A356840 Largest most common prime factor of n.

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

Offset: 2

Author

Jens Ahlström, Aug 31 2022

Keywords

• nonn
• easy

Comments

Pick the prime factors of n with the largest exponent. a(n) is the largest one of those prime factors. If the prime factorization of n has a unique largest exponent, then a(n) = A356838(n). Otherwise, a(n) is the largest of those most common prime factors, while A356838(n) is the smallest of them.

Examples

			a(180) = 3, since 180 = 2^2 * 3^2 * 5 and the largest of the most common prime factor is 3.
		

Links

• Jens Ahlström, Table of n, a(n) for n = 2..9999

Crossrefs

Cf. A051903, A356838, A356862.

Programs

Mathematica

a[n_] := Module[{f = FactorInteger[n], p, e}, p = f[[;; , 1]]; e = f[[;; , 2]]; p[[Position[e, Max[e]][[-1,1]]]]]; Array[a, 100, 2] (* Amiram Eldar, Sep 01 2022 *)

PARI

a(n) = my(f=factor(n), m=vecmax(f[,2]), w=select(x->(f[x,2] == m), [1..#f~])); vecmax(vector(#w, k, f[w[k], 1])); \\ Michel Marcus, Sep 01 2022

Python

from sympy import factorint
from collections import Counter
def reversed_factors(n):
    return dict(reversed(list(factorint(n).items())))
def a(n):
    return Counter(reversed_factors(n)).most_common(1)[0][0]

Python

from sympy import factorint
def A356840(n): return max(factorint(n).items(),key=lambda x:(x[1],x[0]))[0] # Chai Wah Wu, Sep 10 2022