A163569 Numbers of the form p^3*q^2*r where p, q and r are three distinct primes.

360, 504, 540, 600, 756, 792, 936, 1176, 1188, 1224, 1350, 1368, 1400, 1404, 1500, 1656, 1836, 1960, 2052, 2088, 2200, 2232, 2250, 2484, 2600, 2646, 2664, 2904, 2952, 3096, 3132, 3348, 3384, 3400, 3500, 3800, 3816, 3996, 4056, 4116, 4248, 4312, 4392

Offset: 1

Author

Vladimir Joseph Stephan Orlovsky, Jul 31 2009

Keywords

• nonn

Comments

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

Examples

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

Links

• T. D. Noe, Table of n, a(n) for n = 1..1000
• Index to sequences related to prime signature

Crossrefs

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

Programs

Mathematica

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