1,1

There is no constraint on which of the three primes is the largest or smallest.

T. D. Noe, Table of n, a(n) for n = 1..1000

360=2^3*3^2*5. 504=2^3*3^2*7. 1188=2^2*3^3*11.

f[n_]:=Sort[Last/@FactorInteger[n]]=={1, 2, 3}; Select[Range[5000], f]

(PARI) list(lim)=my(v=List(), t1, t2); forprime(p=2, (lim\12)^(1/3), t1=p^3; forprime(q=2, sqrt(lim\t1), if(p==q, next); t2=t1*q^2; forprime(r=2, lim\t2, if(p==r||q==r, next); listput(v, t2*r)))); vecsort(Vec(v)) \\ Charles R Greathouse IV, Jul 20 2011

(Python)

from math import isqrt

from sympy import primepi, primerange, integer_nthroot

def A163569(n):

def bisection(f, kmin=0, kmax=1):

while f(kmax) > kmax: kmax <<= 1

kmin = kmax >> 1

while kmax-kmin > 1:

kmid = kmax+kmin>>1

if f(kmid) <= kmid:

kmax = kmid

else:

kmin = kmid

return kmax

def f(x): return n+x-sum(primepi(x//(p**3*q**2)) for p in primerange(integer_nthroot(x, 3)[0]+1) for q in primerange(isqrt(x//p**3)+1))+sum(primepi(integer_nthroot(x//p**3, 3)[0]) for p in primerange(integer_nthroot(x, 3)[0]+1))+sum(primepi(isqrt(x//p**4)) for p in primerange(integer_nthroot(x, 4)[0]+1))+sum(primepi(x//p**5) for p in primerange(integer_nthroot(x, 5)[0]+1))-(primepi(integer_nthroot(x, 6)[0])<<1)

return bisection(f, n, n) # Chai Wah Wu, Mar 27 2025

Subsequence of A137487. - R. J. Mathar, Aug 01 2009

Sequence in context: A323024 A072414 A375075 * A063067 A076205 A048978

Adjacent sequences: A163566 A163567 A163568 * A163570 A163571 A163572

Vladimir Joseph Stephan Orlovsky, Jul 31 2009

approved