A276378 Numbers k such that 6*k is squarefree.

1, 5, 7, 11, 13, 17, 19, 23, 29, 31, 35, 37, 41, 43, 47, 53, 55, 59, 61, 65, 67, 71, 73, 77, 79, 83, 85, 89, 91, 95, 97, 101, 103, 107, 109, 113, 115, 119, 127, 131, 133, 137, 139, 143, 145, 149, 151, 155, 157, 161, 163, 167, 173, 179, 181, 185, 187, 191, 193, 197, 199, 203, 205, 209, 211

Offset: 1

Author

Juri-Stepan Gerasimov, Sep 02 2016

Keywords

• nonn
• easy

Comments

These are the numbers from A005117 that are not divisible by 2 and 3.

Squarefree numbers coprime to 6. - Robert Israel, Sep 02 2016

Numbers k such that A008588(k) is in A005117. - Felix Fröhlich, Sep 02 2016

The asymptotic density of this sequence is 3/Pi^2 (A104141). - Amiram Eldar, May 22 2020

From Peter Munn, Nov 20 2020: (Start)

The products generated from each subset of A215848 (primes greater than 3).

Closed under the commutative binary operation A059897(.,.), forming a subgroup of the positive integers under A059897. (End)

Multiplied by 6 we have 6, 30, 42, 66, 78, 102, ..., the values that may appear in A076978 after the 1, 2. [Don Reble, Dec 02 2020] - R. J. Mathar, Dec 15 2020

By the von Staudt-Clausen theorem, denominators of Bernoulli numbers are of the form 6*a(n) for some n. - Charles R Greathouse IV, May 16 2024

Examples

			5 is in this sequence because 6*5 = 30 = 2*3*5 is squarefree.
		

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Numbers m such that k*m is squarefree: A005117 (k = 1), A056911 (k = 2), A261034 (k = 3), A274546 (k = 5).

Subsequence of A007310, A300957, and A339690.

Cf. A003961, A008588, A059897, A076978, A104141, A215848.

Programs

Magma

[n: n in [1..230] | IsSquarefree(6*n)];

Maple

select(numtheory:-issqrfree, [seq(seq(6*i+j,j=[1,5]),i=0..100)]); # Robert Israel, Sep 02 2016

Mathematica

Select[Range@ 212, SquareFreeQ[6 #] &] (* Michael De Vlieger, Sep 02 2016 *)

PARI

is(n) = issquarefree(6*n) \\ Felix Fröhlich, Sep 02 2016

Formula

{a(n) : n >= 1} = {A003961(A003961(A005117(n))) : n >= 1} = {A003961(A056911(n)) : n >= 1}. - Peter Munn, Nov 20 2020

Sum_{n>=1} 1/a(n)^s = (6^s)*zeta(s)/((1+2^s)*(1+3^s)*zeta(2*s)), s>1. - Amiram Eldar, Sep 26 2023