# Do Sierpiński numbers of Izotov type have a covering set?

Sierpiński number is an odd integer $$k$$ such that $$2^nk+1$$ is composite for all $$n\in{\mathbb N}$$. In the paper Sur un probleme concernant les nombres $$k\cdot2^n+1$$, zbl:0093.04602 (1960), Sierpiński proved that there are infinitely many $$k$$ with this property. All integers $$2^nk+1$$ constructed by Sierpiński are divisible by at least one of the primes in an explicit small set (a covering set).

Then, in A note on Sierpinski numbers, zbl:0849.11016 (1995), Izotov has given infinitely many Sierpiński numbers of a new type $$(*)$$:

• for $$n=4m+2$$, the number $$2^nk+1$$ has an easy algebraic factorization;

• for all other $$n$$, the number $$2^nk+1$$ is divisible by at least one of the primes $$\{3,17,257,641,65537,6700417\}$$ (a covering set).

(The author notes that the above covering set does not work when $$n=4m+2$$.)

Two decades later, in the paper Sierpiński and Carmichael numbers, zbl:1325.11010 (2015), Banks et al. write:

Every currently known Sierpiński number $$k$$ possesses at least one covering set $$\cal P$$, which is a finite set of prime numbers with the property that $$2^nk+1$$ is divisible by some prime in $$\cal P$$ for every $$n\in{\mathbb N}$$.

So, for Sierpiński numbers of the Izotov type $$(*)$$, was a bigger covering set found between 1995 and 2015?

• – Alex
May 31 '20 at 20:22
• Full reference: W.Banks, C.Finch, F.Luca, C.Pomerance, P.Stănică, Sierpiński and Carmichael numbers. Trans. Am. Math. Soc. 367, No. 1, 355-376 (2015).
– Alex
May 31 '20 at 21:07
• See math.stackexchange.com/questions/1683082/… where the question of Sierpinski without covering was discussed (but not settled). May 31 '20 at 23:29

So, for Sierpiński numbers of the Izotov type $$(*)$$, was a bigger covering set found between 1995 and 2015?
No, it was not. And it is conjectured that none exists. In the Math Stack Exchange thread linked by Gerry Myerson in a comment to the question, I give other examples of Sierpiński numbers for which it is not likely that they should possess such a covering set $$\cal P$$.