A352274 Numbers whose squarefree part is congruent to 1 modulo 6 or 3 modulo 18.

1, 3, 4, 7, 9, 12, 13, 16, 19, 21, 25, 27, 28, 31, 36, 37, 39, 43, 48, 49, 52, 55, 57, 61, 63, 64, 67, 73, 75, 76, 79, 81, 84, 85, 91, 93, 97, 100, 103, 108, 109, 111, 112, 115, 117, 121, 124, 127, 129, 133, 139, 144, 145, 147, 148, 151, 156, 157, 163, 165, 169, 171, 172

Offset: 1

Author

Peter Munn, Mar 10 2022

Keywords

• nonn
• easy

Comments

Numbers of the form 4^i * 3^j * (6k+1), i, j, k >= 0. Numbers whose prime factorization has an even number of factors of 2 and an even number of factors of the form 6k+5 (therefore also an even number of factors of the form 3k+2).

Closed under multiplication.

Includes the nonzero Loeschian numbers (A003136). The two sequences have few early differences (the first extra number here is a(22) = 55, followed by 85, 115, 145, ...), but their densities diverge progressively, driven by the presence here - and absence from A003136 - of the nonsquare terms of A108166. Asymptotic densities are 1/3 and 0 respectively.

Term by term, the sequence is one half of its complement within A225837.

Examples

			4 = 2^2 has square part 2^2, therefore squarefree part 4/2^2 = 1, which is congruent to 1 mod 6, so 4 is in the sequence.
63 = 3^2 * 7 has square part 3^2, therefore squarefree part 63/3^2 = 7, which is congruent to 1 mod 6, so 63 is in the sequence.
21 = 3*7 has square part 1^2 and squarefree part 21, which is congruent to 3 mod 18, so 21 is in the sequence.
72 = 2^3 * 3^2 has square part 2^2 * 3^2 = 6^2, therefore squarefree part 72/6^2 = 2, which is congruent to 2 mod 6 and to 2 mod 18, so 72 is not in the sequence.
		

Links

• Eric Weisstein's World of Mathematics, Squarefree Part.

Crossrefs

Intersection of any two of A003159, A026225 and A225837.

Closure of A084089 under multiplication by 3.

Cf. A007913.

Subsequences: A003136\{0}, A108166, A352272.

Programs

PARI

isok(m) = core(m) % 6 == 1 || core(m) % 18 == 3;

Python

from itertools import count
def A352274(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(0,2):
            i2 = 1<x:
                break
            for j in count(0):
                k = i2*3**j
                if k>x:
                    break
                c -= (x//k-1)//6+1
        return c
    return bisection(f,n,n) # Chai Wah Wu, Feb 14 2025

Formula

{a(n): n >= 1} = {m >= 1 : A007913(m) == 1 (mod 6)} U {m >= 1 : A007913(m) == 3 (mod 18)} = {A352272(m): m >= 1} U {3*A352272(m): m >= 1}.

{A225837(n): n >= 1} = {a(m): m >= 1} U {2*a(m): m >= 1}.