

Offset: 1

Author

Reinhard Zumkeller, Jul 15 2004

Keywords

• nonn

Links

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

Crossrefs

Cf. A020639, A007304, A096918, A096919.

Programs

Mathematica

f[n_]:=Last/@FactorInteger[n]=={1,1,1};f1[n_]:=Min[First/@FactorInteger[n]];f2[n_]:=Max[First/@FactorInteger[n]];lst={};Do[If[f[n],AppendTo[lst,f1[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 A096917(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 min(primefactors(bisection(f))) # Chai Wah Wu, Aug 30 2024

Formula

a(n)*A096918(n)*A096919(n) = A007304(n).

a(n) < A096918(n) < A096919(n).

a(n) = A020639(A007304(n)).