

A318291


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


3



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
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OFFSET

1,1


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^k3 for k == 2 (mod 3) if n == 6 (mod 7),
5  n*2^k3 for k == 1 (mod 4) if n == 4 (mod 5), ...,
97  n*2^k3 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^k3 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 nonmultiples 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


MAPLE

f:= proc(n) local k;
if n mod 3 = 0 then return 0 fi;
for k from 1 do if isprime(n*2^k3) 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 *)


PROG

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


CROSSREFS

Cf. A000040, A050412, A078680.
Sequence in context: A107889 A138384 A129172 * A251482 A172083 A337199
Adjacent sequences: A318288 A318289 A318290 * A318292 A318293 A318294


KEYWORD

nonn


AUTHOR

Martin Michael Musatov, Aug 23 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


STATUS

approved



