A133466 Positive integers k for which there is exactly one integer i in {1,2,3,...,k-1} such that i*k is a square.

4, 8, 12, 20, 24, 28, 40, 44, 52, 56, 60, 68, 76, 84, 88, 92, 104, 116, 120, 124, 132, 136, 140, 148, 152, 156, 164, 168, 172, 184, 188, 204, 212, 220, 228, 232, 236, 244, 248, 260, 264, 268, 276, 280, 284, 292, 296, 308, 312, 316, 328, 332, 340, 344, 348, 356

Offset

1,1

Comments

It appears that all terms of this sequence are exactly four times those of the squarefree integers (A005117).

The observed behavior is true for all n. All positive integers n are written uniquely as k*m^2 where k is squarefree, k >=1, m >= 1. The square multiples of n are j^2*k*n, j >= 1. We seek n with exactly 1 multiple that is square and less than n^2. If m = 1, there are no such multiples as we have k = n, so the least square multiple is n^2. If m >= 2, k*n is square and less than n^2. However, 4*k*n also qualifies as square and less than n^2 if m > 2. So the qualifying values of n are those with m=2. - Peter Munn, Nov 28 2019

The asymptotic density of this sequence is 3/(2*Pi^2). - Amiram Eldar, Mar 08 2021

Links

Reinhard Zumkeller, Table of n, a(n) for n = 1..1000

Formula

A057918(a(n)) = 1. - Reinhard Zumkeller, Mar 27 2012

From Peter Munn, Nov 28 2019: (Start)

a(n) = 4 * A005117(n).

{a(n)} = {A225546(A007283(n)) : n >= 0}, where {a(n)} denotes the set of integers in the sequence.

(End)

Sum_{n>=1} 1/a(n)^s = zeta(s)/(4^s*zeta(2*s)), s>1. - Amiram Eldar, Sep 26 2023

Example

4 is in the sequence because among the products 1*4,2*4,3*4 = 4,8,12 there is exactly one square.

Mathematica

eoiQ[n_]:=Count[n*Range[n-1], _?(IntegerQ[Sqrt[#]]&)]==1; Select[Range[ 400], eoiQ] (* Harvey P. Dale, Mar 14 2015 *)

Prog

(Haskell)
a133466 n = a133466_list !! (n-1)
a133466_list = map (+ 1) $ elemIndices 1 a057918_list
-- Reinhard Zumkeller, Mar 27 2012
(PARI) isok(n) = sum(k=1, n-1, issquare(k*n)) == 1; \\ Michel Marcus, Nov 29 2019
(Magma) [k:k in [1..350]|#[m:m in [1..k-1]| IsSquare(m*k)] eq 1]; // Marius A. Burtea, Dec 03 2019
(Python)
from math import isqrt
from sympy import mobius
def A133466(n):
    def f(x): return n+x-sum(mobius(k)*(x//k**2) for k in range(1, isqrt(x)+1))
    m, k = n, f(n)
    while m != k:
        m, k = k, f(k)
    return int(m)<<2 # Chai Wah Wu, Aug 15 2024

Crossrefs

Cf. A005117, A007283, A225546.

Cf. A057918, A195085.

Sequence in context: A381122 A171949 A217319 * A311654 A311655 A311656

Adjacent sequences: A133463 A133464 A133465 * A133467 A133468 A133469

Keyword

nonn

Author

John W. Layman, Nov 28 2007

Status

approved