A385652 Maximum frequency of gpf(k) for 2 <= k <= n, where gpf(k) = A006530(k) is the greatest prime factor of k.

1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 6, 6, 6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 10, 11, 11, 11, 11, 11, 11, 11, 11, 12

Offset: 2

Author

Pontus von Brömssen, Jul 06 2025

Keywords

• nonn

Comments

The prime p is popular on the interval [2,n] if it is the greatest prime factor of a(n) numbers in that interval; see A385503.

Examples

			     |     | cumulative frequencies for gpf's |
   n | gpf |    2  3  5  7 11 13 17 19 23     | a(n)
  ---+-----+----------------------------------+-----
   2 |   2 |    1  0  0  0  0  0  0  0  0     |  1
   3 |   3 |    1  1  0  0  0  0  0  0  0     |  1
   4 |   2 |    2  1  0  0  0  0  0  0  0     |  2
   5 |   5 |    2  1  1  0  0  0  0  0  0     |  2
   6 |   3 |    2  2  1  0  0  0  0  0  0     |  2
   7 |   7 |    2  2  1  1  0  0  0  0  0     |  2
   8 |   2 |    3  2  1  1  0  0  0  0  0     |  3
   9 |   3 |    3  3  1  1  0  0  0  0  0     |  3
  10 |   5 |    3  3  2  1  0  0  0  0  0     |  3
  11 |  11 |    3  3  2  1  1  0  0  0  0     |  3
  12 |   3 |    3  4  2  1  1  0  0  0  0     |  4
  13 |  13 |    3  4  2  1  1  1  0  0  0     |  4
  14 |   7 |    3  4  2  2  1  1  0  0  0     |  4
  15 |   5 |    3  4  3  2  1  1  0  0  0     |  4
  16 |   2 |    4  4  3  2  1  1  0  0  0     |  4
  17 |  17 |    4  4  3  2  1  1  1  0  0     |  4
  18 |   3 |    4  5  3  2  1  1  1  0  0     |  5
  19 |  19 |    4  5  3  2  1  1  1  1  0     |  5
  20 |   5 |    4  5  4  2  1  1  1  1  0     |  5
  21 |   7 |    4  5  4  3  1  1  1  1  0     |  5
  22 |  11 |    4  5  4  3  2  1  1  1  0     |  5
  23 |  23 |    4  5  4  3  2  1  1  1  1     |  5
  24 |   3 |    4  6  4  3  2  1  1  1  1     |  6
		

Links

• Pontus von Brömssen, Table of n, a(n) for n = 2..10000

Crossrefs

Cf. A006530, A078899, A385503, A385653, A385654.

Programs

PARI

gpf(n) = if (n==1,1, vecmax(factor(n)[,1])); \\ A006530
a(n) = my(v=vector(n, k, gpf(k)), s=Set(v)); vecmax(apply(x->#x, vector(#s, i, select(x->(x==s[i]), v)))); \\ Michel Marcus, Jul 06 2025

Python

from collections import Counter
from itertools import count
from sympy import factorint
def A385652_generator():
    c = Counter()
    M = 0
    for n in count(2):
        gpf = max(factorint(n))
        c[gpf] += 1
        if c[gpf] > M: M += 1
        yield M

Formula

a(n) = max_{k=2..n} A078899(k).