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%I #74 Nov 14 2018 14:27:05
%S 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,
%T 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,
%U 0,1,1,0,5,2,0,1,1,0,2,1,0,3,1,0,1,4,0
%N a(n) is the minimum k > 0 such that n*2^k - 3 is prime, or 0 if no such k exists.
%C Question: Other than multiples of 3, do there exist any numbers n > 3 such that a(n) = 0?
%C From _Robert Israel_, Aug 24 2018: (Start)
%C The answer is yes. The situation is similar to that of Riesel or Sierpinski numbers.
%C Every integer k is in at least one of the following residue classes:
%C 2 (mod 3)
%C 1 (mod 4)
%C 4 (mod 5)
%C 3 (mod 8)
%C 4 (mod 9)
%C 8 (mod 10)
%C 6 (mod 12)
%C 10 (mod 15)
%C 7 (mod 16)
%C 16 (mod 18)
%C 12 (mod 20)
%C 12 (mod 24)
%C 16 (mod 25)
%C 1 (mod 25)
%C 0 (mod 30)
%C 10 (mod 36)
%C 27 (mod 36)
%C 16 (mod 40)
%C 1 (mod 45)
%C 33 (mod 45)
%C 15 (mod 48)
%C 31 (mod 48)
%C 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.
%C Now 7 | n*2^k-3 for k == 2 (mod 3) if n == 6 (mod 7),
%C 5 | n*2^k-3 for k == 1 (mod 4) if n == 4 (mod 5), ...,
%C 97 | n*2^k-3 for k == 31 (mod 48) if n == 75 (mod 97).
%C 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.
%C One such n is 72726958979572419805016319140106929109473069209 (which is not divisible by 3). (End)
%C 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
%C 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
%H Robert Israel, <a href="/A318291/b318291.txt">Table of n, a(n) for n = 1..10000</a>
%p f:= proc(n) local k;
%p if n mod 3 = 0 then return 0 fi;
%p for k from 1 do if isprime(n*2^k-3) then return k fi od
%p end proc:
%p f(3):= 1:
%p map(f, [$1..100]); # _Robert Israel_, Sep 03 2018
%t 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 *)
%o (PARI) a(n)={my(k=0); if(n%3||n==3, k++; while(!isprime((n<<k)-3), k++)); k} \\ _Andrew Howroyd_, Aug 24 2018
%Y Cf. A000040, A050412, A078680.
%K nonn
%O 1,1
%A _Martin Michael Musatov_, Aug 23 2018
%E a(3) corrected and a(19)-a(87) from _Andrew Howroyd_, Aug 25 2018
%E a(47), a(62), and a(86) corrected by _Jon E. Schoenfield_, Aug 29 2018