Henry has already argued that if p and q are consecutive primes then
(p+q)%2 can not be prime.  I just want to say that reasoning of the sort:

While it might be possible for the larger primes, I'm thinking not - just
by induction.

is unreliable, at best.  That is, it is unreliable to come a conclusion by
trying a few cases, or a few billion cases, and think that a property is
true in general. http://en.wikipedia.org/wiki/Prime_number_theorem tells of
one such instance:

This suggests that Li(*x*) should usually be larger than π(*x*) by roughly
Li(*x*1/2)/2, and in particular should usually be larger than π(*x*).
However, in 1914, J. E. Littlewood proved that this is not always the case.
The first value of *x* where π(*x*) exceeds Li(*x*) is probably around *x* =
10316; see the article on Skewes' number for more details.




On Sun, May 12, 2013 at 3:07 AM, Alan Stebbens <[email protected]> wrote:

> ProgrammingPraxis (at http://programmingpraxis.com/2013/05/10/mindcipher)
> offered a problem asking, given p, q as two consecutive pairs of primes, if
> (p+2)%2 could be prime.
>
> Since both p & q (> 2) are prime, their sum is an even number and not
> prime, but could the half of their sum be a prime?
>
> I'm not much of a mathematician, but I figured I could brute-force an
> approximation with J.
>
> The gist below is my experiment showing that the answer is no, for the
> consecutive pairs of primes in the set of the first million primes.  While
> it might be possible for the larger primes, I'm thinking not - just by
> induction.
>
> Probably some of you could show a proof, but it was more fun for me to
> cobble up this in J, and also demonstrate J to the non-J audience at
> Programming Praxis (which has been mostly scheme, Haskell, python, ruby).
>
> https://gist.github.com/aks/5563008
>
> Here's my experiment:
>
>    i. 20
> 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19
>
>    NB. generate the first 20 primes
>    p: i. 20
> 2 3 5 7 11 13 17 19 23 29 31 37 41 43 47 53 59 61 67 71
>
>    NB. box up consecutive pairs of those primes
>    (2 <\ ]) p: i. 20
>
> ┌───┬───┬───┬────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┬─────┐
> │2 3│3 5│5 7│7 11│11 13│13 17│17 19│19 23│23 29│29 31│31 37│37 41│41 43│43
> 47│47 53│53 59│59 61│61 67│67 71│
>
> └───┴───┴───┴────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┴─────┘
>
>    NB. sum up each pair of primes
>    +/ each (2 <\ ])p: i. 20
> ┌─┬─┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬───┬───┬───┬───┬───┐
> │5│8│12│18│24│30│36│42│52│60│68│78│84│90│100│112│120│128│138│
> └─┴─┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴───┴───┴───┴───┴───┘
>
>    NB. divide each sum by 2
>    2 %~ each +/ each (2 <\ ])p: i.
> 20
>
> ┌───┬─┬─┬─┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┬──┐
> │2.5│4│6│9│12│15│18│21│26│30│34│39│42│45│50│56│60│64│69│
> └───┴─┴─┴─┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┴──┘
>
>    NB. now, test each of those results for being prime.  1 p: y -- tests y
> for being prime
>
>    1&p: each 2 %~ each +/ each (2 <\ ])p: i.
> 20
>
> ┌─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┬─┐
> │0│0│0│0│0│0│0│0│0│0│0│0│0│0│0│0│0│0│0│
> └─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┴─┘
>
>    NB. open the boxed results, so we can add them up
>    >1&p: each 2 %~ each +/ each (2 <\ ])p: i. 20
> 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
>
>    NB. sum/reduce the vector of booleans.  If there's a prime, the sum will
> be > 0
>    +/>1&p: each 2 %~ each +/ each (2 <\ ])p: i. 20
> 0
>
>    NB. ok. No primes.  Let's keep checking for larger groups
>
>    +/>1&p: each 2 %~ each +/ each (2 <\ ])p: i. 1000
> 0
>    +/>1&p: each 2 %~ each +/ each (2 <\ ])p: i. 10000
> 0
>    +/>1&p: each 2 %~ each +/ each (2 <\ ])p: i. 100000
> 0
>
>    NB. the previous output took a few seconds.  The next will take a few
> minutes
>
>    +/>1&p: each 2 %~ each +/ each (2 <\ ])p: i. 1000000
> 0
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