[seqfan] Re: [a(n)+a(n1)+a(n+2)] is prime BUT [a(n)+a(n+2]) is not

[israel at math.ubc.ca]

Given positive integers a,b, there will always be infinitely many positive 
integers c such that a+b+c is prime but a+c is composite. In fact, let p,q 
be any distinct primes coprime to a+b and b respectively. Let d == - a 
p^(-1) mod q, and consider the numbers f(k) = d p + k p q for k >= 0. Since 
a + b + d p is coprime to p q, Dirichlet's theorem says a + b + f(k) is 
prime for infinitely many k. On the other hand, a + f(k) is divisible by q 
for all k.

Cheers,
Robert

On Jan 29 2016, Lars Blomberg wrote:

>Hello,
>
> I get 1, 2, 8, 7, 4, 18, 21, 14, 6, 11, 12, 24, 23, 20, 10, 13, 30, 36, 
> 5, 26, 16, 19, 32, 38, 3, 42, 22, 9, 28, 46, 27, 34, 48, 15, 40, 54, 37, 
> 58, 44, 29, 64, 56, 17, 66, 68, 33, 50, 84, 45, 52, 60, 25, 72, 70, 39, 
> 82, 76, 35, 62, 94, 43, 74, 80, 69, 78, 86, 47, ...
>
>And the sequence extends to 10^5 terms without problem.
>
>/Lars
>
>-----Ursprungligt meddelande-----
>Från: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] För Eric Angelini
>Skickat: den 29 januari 2016 19:47
>Till: Sequence Fanatics Discussion list <seqfan at list.seqfan.eu>
>Ämne: [seqfan] [a(n)+a(n1)+a(n+2)] is prime BUT [a(n)+a(n+2]) is not
>
>Hello SeqFans,
>Does this seq S extend smoothly without backtracking?
>
> Start S with a(1)=1 and extend S with the smallest unused integer such 
> that [a(n)+a(n1)+a(n+2)] is prime BUT [a(n)+a(n+2]) is not.
>
>  
> S=1,2,8,7,4,18,21,14,12,11,6,24,29,20,10,13,30,36,5,26,16,19,32,38,3,42,22,...
>
>I guess S is the lexico-first permutation of the integers >0.
>Best,
>É.
>
>
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