A228649 Numbers n such that n-1, n and n+1 are all squarefree.

2, 6, 14, 22, 30, 34, 38, 42, 58, 66, 70, 78, 86, 94, 102, 106, 110, 114, 130, 138, 142, 158, 166, 178, 182, 186, 194, 202, 210, 214, 218, 222, 230, 238, 254, 258, 266, 282, 286, 302, 310, 318, 322, 330, 346, 354, 358, 366, 382, 390, 394, 398, 402, 410, 418, 430, 434, 438, 446, 454, 462, 466, 470, 482, 498

Offset: 1

Author

Adam P. Goucher, Aug 29 2013

Keywords

• easy
• nonn

Comments

Equivalently, a positive integer n is comfortably squarefree if and only if n^3 - n is squarefree. The 'if' direction is obvious from the factorization n(n-1)(n+1), and the converse follows from the coprimality of n, n - 1 and n + 1.

The asymptotic density of comfortably squarefree numbers is the product over all primes of 1 - 3/p^2, which is A206256 = 0.125486980905....

See also comments in A007675.

Links

• Harvey P. Dale, Table of n, a(n) for n = 1..1000
• Dan Asimov, Interesting sequence on MathOverflow, Math-fun mailing list, Mar 28 2023.
• Ewan Delanoy, Are there infinitely many triples of consecutive square-free integers?, Math Overflow.
• A. P. Goucher, Comfortably squarefree numbers, Complex Projective 4-Space.
• Fredrick M. Nelson, Does a(0)=6, a(n+1)=a(n)^3-a(n), define a square-free sequence?, MathOverflow, Mar 24 2023.

Programs

Maple

with(numtheory):
a := n -> `if`(issqrfree(n-1) and issqrfree(n) and issqrfree(n+1), n, NULL);
seq(a(n), n = 1..500); # Peter Luschny, Jan 18 2014

Mathematica

Select[Range[500], (SquareFreeQ[# - 1] && SquareFreeQ[#] && SquareFreeQ[# + 1]) &] (* Adam P. Goucher *)
Select[Range[2, 500, 2], (MoebiusMu[# - 1] MoebiusMu[#] MoebiusMu[# + 1]) != 0 &] (* Alonso del Arte, Jan 16 2014 *)
Flatten[Position[Partition[Boole[SquareFreeQ/@Range[500]],3,1],{1,1,1}]]+1 (* Harvey P. Dale, Jan 14 2015 *)
SequencePosition[Table[If[SquareFreeQ[n],1,0],{n,500}],{1,1,1}][[All,1]]+1 (* Requires Mathematica version 10 or later *) (* Harvey P. Dale, Dec 02 2018 *)

PARI

is(n)=issquarefree(n-1)&&issquarefree(n)&&issquarefree(n+1) \\ Charles R Greathouse IV, Aug 29 2013

Formula

a(n) = A007675(n) + 1. - Giovanni Resta, Aug 29 2013