OFFSET
0,1
COMMENTS
The length of a run of consecutive squarefree positive integers is limited to 3 because no multiple of 4 is squarefree; thus, a(0) <= 3.
But suppose we consider runs of consecutive positive integers that are either squarefree or divisible by 4. No such set of 18 numbers can exist, since it would include two multiples of 3^2 = 9, at most one of which would be divisible by 4, so a(1) <= 17.
No set of 75 consecutive positive integers can have every member squarefree or divisible by 4 or 9, since any such set would include three multiples of 5^2 = 25, at most two of which would be divisible by 4 or 9, so a(2) <= 74.
By a similar argument, a(3) <= 4*7^2 - 1 = 195.
However, it does not follow in general that a(n) <= (n+1)*prime(n+1)^2 - 1; e.g., a(4) = 675 > 605 = 5*11^2, and the five multiples of 121 in the interval [2070582452238, 2070582452912], i.e., 121*{17112251672..17112251676}, are also divisible by 4, 9, 49, 25, and 4, respectively.
For n = 0..4, the smallest positive k that begins a run of a(n) consecutive integers of which each is squarefree or divisible by the square of at least one of the first n primes is 1, 28, 28976, 64779, and 2070582452238, respectively.
EXAMPLE
1, 2, and 3 are all squarefree, and no four consecutive positive integers can be squarefree, so a(0) = 3.
Each of the 17 integers in {28, ..., 44} is either squarefree or divisible by 2^2 = 4, and no interval of more than 17 consecutive positive integers has this property, so a(1) = 17.
Each of the 74 consecutive positive integers in {28976, ..., 29049} is squarefree or divisible by 4 or 9, and no larger set has this property, so a(2) = 74.
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Jon E. Schoenfield, Jul 30 2017
EXTENSIONS
a(5)-a(7) from Jinyuan Wang, Jun 24 2022
STATUS
approved