

A264097


Smallest odd number k divisible by 3 such that k*2^n1 is a prime.


3



3, 3, 3, 3, 3, 15, 3, 3, 27, 45, 15, 3, 87, 9, 15, 9, 45, 15, 3, 51, 57, 9, 33, 69, 39, 57, 57, 21, 27, 45, 213, 15, 57, 147, 3, 33, 45, 21, 3, 63, 117, 15, 33, 3, 57, 165, 33, 213, 117, 69, 87, 21, 183, 147, 45, 3, 33, 51, 111, 45, 93, 69, 57, 9, 3, 99, 63
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OFFSET

0,1


COMMENTS

As N increases, (Sum_{n=1..N} a(n))/(Sum_{n=1..N} n) appears to tend to 2*log(2), as can be seen by plotting the first 31000 terms.
This observation is consistent with the prime number theorem as the probability that k*2^n1 is prime where k is a multiple of 3 is 1/(2*(n*log(2)+log(k))) ~ 1/(2*n*log(2)).


LINKS



EXAMPLE

3*2^01=2 prime so a(0)=3.
3*2^11=5 prime so a(1)=3.
3*2^21=11 prime so a(2)=3.
3*2^31=23 prime so a(3)=3.


MATHEMATICA

Table[k = 3; While[! PrimeQ[k 2^n  1], k += 6]; k, {n, 0, 68}] (* Michael De Vlieger, Nov 03 2015 *)


PROG

(PFGW & SCRIPT)
Command: pfgw64 f e500000 in.txt
in.txt SCRIPT FILE:
SCRIPT
DIM k
DIM n, 0
DIMS t
OPENFILEOUT myf, a(n).txt
LABEL loop1
SET n, n+1
SET k, 3
LABEL loop2
SET k, k+6
SETS t, %d, %d\,; n; k
PRP k*2^n1, t
IF ISPRP THEN WRITE myf, t
IF ISPRP THEN GOTO loop1
GOTO loop2
(PARI) a(n) = {k = 3; while (!isprime(k*2^n1), k += 6); k; } \\ Michel Marcus, Nov 03 2015


CROSSREFS



KEYWORD

nonn


AUTHOR



STATUS

approved



