A087050 Square root of the largest square >1 dividing the n-th nonsquarefree number.

2, 2, 3, 2, 4, 3, 2, 2, 5, 3, 2, 4, 6, 2, 2, 3, 4, 7, 5, 2, 3, 2, 2, 3, 8, 2, 6, 5, 2, 4, 9, 2, 2, 3, 2, 4, 7, 3, 10, 2, 6, 4, 2, 3, 2, 11, 2, 5, 3, 8, 2, 3, 2, 2, 12, 7, 2, 5, 2, 3, 2, 4, 9, 2, 2, 13, 3, 2, 5, 4, 6, 2, 2, 3, 8, 14, 3, 10, 2, 3, 4, 2, 6, 2, 4, 15, 2, 2, 3, 2, 4, 11, 9, 2, 7, 2, 5, 6, 16, 2, 3

Offset: 1

Author

Wolfdieter Lang, Sep 08 2003

Keywords

• nonn
• easy

Examples

			n=10, A013929(10) = 27, a(10)^2 = 3^2 = 9. 27 = 9*3.
n=39, A013929(39) = 100, a(39)^2 = 10^2 = 100.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A087049, A013929, A000188, A000290.

Cf. A001610, A013661, A306016.

Programs

Mathematica

f[p_, e_] := p^Floor[e/2]; s[1] = 1; s[n_] := Times @@ f @@@ FactorInteger[n]; s /@ Select[Range[300], !SquareFreeQ[#] &] (* Amiram Eldar, Feb 11 2021 *)

Python

from math import isqrt, prod
from sympy import mobius, factorint
def A087050(n):
    def f(x): return n+sum(mobius(k)*(x//k**2) for k in range(1, isqrt(x)+1))
    m, k = n, f(n)
    while m != k:
        m, k = k, f(k)
    return prod(p**(e>>1) for p, e in factorint(m).items() if e>1) # Chai Wah Wu, Jul 22 2024

Formula

a(n)^2 is the largest square factor (from A000290) of the nonsquarefree number A013929(n), n>=1.

a(n) = A000188(A013929(n)). - Amiram Eldar, Feb 11 2021

Sum_{k=1..n} a(k) ~ (n/(2*(zeta(2)-1))) * (log(n) + 3*gamma - 3 - 2*zeta'(2)/zeta(2) - log(1-1/zeta(2))), where gamma is Euler's constant (A001620). - Amiram Eldar, Jan 14 2024