Problems & Puzzles:
Puzzles
Puzzle 150.
Primes type n+/- Fm, for m=0 to 4
As a follow up of our invitation in the Puzzle
149 to suggest new puzzles about the Fermat' issues Felice Russo proposes the following
one:
"Carlos by reading your puzzle 149 and
some pages of fantastic Guy's book "Unsolved problmes in
Number Theory" my attention has been driven by following problem:
1) Find the smallest n such that (n+3, n+5, n+17, n+257, n+65537)
are all primes.
With a simple ubasic code I have found that 14 is the smallest of
such numbers. Then I calculated for N<10^9 the counting function p(N)
for those numbers and seems that it follow the law: p(N)~c*N^(gamma)
where c is a positive constant and gamma the Euler-Mascheroni constant.
If this law is true for all N then the sequence of numbers n is infinite
(see EIS A063799).
2) Find the smallest n such that (n-3, n-5, n-17, n-257, n-65537)
are all primes. As above I find that the smallest of such numbers is
65704 and again the counting function p(N) seems to grow like p(N)~c*N^(4/3*gamma).
If so then the sequence of n is infinite (see EIS A063825).
3) Find the smallest n such that ([n+3, n-3], [n+5, n-5],
[n+17,n-17], [n+257,n-257], [n+65537, n-65537]) are all primes. Up to
10^9 no such n has been found.
Then the following questions can be posed:
1. Is there any number n such that the
question 3) is satisfied? If yes is the sequence of those numbers
infinite?
2. Are the counting functions found for
the case 1 and 2 reasonable? If not which is a better approximation for
p(N)?"
Solution:
Jean Brette immediately and the first saw the
impossibility to find any n>65537 such that all the following [n+3, n-3], [n+5, n-5],
[n+17,n-17], [n+257,n-257], [n+65537, n-65537] are primes. Why? Because
it's impossible that the following four [n+3, n-3], [n+5, n-5] are primes.
Why?
In his own Brette's words:
Since (n-3) must be prime (n-3) = 1 or 2 (mod 3)
a) if (n-3) = 2 (mod 3) then (n-5) = 2-2 = 0 (mod 3),
so (n-5) is not prime (except if n=8)
b) if (n-3)=1 (mod 3) then (n+5) = 1+8 = 0 (mod 3) so
(n+5) is not prime.
***
Of course that the situation is reciprocal between the
asked prime (n+3) and the asked primes (n-5) and (n+5). Then the most of
primes in these 10 numbers is 8.
But I have a more intuitive way of arriving to the same
conclusion than Brette:
As we know all primes are in the strip 6k-1 or 6k-/+1. How close may be two pair of
twins (as twins are n-5 & n-3 and n+3 & n+5)? Then
- observing the strip situation - the most close they can be is such that the lower prime of the upper twin is 4
units under the upper prime of the lower twin.
Well the pair of twins (n-5, n-3) and (n+3, n+5) are at 6
units of distance. Then necessarily at least one of the extremes numbers (or n-5 or n+5) can not
be prime.
***
Here are some more ideas from Jean Brette sent the
25/9/01:
Question 1,
Some days after, using a similar argument with congruences, Jean
shows that the best which can be hoped would to have all the N+/- Fk
primes, excepted N+/-Fo, which was confirmed by Carlos Rivera .
Here are the ten first such N's, from Carlos : 234786 , 247596,
646164, 1423974 , 2269344 , 2323566 , 3268224 , 3837996 , 5217864 ,
5792424
Question 2.
In the same time (08/28/01), Jean suggested that the laws, for the
two counting functions computed by Felice Russo, cannot be
"power laws", which seemed to him an "artifact".
During the last month, he gave some heuristic arguments to support
his claim and, with the help of Felice Russo for extended tables
for the two counting functions, he has gotten the following results :
1) For x large enough, the two counting functions must be almost
equal. The relative difference between them, for small x (x<1000000)
are due to the "initial gap" : when you are counting the N's, in
the first case (N+Fk) , N begins at zero. In the second case (N-Fk) , the
first N must be > 65537.
Here are some values of the two counting functions, from Felice :
x..........................P(x) for (N+Fk) ........P(x) for (N-Fk)
100000.........................54...........................18
1000000.....................210.........................191
10000000...................915.........................908
100000000...............4478.......................4409
2) With his heuristic argument, the law expected by Jean, for the
Felice's counting functions, can be written : P(x) = C * x / (ln x)^5 ,
where C is a "varying constant" which decreases when x grows : C
= C1 + C2/ln x, which can be readen as a "truly constant" C1
(for x infinite) and an "error term" C2/ln x, which decreases,
and vanishes for x infinte.
With the datas from Felice, he finds (08/21/01) that C1 = 63 and C2
= 589 give the rather sharp approximations (for N+Fk) :
x.............................P(x)......................P'(x) from
JB
100000..................54........................56,43
1000000..............210......................209,87
10000000............915......................915,04
100000000........4478....................4477,96
Note :
1) the Jean 's law has a form very similar to the prime law, or the
conjectured counting function for twin primes (Hardy-Littlewood),
or some other conjectured counting functions. It is very natural because,
as Felice said : The probability that five random numbers of same
magnitude x are all primes is , roughly, 1/(ln x)^5.
2) So, may be, the main term x / ln(x)^5 may be refined and replaced
by : integral, from 2 to x, of dx / (ln x)^5; which is known to be sharper
in many other cases (and of course with two other constants C'1 and C'2.)
3) Anyway, a new question is now : Is the constant C1 (at infinity)
really equal (or very close) to 63? If yes, why ? , i.e. is there a
theoretical meaning , or / and a (relatively ) closed formula for C1 and
C2 ? Of course, C1 is more important than C2 since C2 appears in an
"error term" which vanishes
***
|
