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A190377
Numbers with prime factorization p^2*q^2*r^2*s^2 where p, q, r, and s are distinct primes.
1
44100, 108900, 152100, 213444, 260100, 298116, 324900, 476100, 509796, 592900, 636804, 736164, 756900, 828100, 864900, 933156, 1232100, 1258884, 1334025, 1416100, 1483524, 1512900, 1572516, 1664100, 1695204, 1758276, 1768900, 1863225
OFFSET
1,1
FORMULA
Sum_{n>=1} 1/a(n) = (P(2)^4 - 6*P(2)^2*P(4) + 8*P(2)*P(6) + 3*P(4)^2 - 6*P(8))/24 = 0.00010511750849230980748..., where P is the prime zeta function. - Amiram Eldar, Mar 07 2024
a(n) = A046386(n)^2. - Chai Wah Wu, Mar 27 2025
MATHEMATICA
f[n_]:=Sort[Last/@FactorInteger[n]]=={2, 2, 2, 2}; Select[Range[3000000], f]
PROG
(PARI) list(lim)=my(v=List(), t1, t2, t3); forprime(p=2, sqrtint(lim\900), t1=p^2; forprime(q=2, sqrtint(lim\(36*t1)), if(q==p, next); t2=q^2*t1; forprime(r=2, sqrtint(lim\(4*t2)), if(r==p || r==q, next); t3=r^2*t2; forprime(s=2, sqrtint(lim\t3), if(s==p || s==q || s==r, next); listput(v, t3*s^2))))); Set(v) \\ Charles R Greathouse IV, Aug 25 2016
(Python)
from math import isqrt
from sympy import primepi, primerange, integer_nthroot
def A190377(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 int(n+x-sum(primepi(x//(k*m*r))-c for a, k in enumerate(primerange(integer_nthroot(x, 4)[0]+1), 1) for b, m in enumerate(primerange(k+1, integer_nthroot(x//k, 3)[0]+1), a+1) for c, r in enumerate(primerange(m+1, isqrt(x//(k*m))+1), b+1)))
return bisection(f, n, n)**2 # Chai Wah Wu, Mar 27 2025
CROSSREFS
KEYWORD
nonn
AUTHOR
STATUS
approved