A076259 - OEIS

A076259

Gaps between squarefree numbers: a(n) = A005117(n+1) - A005117(n).

1, 1, 2, 1, 1, 3, 1, 2, 1, 1, 2, 2, 2, 1, 1, 3, 3, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 3, 1, 4, 2, 2, 2, 1, 1, 2, 1, 3, 1, 1, 2, 1, 1, 2, 1, 3, 1, 1, 3, 1, 2, 1, 1, 2, 2, 2, 1, 1, 2, 4, 1, 1, 2, 1, 1, 2, 1, 1, 2, 1, 1, 3, 1, 3, 1, 4, 2, 1, 1, 2, 1, 3, 1, 1, 2, 1, 1, 2, 1, 3, 2, 3, 1, 2, 1, 1, 2, 2, 2, 1, 1, 3, 3, 1

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OFFSET

1,3

COMMENTS

This sequence is unbounded, as a simple consequence of the Chinese remainder theorem. - Thomas Ordowski, Jul 22 2015

Conjecture: lim sup_{n->oo} a(n)/log(A005117(n)) = 1/2. - Thomas Ordowski, Jul 23 2015

a(n) = 1 infinitely often since the density of the squarefree numbers, 6/Pi^2, is greater than 1/2. In particular, at least 2 - Pi^2/6 = 35.5...% of the terms are 1. - Charles R Greathouse IV, Jul 23 2015

From Amiram Eldar, Mar 09 2021: (Start)

The asymptotic density of the occurrences of 1 in this sequence is density(A007674)/density(A005117) = A065474/A059956 = 0.530711... (A065469).

The asymptotic density of the occurrences of 2 in this sequence is (density(A069977)-density(A007675))/density(A005117) = (A065474-A206256)/A059956 = 0.324294... (End)

LINKS

Reinhard Zumkeller, Table of n, a(n) for n = 1..10000

FORMULA

Asymptotic mean: lim_{n->oo} (1/n) Sum_{k=1..n} a(k) = Pi^2/6 (A013661). - Amiram Eldar, Oct 21 2020

EXAMPLE

As 24 = 3*2^3 and 25 = 5^2, the next squarefree number greater A005117(16) = 23 is A005117(17) = 26, therefore a(16) = 26-23 = 3.

MAPLE

A076259 := proc(n) A005117(n+1)-A005117(n) ; end proc: # R. J. Mathar, Jan 09 2013

MATHEMATICA

Select[Range[200], SquareFreeQ] // Differences (* Jean-François Alcover, Mar 10 2019 *)

PROG

(Haskell)