A334832 Numbers whose squarefree part is congruent to 1 (mod 24).

1, 4, 9, 16, 25, 36, 49, 64, 73, 81, 97, 100, 121, 144, 145, 169, 193, 196, 217, 225, 241, 256, 265, 289, 292, 313, 324, 337, 361, 385, 388, 400, 409, 433, 441, 457, 481, 484, 505, 529, 553, 576, 577, 580, 601, 625, 649, 657, 673, 676, 697, 721, 729, 745, 769, 772, 784, 793, 817, 841

Offset: 1

Author

Peter Munn, Jun 15 2020

Keywords

• nonn
• easy

Comments

Closed under multiplication.

The sequence forms a subgroup of the positive integers under the commutative operation A059897(.,.). A059897 has a relevance to squarefree parts that arises from the identity A007913(k*m) = A059897(A007913(k), A007913(m)), where A007913(n) is the squarefree part of n.

The subgroup is one of 8 A059897(.,.) subgroups of the form {k : A007913(k) == 1 (mod m)}. The list seems complete, in anticipation of proof that such sets form subgroups only when m is a divisor of 24 (based on the property described by A. G. Astudillo in A018253). This sequence might be viewed as primitive with respect to the other 7, as the latter correspond to subgroups of its quotient group, in the sense that each one (as a set) is also a union of cosets described below. The 7 include A003159 (m=2), A055047 (m=3), A277549 (m=4), A234000 (m=8) and the trivial A000027 (m=1).

The subgroup has 32 cosets. For each i in {1, 5, 7, 11, 13, 17, 19, 23}, j in {1, 2, 3, 6} there is a coset {n : n = k^2 * (24m + i) * j, k >= 1, m >= 0}. The divisors of 2730 = 2*3*5*7*13 form a transversal. (11, clearly not such a divisor, is in the same coset as 35 = 11 + 24; 17, 19, 23 are in the same cosets as 65, 91, 455 respectively.)

The asymptotic density of this sequence is 1/16. - Amiram Eldar, Mar 08 2021

Examples

			The squarefree part of 26 is 26, which is congruent to 2 (mod 24), so 26 is not in the sequence.
The squarefree part of 292 = 2^2 * 73 is 73, which is congruent to 1 (mod 24), so 292 is in the sequence.
		

Links

• Amiram Eldar, Table of n, a(n) for n = 1..10000
• Eric Weisstein's World of Mathematics, Quotient Group.
• Eric Weisstein's World of Mathematics, Right Transversal.

Crossrefs

A subgroup under A059897, defined using A007913.

Subsequences: A000290\{0}, A103214, A107008.

Equivalent sequences modulo other members of A018253: A000027 (1), A003159 (2), A055047 (3), A277549 (4), A352272(6), A234000 (8).

Programs

Mathematica

Select[Range[850], Mod[Sqrt[#] /. (c_ : 1)*a_^(b_ : 0) :> (c*a^b)^2, 24] == 1 &] (* Michael De Vlieger, Jun 24 2020 *)

PARI

isok(m) = core(m) % 24 == 1; \\ Peter Munn, Jun 21 2020

Python

from sympy import integer_log
def A334832(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): return n+x-sum((x//(i<<(j<<1))-1)//24+1 for i in (9**k for k in range(integer_log(x,9)[0]+1)) for j in range((x//i>>1).bit_length()+1))
    return bisection(f,n,n) # Chai Wah Wu, Mar 21 2025

Formula

{a(n)} = {n : n = k^2 * (24m + 1), k >= 1, m >= 0}.