--- Mark Underwood <

mark.underwood@...> wrote:

> Note for below: negatives included as primes.

>

> Given positive a,b, a<b, we want b+a and b-a both prime. First

> solution: (a,b): (1,4)

>

> Given positive a,b,c, a<b<c, we want c+b+a, c+b-a, c-b+a,-c+b+a all

> prime. First solution: (a,b,c) :(1,4,8)

>

> Given positive a,b,c,d, a<b<c<d we want d+c+b+a, d+c+b-a, d+c-b+a, d-

> c+b+a, -d+c+b+a, d+c-b-a, d-c+b-a,-d+c+b-a, d-c-b+a, -d+c-b+a, -d-c+b+a

> all prime. (11 combinations.) First solution: (a,b,c,d): (3,5,8,13)

> (No "fibbing"!)

-d+c+b-a and d-c-b+a are the same prime

-d+c-b+a and d-c+b-a are the same prime

-d-c+b+a and d+c-b-a are the same prime

> Given positive a,b,c,d,e, a<b<c<d<e, we want all 16 combinations to be

> prime. First solution: (a,b,c,d,e): (3,10,12,15,27)

>

> What about for 6 terms? That would be 1 + 6 + 15 + 20 = 42

> combinations. All prime? Eeek! If it is solved it won't be on my

> computer....

Just line up the sums/differences, so that many combinations evaluate to the

same thing.

I think your count of combinations is incorrect. There are 2^(n-1) ways to

chose the signs. See above examples of miscounting.

Phil

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