[seqfan] Re: Christmas puzzle, reformulated :)
franktaw at netscape.net
franktaw at netscape.net
Fri Dec 25 19:12:50 CET 2009
See A002374, A078587. You are requiring that the prime be at least 2
away from n, which is not in the database; but this seems like an
arbitrary condition.
That the values for the smaller prime remain positive is in the
category of "minor extensions to Goldbach's conjecture"; it is almost
certainly true but not proved. That there is always a solution follows
from the conjecture that every even number is the difference of two
primes [infinitely many ways], which with the infinitely many ways
condition is a minor extension to the twin prime conjecture; even
without it this remains unproved as far as I know; but this falls into
the "there's no way this can be false" category.
Franklin T. Adams-Watters
-----Original Message-----
From: Creighton Kenneth Dement
<creighton.k.dement at mail.uni-oldenburg.de>
Note this is a continuation of the thought behind A082467!
I tried to "reformulate" the question a bit:
Given an n, find m such that
if n is odd, n + 2*m and n - 2*m are both prime
if n is even, n + 2*m+1 and n - (2*m+1) are both prime
where in this case I consider n - (2*m+1) to be prime if
|n - (2*m+1)| is prime.
For odd n:
n: 1, m: 2, (n+2m): 5, (n-2m): -3
n: 3, m: 4, (n+2m): 11, (n-2m): -5
n: 5, m: 1, (n+2m): 7, (n-2m): 3
n: 7, m: 2, (n+2m): 11, (n-2m): 3
n: 9, m: 1, (n+2m): 11, (n-2m): 7
n: 11, m: 3, (n+2m): 17, (n-2m): 5
n: 13, m: 3, (n+2m): 19, (n-2m): 7
n: 15, m: 1, (n+2m): 17, (n-2m): 13
n: 17, m: 3, (n+2m): 23, (n-2m): 11
n: 19, m: 6, (n+2m): 31, (n-2m): 7
n: 21, m: 1, (n+2m): 23, (n-2m): 19
n: 23, m: 3, (n+2m): 29, (n-2m): 17
n: 25, m: 3, (n+2m): 31, (n-2m): 19
n: 27, m: 2, (n+2m): 31, (n-2m): 23
n: 29, m: 6, (n+2m): 41, (n-2m): 17
n: 31, m: 6, (n+2m): 43, (n-2m): 19
n: 33, m: 2, (n+2m): 37, (n-2m): 29
n: 35, m: 3, (n+2m): 41, (n-2m): 29
n: 37, m: 3, (n+2m): 43, (n-2m): 31
n: 39, m: 1, (n+2m): 41, (n-2m): 37
For even n:
n: 2, m: 2, (n+2m+1): 7, (n-2m-1): -3
n: 4, m: 3, (n+2m+1): 11, (n-2m-1): -3
n: 6, m: 5, (n+2m+1): 17, (n-2m-1): -5
n: 8, m: 1, (n+2m+1): 11, (n-2m-1): 5
n: 10, m: 1, (n+2m+1): 13, (n-2m-1): 7
n: 12, m: 2, (n+2m+1): 17, (n-2m-1): 7
n: 14, m: 1, (n+2m+1): 17, (n-2m-1): 11
n: 16, m: 1, (n+2m+1): 19, (n-2m-1): 13
n: 18, m: 2, (n+2m+1): 23, (n-2m-1): 13
n: 20, m: 1, (n+2m+1): 23, (n-2m-1): 17
n: 22, m: 4, (n+2m+1): 31, (n-2m-1): 13
n: 24, m: 2, (n+2m+1): 29, (n-2m-1): 19
n: 26, m: 1, (n+2m+1): 29, (n-2m-1): 23
n: 28, m: 4, (n+2m+1): 37, (n-2m-1): 19
n: 30, m: 3, (n+2m+1): 37, (n-2m-1): 23
n: 32, m: 4, (n+2m+1): 41, (n-2m-1): 23
n: 34, m: 1, (n+2m+1): 37, (n-2m-1): 31
n: 36, m: 2, (n+2m+1): 41, (n-2m-1): 31
I don't see the sequence of m's for odd or even n
or the combining both: (2, 2, 4, 3, 1, 5, 2, 1, 1,...) in the OEIS.
My "reformulated Christmas question" is:
In the "for even n" case, we see that (n-2m) jumps from being negative
to
positive- does it ever go back to being negative?
Similarly, (n-2m-1) in the odd case also jumps from negative to positive
and apparently never again.
Of course, there is also the looming question of whether an m always
exists, which is perhaps the most difficult question of all.
Happy New Year (whatever calendar you may be using),
Creighton
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