A030626 Numbers with exactly 8 divisors.

24, 30, 40, 42, 54, 56, 66, 70, 78, 88, 102, 104, 105, 110, 114, 128, 130, 135, 136, 138, 152, 154, 165, 170, 174, 182, 184, 186, 189, 190, 195, 222, 230, 231, 232, 238, 246, 248, 250, 255, 258, 266, 273, 282, 285, 286, 290, 296, 297, 310, 318, 322, 328, 344, 345, 351, 354, 357, 366, 370, 374, 375, 376, 385, 399, 402

Offset: 1

Author

Jeff Burch

Keywords

• nonn

Comments

Since A119479(8)=7, there are never more than 7 consecutive terms. Runs of 7 consecutive terms start at 171897, 180969, 647385, ... (subsequence of A049053). - Ivan Neretin, Feb 08 2016

Links

• Jon E. Schoenfield, Table of n, a(n) for n = 1..10000 (first 1000 terms from R. J. Mathar)
• Jérôme Germoni, Nombres à huit diviseurs, Images des Mathématiques, CNRS, 2017 (in French).
• Eric Weisstein's World of Mathematics, Divisor Product.

Crossrefs

Essentially the same as A111398.

Cf. A000005, A007304, A049053, A065036, A092759, A119479.

Programs

Magma

[n: n in [1..400] | DivisorSigma(0, n) eq 8]; // Vincenzo Librandi, Oct 05 2017

Maple

select(numtheory:-tau=8, [$1..1000]); # Robert Israel, Dec 17 2014

Mathematica

Select[Range[400], DivisorSigma[0, #]== 8 &] (* Vincenzo Librandi, Oct 05 2017 *)

PARI

Vec(select(x->x==8,vector(500, i, numdiv(i)),1)) \\ Michel Marcus, Dec 17 2014

Python

from sympy import divisor_count
isok = lambda n: divisor_count(n) == 8
print([n for n in range(1, 400) if isok(n)]) # Darío Clavijo, Oct 17 2023

Python

from math import isqrt
from sympy import primepi, primerange, integer_nthroot
def A030626(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))-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))-sum(primepi(x//p**3) for p in primerange(integer_nthroot(x,3)[0]+1))+primepi(integer_nthroot(x,4)[0])-primepi(integer_nthroot(x,7)[0]))
    return bisection(f,n,n) # Chai Wah Wu, Feb 21 2025

Formula

A000005(a(n))=8. - Juri-Stepan Gerasimov, Oct 10 2009

Equals A065036 (p*q^3) U A007304 (p*q*r) U A092759 (p^7). - Amarnath Murthy, Apr 21 2001