

A285808


a(n) = smallest k such that (6*k3)*2^n1 is prime.


4



1, 1, 1, 1, 3, 1, 1, 5, 8, 3, 1, 15, 2, 3, 2, 8, 3, 1, 9, 10, 2, 6, 12, 7, 10, 10, 4, 5, 8, 36, 3, 10, 25, 1, 6, 8, 4, 1, 11, 20, 3, 6, 1, 10, 28, 6, 36, 20, 12, 15, 4, 31, 25, 8, 1, 6, 9, 19, 8, 16, 12, 10, 2, 1, 17, 11, 19, 11, 9, 5, 21, 22, 3, 4, 13, 1
(list;
graph;
refs;
listen;
history;
text;
internal format)



OFFSET

1,5


COMMENTS

As N increases, (Sum_{n=1..N} a(n)) / (Sum_{n=1..N} n) tends to log(2)/3 as seen by plotting data; this is consistent with the prime number theorem as the probability that (6*x3)*2^n  1 is prime is ~ 3/(n*log(2)) if n is great enough, so after n*log(2)/3 try n*log(2)/3*(3/n*log(2))=1.
For n=1 to 14000, a(n)/n is always < 3.


LINKS



MATHEMATICA

Table[k = 1; While[! PrimeQ[(6 k  3)*2^n  1], k++]; k, {n, 76}] (* Michael De Vlieger, Apr 27 2017 *)


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



