A059404 Numbers with different exponents in their prime factorizations.

12, 18, 20, 24, 28, 40, 44, 45, 48, 50, 52, 54, 56, 60, 63, 68, 72, 75, 76, 80, 84, 88, 90, 92, 96, 98, 99, 104, 108, 112, 116, 117, 120, 124, 126, 132, 135, 136, 140, 144, 147, 148, 150, 152, 153, 156, 160, 162, 164, 168, 171, 172, 175, 176, 180, 184, 188, 189, 192, 198, 200

Offset: 1

Author

Labos Elemer, Jul 18 2001

Keywords

• nonn
• easy

Comments

Former name: Numbers k such that k/(largest power of squarefree kernel of k) is larger than 1.

Also numbers k = p(1)^alpha(1)* ... * p(r)^alpha(r) such that RootMeanSquare(alpha(1), ..., alpha(r)) is not an integer. - Ctibor O. Zizka, Sep 19 2008

Examples

			440 is in the sequence because 440 = 2^3*5*11 and it has two distinct exponents, 3 and 1.
		

Links

• Donald Alan Morrison, Table of n, a(n) for n = 1..10000
• Donald Alan Morrison, Sage program

Crossrefs

Complement of A072774 (powers of squarefree numbers).

Cf. A003557, A007947, A051904, A062759, A062760, A071625.

Programs

Maple

isA := n -> 1 < nops({seq(padic:-ordp(n, p), p in NumberTheory:-PrimeFactors(n))}): select(isA, [seq(1..190)]);  # Peter Luschny, Apr 14 2025

Mathematica

A059404Q[n_] := Length[Union[FactorInteger[n][[All, 2]]]] > 1;
Select[Range[200], A059404Q] (* Paolo Xausa, Jan 07 2025 *)

PARI

is(n)=#Set(factor(n)[,2])>1 \\ Charles R Greathouse IV, Sep 18 2015

Python

from sympy import factorint
def ok(n): return len(set(factorint(n).values())) > 1
print([k for k in range(190) if ok(k)]) # Michael S. Branicky, Sep 01 2022

Python

from math import isqrt
from sympy import mobius, integer_nthroot
def A059404(n):
    def g(x): return int(sum(mobius(k)*(x//k**2) for k in range(1, isqrt(x)+1)))
    def f(x): return n+1-(y:=x.bit_length())+sum(g(integer_nthroot(x,k)[0]) for k in range(1,y))
    kmin, kmax = 1,2
    while f(kmax) >= kmax:
        kmax <<= 1
    while True:
        kmid = kmax+kmin>>1
        if f(kmid) < kmid:
            kmax = kmid
        else:
            kmin = kmid
        if kmax-kmin <= 1:
            break
    return kmax # Chai Wah Wu, Aug 19 2024

SageMath

def isA(n): return 1 < len(set(valuation(n, p) for p in prime_divisors(n)))
print([n for n in range(1, 190) if isA(n)])  # Peter Luschny, Apr 14 2025

Formula

A062760(a(n)) > 1, i.e., a(n)/(A007947(a(n))^A051904(a(n))) = a(n)/A062759(a(n)) > 1.

A071625(a(n)) > 1. - Michael S. Branicky, Sep 01 2022

Sum_{n>=1} 1/a(n)^s = zeta(s) - 1 - Sum_{k>=1} (zeta(k*s)/zeta(2*k*s)-1) for s > 1. - Amiram Eldar, Mar 20 2025