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A356840
Largest most common prime factor of n.
5
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
2,1
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
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
KEYWORD
nonn,easy
AUTHOR
Jens Ahlström, Aug 31 2022
STATUS
approved