A143610 Numbers of the form p^2 * q^3, where p,q are distinct primes.
72, 108, 200, 392, 500, 675, 968, 1125, 1323, 1352, 1372, 2312, 2888, 3087, 3267, 4232, 4563, 5324, 6125, 6728, 7688, 7803, 8575, 8788, 9747, 10952, 11979, 13448, 14283, 14792, 15125, 17672, 19652, 19773, 21125, 22472, 22707, 25947, 27436
Offset: 1
Examples
The first three terms of this sequence are 3^2 * 2^3 = 72, 2^2 * 3^3 = 108, 5^2 * 2^3 = 200.
Links
Programs
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Mathematica
f[n_] := Sort[Last/@FactorInteger[n]] == {2, 3}; Select[Range[30000], f] (* Vladimir Joseph Stephan Orlovsky, Oct 09 2009 *)
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PARI
for(n=1, 10^5, omega(n)==2 || next; vecsort(factor(n)[,2])==[2,3]~ && print1(n","))
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PARI
list(lim)=my(v=List(),t);forprime(p=2, (lim\4)^(1/3), t=p^3;forprime(q=2, sqrt(lim\t), if(p==q, next);listput(v,t*q^2))); vecsort(Vec(v)) \\ Charles R Greathouse IV, Jul 20 2011
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Python
from math import isqrt from sympy import primepi, primerange, integer_nthroot def A143610(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(isqrt(x//p**3)) for p in primerange(integer_nthroot(x,3)[0]+1))+primepi(integer_nthroot(x,5)[0]) return bisection(f,n,n) # Chai Wah Wu, Feb 21 2025
Formula
Sum_{n>=1} 1/a(n) = P(2)*P(3) - P(5) = A085548 * A085541 - A085965 = 0.043280..., where P is the prime zeta function. - Amiram Eldar, Jul 06 2020
Comments