A356840 - OEIS

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

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OFFSET

2,1

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.

LINKS

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

EXAMPLE

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

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 *)

PROG

(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

(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

CROSSREFS

Cf. A051903, A356838, A356862.

Sequence in context: A273289 A365521 A361725 * A346152 A090662 A088387

Adjacent sequences: A356837 A356838 A356839 * A356841 A356842 A356843

KEYWORD

nonn,easy

AUTHOR

Jens Ahlström, Aug 31 2022

STATUS

approved