A356838 The smallest of the most common prime factors of n.

2, 3, 2, 5, 2, 7, 2, 3, 2, 11, 2, 13, 2, 3, 2, 17, 3, 19, 2, 3, 2, 23, 2, 5, 2, 3, 2, 29, 2, 31, 2, 3, 2, 5, 2, 37, 2, 3, 2, 41, 2, 43, 2, 3, 2, 47, 2, 7, 5, 3, 2, 53, 3, 5, 2, 3, 2, 59, 2, 61, 2, 3, 2, 5, 2, 67, 2, 3, 2, 71, 2, 73, 2, 5, 2, 7, 2, 79, 2, 3, 2, 83, 2, 5, 2, 3, 2, 89

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 smallest one of those prime factors.

If the prime factorization of n has a unique largest exponent (A356862), then a(n) = A356840(n). Otherwise a(n) is the smallest of the most common prime factors, while A356840(n) is the largest of them.

a(n) differs from A020639(n) in case the smallest prime factor is not the most common. For example: a(90) = 3, since 90 = 2 * 3^2 * 5 and 3 has the largest exponent. This is different from A020639(90) = 2, since 2 is the smallest prime factor.

Examples

			a(18) = 3, since 18 = 2 * 3^2 and 3 is the most common prime factor.
a(450) = 3, since 450 = 2 * 3^2 * 5^2 and 3 is the smallest of the most common prime factors.
		

Links

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

Crossrefs

Cf. A051903, A356840, A020639, A356862.

Programs

Mathematica

a[n_] := Module[{f = FactorInteger[n], p, e}, p = f[[;; , 1]]; e = f[[;; , 2]]; p[[FirstPosition[e, Max[e]][[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~])); vecmin(vector(#w, k, f[w[k], 1])); \\ Michel Marcus, Sep 01 2022

Python

from sympy import factorint
from collections import Counter
def a(n):
    return Counter(factorint(n)).most_common(1)[0][0]

Python

from sympy import factorint
def A356838(n): return max(factorint(n).items(),key=lambda x:(x[1],-x[0]))[0] # Chai Wah Wu, Sep 10 2022