A318291 a(n) is the minimum k > 0 such that n*2^k - 3 is prime, or 0 if no such k exists.

3, 2, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 2, 0, 1, 1, 0, 2, 1, 0, 1, 1, 0, 1, 2, 0, 1, 2, 0, 1, 1, 0, 3, 1, 0, 1, 1, 0, 2, 1, 0, 1, 2, 0, 1, 3, 0, 2, 1, 0, 1, 1, 0, 1, 1, 0, 1, 2, 0, 2, 7, 0, 3, 1, 0, 1, 2, 0, 1, 1, 0, 5, 2, 0, 1, 1, 0, 2, 1, 0, 3, 1, 0, 1, 4, 0

Offset: 1

Author

Martin Michael Musatov, Aug 23 2018

Keywords

• nonn

Comments

Question: Other than multiples of 3, do there exist any numbers n > 3 such that a(n) = 0?

From Robert Israel, Aug 24 2018: (Start)

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). (End)

For the record high values in this sequence, see A316493; for the indices at which those values occur, see A318561. - Jon E. Schoenfield, Aug 26 2018

Conjecture: For every odd prime p, there exist infinitely many numbers j that are non-multiples of p and have the property that j*2^k - p is composite for every k > 0. - Martin Michael Musatov, Sep 04 2018

Links

• Robert Israel, Table of n, a(n) for n = 1..10000

Crossrefs

Cf. A000040, A050412, A078680.

Programs

Maple

f:= proc(n) local k;
  if n mod 3 = 0 then return 0 fi;
  for k from 1 do if isprime(n*2^k-3) then return k fi od
end proc:
f(3):= 1:
map(f, [$1..100]); # Robert Israel, Sep 03 2018

Mathematica

Array[If[And[Mod[#, 3] == 0, # > 3], 0, Block[{k = 1}, While[! PrimeQ[# 2^k - 3], k++]; k]] &, 105] (* Michael De Vlieger, Sep 04 2018 *)

PARI

a(n)={my(k=0); if(n%3||n==3, k++; while(!isprime((n<Andrew Howroyd, Aug 24 2018

Extensions

a(3) corrected and a(19)-a(87) from Andrew Howroyd, Aug 25 2018

a(47), a(62), and a(86) corrected by Jon E. Schoenfield, Aug 29 2018