A340682 - OEIS

A340682

The closure under squaring of the nonunit squarefree numbers.

2, 3, 4, 5, 6, 7, 9, 10, 11, 13, 14, 15, 16, 17, 19, 21, 22, 23, 25, 26, 29, 30, 31, 33, 34, 35, 36, 37, 38, 39, 41, 42, 43, 46, 47, 49, 51, 53, 55, 57, 58, 59, 61, 62, 65, 66, 67, 69, 70, 71, 73, 74, 77, 78, 79, 81, 82, 83, 85, 86, 87, 89, 91, 93, 94, 95, 97, 100, 101, 102, 103, 105, 106, 107, 109, 110, 111, 113

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OFFSET

1,1

COMMENTS

Numbers of the form s^(2^e), where s is a nonunit squarefree number, and e >= 0.

The categorization provided by this sequence and its complement, A340681, is an alternative extension (to all integers greater than 1) of the 2-way distinction between squarefree and nonsquarefree as it applies to nonsquares.

All positive integers have a unique factorization into powers of nonunit squarefree numbers with distinct exponents that are powers of 2. This sequence lists the numbers where this factorization has only one term, that is numbers m such that A331591(m) = 1.

Presence in the sequence is determined by prime signature. The set of represented signatures starts: {{1}, {2}, {1,1}, {1,1,1}, {4}, {2,2}, {1,1,1,1}, {1,1,1,1,1}, {2,2,2}, {1,1,1,1,1,1}, {1,1,1,1,1,1,1}, {8}, {4,4}, {2,2,2,2}, {1,1,1,1,1,1,1,1}, ...}. Representing each signature in the set by the least number with that signature, we get the set A133492.

Positions of terms > 1 in A340675.

LINKS

Table of n, a(n) for n=1..78.

Index entries for sequences related to prime signature

EXAMPLE

12 = 3 * 4 = 3^1 * 2^2 = 3^(2^0) * 2^(2^1). This is the (unique) factorization into powers of nonunit squarefree numbers with distinct exponents that are powers of 2. As this factorization has 2 terms, 12 is not in the sequence.

The equivalent factorization for 36 is 36 = 6^2 = 6^(2^1). As this factorization has only 1 term, 36 is in the sequence.

MATHEMATICA

Select[Range[2, 120], Length[(u = Union[FactorInteger[#][[;; , 2]]])] == 1 && u[[1]] == 2^IntegerExponent[u[[1]], 2] &] (* Amiram Eldar, Feb 13 2021 *)

PROG

(PARI) isA340682(n) = if(!issquare(n), issquarefree(n), (n>1)&&isA340682(sqrtint(n)));

(Python)

from math import isqrt

from sympy import mobius, integer_nthroot

def A340682(n):

def g(x): return sum(mobius(k)*(x//k**2) for k in range(1, isqrt(x)+1))

def f(x): return int(n+x-sum(g(integer_nthroot(x, 1<<e)[0])-1 for e in range(x.bit_length().bit_length())))

m, k = n, f(n)

while m != k: m, k = k, f(k)

return m # Chai Wah Wu, Jun 01 2025

CROSSREFS