OFFSET
1,1
COMMENTS
Earls (2005) noted that if m > 2 is a solution to abs(K(m+1) - K(m)) = 1 (A346211) then either m or m+1 is nonsquarefree (A013929), and asked whether there are any solutions with both m and m+1 being squarefree.
There are no such solutions below 10^9.
Since there are also no solutions to K(m) = K(m+1) below 10^9, it can be conjectured that the minimal difference abs(K(m+1) - K(m)) between consecutive numbers m and m+1 that are both squarefree is 2.
a(1)-a(40) were calculated by Earls (2005).
LINKS
Amiram Eldar, Table of n, a(n) for n = 1..1000
Jason Earls, On consecutive values of the Smarandache function, Scientia Magna, Vol. 1, No. 1 (2005), pp. 129-130.
Wikipedia, Kempner function.
EXAMPLE
5 is a term since abs(K(6) - K(5)) = abs(3 - 5) = 2, and both 5 and 6 are squarefree.
MATHEMATICA
kempner[n_] := Module[{m = 1}, While[! Divisible[m!, n], m++]; m]; p = Position[Abs @ Differences @ Array[kempner, 500], 2] // Flatten; Select[p, SquareFreeQ[#] && SquareFreeQ[# + 1] &]
CROSSREFS
KEYWORD
nonn
AUTHOR
Amiram Eldar, Jul 10 2021
STATUS
approved