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Recently a new prime number has been discovered, which eliminates one of the six remaining candidates for the smallest Sierpinski numbers. So I was reading the wikipedia article about the Sierpinski number, where I came across what is called a covering set of primes for a Sierpinski number. Different Sierpinski numbers has different covering sets. I understood that the elements belongs to the covering set divides the Sierpinski number, associated with the covering set. But what a covering set do? How it helps in finding smallest Sierpinski number ? Can anyone guide me through this? Thanks.

Xander Henderson
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Kushal Bhuyan
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    As I understood it, no matter what $n$ is, $78,557\cdot 2^n + 1$ has at least one prime factor which is either $3, 5, 7, 13, 19, 37$ or $73$. Thus we say that these primes form a prime covering set for the Sierpinski number $78,557$. – Arthur Dec 04 '16 at 12:04

1 Answers1

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A covering set doesn't help in "finding smallest Sierpinski number".

It is merely used in order to show that a given $k\in\mathbb{N}$ is a Sierpinski number, as part of proving that the expression $k\cdot2^n+1$ is composite for every $n\in\mathbb{N}$ (becuse it is divisible by one of the values in the covering set).

In other words, the covering set is inferred during the process of proving that $k$ a Sierpinski number.

You could say that a covering set is part of the proof's output rather than input:

We take a $k$ and prove that it has a covering set, not vice-versa.

For the record, allow me to emphasize that I became familiar with these numbers only a few days ago while reading about this on the news, so the answer above is based solely on my understanding of the same Wikipedia article that you mention.

barak manos
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    Note that (if I read Wikipedia correctly) it is not known whether all Sierpinski numbers have a finite covering set. It's just that every Sierpinski number we've found so far happens to have one. – Arthur Dec 04 '16 at 12:10
  • @Arthur: Yes, I have added another statement which perhaps clarifies that the covering set belongs to the "output" of the proof rather than to the "input" of the proof. – barak manos Dec 04 '16 at 12:11
  • So how does one prove a number to be Sierpinski? I mean is there any theoretical approach or just numerical ones? – Kushal Bhuyan Dec 04 '16 at 12:15
  • @KushalBhuyan: As Arther has mentioned in the comment above, the covering-set method works for some values of $k$ but not necessarily for all of them. I guess that if there was an answer to your question, then the Sierpinski problem would have been declared solved by now. – barak manos Dec 04 '16 at 12:19
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    Some (rare) numbers have been proven Sierpiński with an algebraic factorization helping. For example take $k=4008735125781478102999926000625$. When the exponent $n$ is $2$ modulo $4$, we have an Aurifeuillean factorization. Then we do not need a full covering set, only a set accounting for the remaining exponents. For that we can use ${ 3,17,97,241,257,673}$ here. This trick was described in a paper by Anatoly S. Izotov. This $k$ is thought not to have a "full" covering set. See my comment to another answer for the explicit factorization. – Jeppe Stig Nielsen Apr 14 '18 at 12:32