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A356862
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Numbers with a unique largest prime exponent.
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41
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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
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OFFSET
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1,1
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COMMENTS
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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
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LINKS
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EXAMPLE
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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.
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MATHEMATICA
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Select[Range[2, 100], Count[(e = FactorInteger[#][[;; , 2]]), Max[e]] == 1 &] (* Amiram Eldar, Sep 01 2022 *)
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PROG
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(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)))
(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
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CROSSREFS
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Subsequence of A319161 (which has additional terms 1, 180, 252, 300, 396, 450, 468, ...).
For factors instead of exponents we have A102750.
Partitions of this type are counted by A362608.
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KEYWORD
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nonn,easy
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AUTHOR
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STATUS
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approved
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