A162143 Numbers that are the squares of the product of three distinct primes.
900, 1764, 4356, 4900, 6084, 10404, 11025, 12100, 12996, 16900, 19044, 23716, 27225, 28900, 30276, 33124, 34596, 36100, 38025, 49284, 52900, 53361, 56644, 60516, 65025, 66564, 70756, 74529, 79524, 81225, 81796, 84100, 96100, 101124, 103684, 119025, 125316, 127449
Offset: 1
Keywords
Examples
900 = 2^2*3^2*5^2, 1764 = 2^2*3^2*7^2, 4356 = 2^2*3^2*11^2, ..
Programs
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Maple
h := proc(n) local P; P := NumberTheory:-PrimeFactors(n); nops(P) = 3 and n = mul(P) end: A162143List := upto -> seq(n^2, n=select(h, [seq(1..upto)])): # Peter Luschny, Apr 14 2025
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Mathematica
fQ[n_]:=Last/@FactorInteger[n]=={2,2,2}; Select[Range[100000], f]
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Python
from math import isqrt from sympy import primepi, primerange, integer_nthroot def A162143(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 bisection(f)**2 # Chai Wah Wu, Aug 29 2024
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SageMath
def is_a(n): P = prime_divisors(n) return len(P) == 3 and prod(P) == n print([n*n for n in range(1, 439) if is_a(n)]) # Peter Luschny, Apr 14 2025
Formula
a(n) = A007304(n)^2.
A050326(a(n)) = 8. - Reinhard Zumkeller, May 03 2013
Sum_{n>=1} 1/a(n) = (P(2)^3 + 2*P(6) - 3*P(2)*P(4))/6 = (A085548^3 + 2*A085966 - 3*A085548*A085964)/6 = 0.0036962441..., where P is the prime zeta function. - Amiram Eldar, Oct 30 2020
Extensions
Edited by N. J. A. Sloane, Jun 27 2009
Comments