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Alex
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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
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