On Friday, August 24, 2018 at 10:59:07 PM UTC-5, Steven D'Aprano wrote:
> On Fri, 24 Aug 2018 14:40:00 -0700, tomusatov wrote:
>
> > I am looking for a program able to output a set of integers meeting the
> > following requirement:
> >
> > a(n) is the minimum k > 0 such that n*2^k - 3 is prime, or 0 if no such
> > k exists
> >
> > Could anyone get me started? (I am an amateur)
>
>
> That's more a maths question than a programming question. Find out how to
> tackle it mathematically, and then we can code it.
>
>
>
> --
> Steven D'Aprano
> "Ever since I learned about confirmation bias, I've been seeing
> it everywhere." -- Jon Ronson
Steven, let me know if this will suffice for the maths:
a(n) is the minimum k > 0 such that n*2^k - 3 is prime, or 0 if no such k exists
Other than multiples of 3, do there exist any numbers n > 3 such that a(n) = 0?
The answer is yes. The situation is similar to that of Riesel or Sierpinski
numbers.
Every integer k is in at least one of the following residue classes:
2 (mod 3)
1 (mod 4)
4 (mod 5)
3 (mod 8)
4 (mod 9)
8 (mod 10)
6 (mod 12)
10 (mod 15)
7 (mod 16)
16 (mod 18)
12 (mod 20)
12 (mod 24)
16 (mod 25)
1 (mod 25)
0 (mod 30)
10 (mod 36)
27 (mod 36)
16 (mod 40)
1 (mod 45)
33 (mod 45)
15 (mod 48)
31 (mod 48)
where 3,4,5,...,48 are the multiplicative orders of 2 modulo the primes 7, 5,
31, 17, 73, 11, 13, 151, 257, 19, 41, 241, 1801, 601, 331, 109, 37, 61681,
23311, 631, 673, 97 respectively.
Now 7 | n*2^k-3 for k == 2 (mod 3) if n == 6 (mod 7),
5 | n*2^k-3 for k == 1 (mod 4) if n == 4 (mod 5), ...,
97 | n*2^k-3 for k == 31 (mod 48) if n == 75 (mod 97).
Using the Chinese remainder theorem, we get infinitely many n for which all
these congruences hold, and thus for which n*2^k-3 is always divisible by at
least one of those 22 primes.
One such n is 72726958979572419805016319140106929109473069209 (which is not
divisible by 3).
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