A052486 Achilles numbers - powerful but imperfect: if n = Product(p_i^e_i) then all e_i > 1 (i.e., powerful), but the highest common factor of the e_i is 1, i.e., not a perfect power.

72, 108, 200, 288, 392, 432, 500, 648, 675, 800, 864, 968, 972, 1125, 1152, 1323, 1352, 1372, 1568, 1800, 1944, 2000, 2312, 2592, 2700, 2888, 3087, 3200, 3267, 3456, 3528, 3872, 3888, 4000, 4232, 4500, 4563, 4608, 5000, 5292, 5324, 5400, 5408, 5488, 6075

Offset: 1

Author

Henry Bottomley, Mar 16 2000

Keywords

• nonn

Comments

Number of terms < 10^n: 0, 1, 13, 60, 252, 916, 3158, 10553, 34561, 111891, 359340, 1148195, 3656246, 11616582, 36851965, ..., A118896(n) - A070428(n). - Robert G. Wilson v, Aug 11 2014

a(n) = (s(n))^2 * f(n), s(n) > 1, f(n) > 1, where s(n) is not a power of f(n), and f(n) is squarefree and gcd(s(n), f(n)) = f(n). - Daniel Forgues, Aug 11 2015

Examples

			a(3)=200 because 200=2^3*5^2, both 3 and 2 are greater than 1, and the highest common factor of 3 and 2 is 1.
Factorizations of a(1) to a(20):
    72 = 2^3  3^2,  108 = 2^2 3^3,  200 = 2^3 5^2,  288 = 2^5  3^2,
   392 = 2^3  7^2,  432 = 2^4 3^3,  500 = 2^2 5^3,  648 = 2^3  3^4,
   675 = 3^3  5^2,  800 = 2^5 5^2,  864 = 2^5 3^3,  968 = 2^3 11^2,
   972 = 2^2  3^5, 1125 = 3^2 5^3, 1152 = 2^7 3^2, 1323 = 3^3  7^2,
  1352 = 2^3 13^2, 1372 = 2^2 7^3, 1568 = 2^5 7^2, 1800 = 2^3  3^2 5^2.
Examples for a(n) = (s(n))^2 * f(n): (see above comment)
s(n) = 6,  6, 10, 12, 14, 12, 10, 18, 15, 20, 12, 22, 18, 15, 24, 21,
f(n) = 2,  3,  2,  2,  2,  3,  5,  2,  3,  2,  6,  2,  3,  5,  2,  3,
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000 (terms 1..1000 from T. D. Noe)
• Project Euler, Problem 302: Strong Achilles Numbers.
• Robert Israel, log-log plot of a(n).
• Eric Weisstein's World of Mathematics, Achilles Number.
• OEIS Wiki, Achilles numbers.

Crossrefs

Cf. A001597, A001694, A007916, A072102, A082695.

Programs

Maple

filter:= proc(n) local E; E:= map(t->t[2], ifactors(n)[2]); min(E)>1 and igcd(op(E))=1 end proc:
select(filter,[$1..10000]); # Robert Israel, Aug 11 2014

Mathematica

achillesQ[n_] := Block[{ls = Last /@ FactorInteger@n}, Min@ ls > 1 == GCD @@ ls]; Select[ Range@ 5500, achillesQ@# &] (* Robert G. Wilson v, Jun 10 2010 *)

PARI

is(n)=my(f=factor(n)[,2]); n>9 && vecmin(f)>1 && gcd(f)==1 \\ Charles R Greathouse IV, Sep 18 2015, replacing code by M. F. Hasler, Sep 23 2010

Python

from math import gcd
from itertools import count, islice
from sympy import factorint
def A052486_gen(startvalue=1): # generator of terms >= startvalue
    return (n for n in count(max(startvalue,1)) if (lambda x: all(e > 1 for e in x) and gcd(*x) == 1)(factorint(n).values()))
A052486_list = list(islice(A052486_gen(),20)) # Chai Wah Wu, Feb 19 2022

Python

from math import isqrt
from sympy import mobius, integer_nthroot
def A052486(n):
    def squarefreepi(n): return int(sum(mobius(k)*(n//k**2) for k in range(1, isqrt(n)+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
    def f(x):
        c, l = n+x+1, 0
        j = isqrt(x)
        while j>1:
            k2 = integer_nthroot(x//j**2,3)[0]+1
            w = squarefreepi(k2-1)
            c -= j*(w-l)
            l, j = w, isqrt(x//k2**3)
        c -= squarefreepi(integer_nthroot(x,3)[0])-l+sum(mobius(k)*(integer_nthroot(x, k)[0]-1) for k in range(2, x.bit_length()))
        return c
    return bisection(f,n,n) # Chai Wah Wu, Sep 10 2024

Formula

a(n) = O(n^2). - Daniel Forgues, Aug 11 2015

a(n) = O(n^2 / log log n). - Daniel Forgues, Aug 12 2015

Sum_{n>=1} 1/a(n) = zeta(2)*zeta(3)/zeta(6) - Sum_{k>=2} mu(k)*(1-zeta(k)) - 1 = A082695 - A072102 - 1 = 0.06913206841581433836... - Amiram Eldar, Oct 14 2020

Extensions

Example edited by Mac Coombe (mac.coombe(AT)gmail.com), Sep 18 2010

Name edited by M. F. Hasler, Jul 17 2019