OFFSET
1,1
COMMENTS
Also, numbers m such that the proportion of squarefree numbers in the interval [1, m] is less than the corresponding proportion for all k > m.
If the condition "greater than" were changed to "greater than or equal to" the sequence would also include the number 396, where the proportion is the same as at 1089, namely, 156/396 = 429/1089 = 39/99. There seems to be no other such coincidence. There is none up to 2*10^11.
All the terms are congruent to 0 or 1 modulo 4. If the modulus 36 is considered, the only possible residue classes are 0, 1, 9, 12, 20, 28, 29, 32 and 33. Similar restrictions hold for larger moduli. Thus, mod 900 there are only 132 possible residues, the least one being 28. Of these, more than half appear in pairs of two consecutive values.
LINKS
Javier Múgica, Table of n, a(n) for n = 1..211
Javier Múgica, C program for finding (not proving) the terms of this sequence. The computations for the proofs can be found in the following spreadsheets: The first four terms. The first 211 terms. The determination of the possible residue classes mod 36 and 900.
EXAMPLE
There are 151 nonsquarefree numbers up to m = 380, for a proportion of 151/380 ~= 0.39737. This proportion is never again reached for larger values of m, so the number 380 belongs to this list.
CROSSREFS
KEYWORD
nonn
AUTHOR
Javier Múgica, Jul 05 2020
STATUS
approved