A352273 Numbers whose squarefree part is congruent to 5 modulo 6.

5, 11, 17, 20, 23, 29, 35, 41, 44, 45, 47, 53, 59, 65, 68, 71, 77, 80, 83, 89, 92, 95, 99, 101, 107, 113, 116, 119, 125, 131, 137, 140, 143, 149, 153, 155, 161, 164, 167, 173, 176, 179, 180, 185, 188, 191, 197, 203, 207, 209, 212, 215, 221, 227, 233, 236, 239, 245, 251

Offset: 1

Author

Peter Munn, Mar 10 2022

Keywords

• nonn
• easy

Comments

Numbers of the form 4^i * 9^j * (6k+5), i, j, k >= 0.

1/5 of each multiple of 5 in A352272.

The product of any two terms is in A352272.

The product of a term of this sequence and a term of A352272 is a term of this sequence.

The positive integers are usefully partitioned as {A352272, 2*A352272, 3*A352272, 6*A352272, {a(n)}, 2*{a(n)}, 3*{a(n)}, 6*{a(n)}}. There is a table in the example section giving sequences formed from unions of the parts.

The parts correspond to the cosets of A352272 considered as a subgroup of the positive integers under the operation A059897(.,.). Viewed another way, the parts correspond to the intersection of the integers with the cosets of the multiplicative subgroup of the positive rationals generated by the terms of A352272.

The asymptotic density of this sequence is 1/4. - Amiram Eldar, Apr 03 2022

Examples

			The squarefree part of 11 is 11, which is congruent to 5 (mod 6), so 11 is in the sequence.
The squarefree part of 15 is 15, which is congruent to 3 (mod 6), so 15 is not in the sequence.
The squarefree part of 20 = 2^2 * 5 is 5, which is congruent to 5 (mod 6), so 20 is in the sequence.
The table below lists OEIS sequences that are unions of the cosets described in the initial comments, and indicates the cosets included in each sequence. A352272 (as a subgroup) is denoted H, and this sequence (as a coset) is denoted H/5, in view of its terms being one fifth of the multiples of 5 in A352272.
             H    2H    3H    6H    H/5  2H/5  3H/5  6H/5
A003159      X           X           X           X
A036554            X           X           X           X
.
A007417      X     X                 X     X
A145204\{0}              X     X                 X     X
.
A026225      X           X                 X           X
A026179\{1}        X           X     X           X
.
A036668      X                 X     X                 X
A325424            X     X                 X     X
.
A055047      X                             X
A055048            X                 X
A055041                  X                             X
A055040                        X                 X
.
A189715      X                 X           X     X
A189716            X     X           X                 X
.
A225837      X     X     X     X
A225838                              X     X     X     X
.
A339690      X                       X
A329575                  X                       X
.
A352274      X           X
(The sequence groupings in the table start with the subgroup of the quotient group of H, followed by its cosets.)
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• Eric Weisstein's World of Mathematics, Squarefree Part.

Crossrefs

Intersection of any three of A003159, A007417, A189716 and A225838.

Intersection of A036668 and A055048.

Complement within A339690 of A352272.

Cf. A007913, A059897, A307151, A334832.

Closure of A084088 under multiplication by 9.

Other subsequences: A033429\{0}, A016969.

Other sequences in the example table: A036554, A145204, A026179, A026225, A325424, A055040, A055041, A055047, A189715, A225837, A329575, A352274.

Programs

Mathematica

q[n_] := Module[{e2, e3}, {e2, e3} = IntegerExponent[n, {2, 3}]; EvenQ[e2] && EvenQ[e3] && Mod[n/2^e2/3^e3, 6] == 5]; Select[Range[250], q] (* Amiram Eldar, Apr 03 2022 *)

PARI

isok(m) = core(m) % 6 == 5;

Python

from itertools import count
def A352273(n):
    def bisection(f,kmin=0,kmax=1):
        while f(kmax) > kmax: kmax <<= 1
        kmin = kmax >> 1
        while kmax-kmin > 1:
            kmid = kmax+kmin>>1
            if f(kmid) <= kmid:
                kmax = kmid
            else:
                kmin = kmid
        return kmax
    def f(x):
        c = n+x
        for i in count(0):
            i2 = 9**i
            if i2>x: break
            for j in count(0,2):
                k = i2<x: break
                c -= (x//k-5)//6+1
        return c
    return bisection(f,n,n) # Chai Wah Wu, Feb 14 2025

Formula

{a(n) : n >= 1} = {m >= 1 : A007913(m) == 5 (mod 6)}.

{a(n) : n >= 1} = A334832/5 U A334832/11 U A334832/17 U A334832/23 where A334832/k denotes {A334832(m)/k : m >= 1, k divides A334832(m)}.

Using the same notation, {a(n) : n >= 1} = A352272/5 = {A307151(A352272(m)) : m >= 1}.

{A225838(n) : n >= 1} = {m : m = a(j)*k, j >= 1, k divides 6}.