Generating a specific list of intsgers

[Musatov]

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).