A370833 a(n) is the greatest prime dividing the n-th cubefree number, for n >= 2; a(1)=1.

1, 2, 3, 2, 5, 3, 7, 3, 5, 11, 3, 13, 7, 5, 17, 3, 19, 5, 7, 11, 23, 5, 13, 7, 29, 5, 31, 11, 17, 7, 3, 37, 19, 13, 41, 7, 43, 11, 5, 23, 47, 7, 5, 17, 13, 53, 11, 19, 29, 59, 5, 61, 31, 7, 13, 11, 67, 17, 23, 7, 71, 73, 37, 5, 19, 11, 13, 79, 41, 83, 7, 17, 43

Offset: 1

Author

Amiram Eldar, Mar 03 2024

Keywords

• nonn
• easy

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• Jean-Marie De Koninck and Rafael Jakimczuk, Summing the largest prime factor over integer sequences, Revista de la Unión Matemática Argentina, Vol. 67, No. 1 (2024), pp. 27-35.

Crossrefs

Cf. A004709, A006530, A073482, A370834, A370835.

Programs

Mathematica

s[n_] := Module[{f = FactorInteger[n]}, If[AllTrue[f[[;; , 2]], # < 3 &], f[[-1, 1]], Nothing]]; Array[s, 200]

PARI

lista(kmax) = {my(f); print1(1, ", "); for(k = 2, kmax, f = factor(k); if(vecmax(f[, 2]) < 3, print1(f[#f~, 1], ", ")));}

Python

from sympy import mobius, integer_nthroot, primefactors
def A370833(n):
    def f(x): return n+x-sum(mobius(k)*(x//k**3) for k in range(1, integer_nthroot(x,3)[0]+1))
    m, k = n, f(n)
    while m != k:
        m, k = k, f(k)
    return max(primefactors(m),default=1) # Chai Wah Wu, Aug 06 2024

Formula

a(n) = A006530(A004709(n)).

Sum_{A004709(n) <= x} a(n) = Sum_{i=1..k} d_i * x^2/log(x)^i + O(x^2/log(x)^(k+1)), for any given positive integer k, where d_i are constants, d_1 = 315/(4*Pi^4) = 0.808446... (De Koninck and Jakimczuk, 2024).