A356862 Numbers with a unique largest prime exponent.

2, 3, 4, 5, 7, 8, 9, 11, 12, 13, 16, 17, 18, 19, 20, 23, 24, 25, 27, 28, 29, 31, 32, 37, 40, 41, 43, 44, 45, 47, 48, 49, 50, 52, 53, 54, 56, 59, 60, 61, 63, 64, 67, 68, 71, 72, 73, 75, 76, 79, 80, 81, 83, 84, 88, 89, 90, 92, 96, 97, 98, 99, 101, 103, 104

Offset: 1

Author

Jens Ahlström, Sep 01 2022

Keywords

• nonn
• easy

Comments

If the prime factorization of k has a unique largest exponent, then k is a term.

Numbers whose multiset of prime factors (with multiplicity) has a unique mode. - Gus Wiseman, May 12 2023

Disjoint union of A246655 and A376250. The asymptotic density of this sequence, 0.3660366524547281232052..., is equal to the density of A376250 since the prime powers have a zero density. - Amiram Eldar, Sep 17 2024

Examples

			Prime powers (A246655) are in the sequence, since they have only one prime exponent in their prime factorization, hence a unique largest exponent.
144 is in the sequence, since 144 = 2^4 * 3^2 and there is the unique largest exponent 4.
225 is not in the sequence, since 225 = 3^2 * 5^2 and the largest exponent 2 is not unique, but rather it is the exponent of both the prime factor 3 and of the prime factor 5.
		

Links

• Jens Ahlström, Table of n, a(n) for n = 1..10000
• Wikipedia, Mode (statistics).

Crossrefs

Subsequence of A319161 (which has additional terms 1, 180, 252, 300, 396, 450, 468, ...).

Cf. A246655, A356838, A356840, A376250.

For factors instead of exponents we have A102750.

For smallest instead of largest we have A359178, counted by A362610.

The complement is A362605, counted by A362607.

The complement for co-mode is A362606, counted by A362609.

Partitions of this type are counted by A362608.

These are the positions of 1's in A362611, for co-modes A362613.

A001221 is the number of prime exponents, sum A001222.

A027746 lists prime factors, A112798 indices, A124010 exponents.

A362614 counts partitions by number of modes, A362615 co-modes.

Cf. A000041, A002865, A051903, A056239, A070003, A247180, A283050, A327473.

Programs

Mathematica

Select[Range[2, 100], Count[(e = FactorInteger[#][[;; , 2]]), Max[e]] == 1 &] (* Amiram Eldar, Sep 01 2022 *)

PARI

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

Python

from sympy import factorint
from collections import Counter
def ok(k):
    c = Counter(factorint(k)).most_common(2)
    return not (len(c) > 1 and c[0][1] == c[1][1])
print([k for k in range(2, 105) if ok(k)])

Python

from sympy import factorint
from itertools import count, islice
def A356862_gen(startvalue=2): # generator of terms >= startvalue
    return filter(lambda n:len(f:=sorted(factorint(n).values(),reverse=True))==1 or f[0]!=f[1],count(max(startvalue,2)))
A356862_list = list(islice(A356862_gen(),30)) # Chai Wah Wu, Sep 10 2022