A112530: p(n)+/ n and p(n)+/-2n are all primes.

[David Wilson]

Let P(n) = (p(n)-2n, p(n)-n, p(n), p(n)+n, p(n)+2n), where p(n) is the nth 
prime.

Lemma:  If P(n) includes only primes, then all elements of P(n) exceed 5.

Proof: Consider the sequence q, indexed starting at 1, which includes, in 
ascending order, 2, 3, 5, and all larger numbers not divisible by 2, 3, or 
5.  We can show that for n >= 4, q(n+8)-q(n) = 30; this fact can be used to 
establish that q(n)-2n > 5 for n >= 10.  Also, every prime number is in q, 
so p(n) >= q(n), and we get p(n)-2n > 5 for n >= 10.  For every n < 10, P(n) 
includes a nonprime, hence n >= 10, and we conclude p(n)-2n > 5.  Thus the 
smallest element of P(n) exceeds 5, hence so do all elements, QED.

Theorem:  If P(n) includes only primes, then 30 divides n.

Let p be a prime <= 5, and suppose that p does not divide n.  This means 
that n is coprime to p, and since p <= 5, the five elements of P(n) will 
form a complete residue class modulo p.  This means that some element e of 
P(n) is divisible by prime p.  But since P(n) includes only primes, e is 
prime, and we get e = p, so e <= 5.  But by our lemma, all elements of P(n), 
including e, exceed 5, a contradiction. Therefore, p must divide n.

That is, if p is a prime <= 5, p divides n.  Thus 2, 3, and 5 divide n, so 
that 30 divides n, QED.

----- Original Message ----- 
From: "zak seidov" <zakseidov at yahoo.com>
To: <seqfan at ext.jussieu.fr>
Sent: Saturday, September 10, 2005 9:15 AM
Subject: A112530: p(n)+/ n and p(n)+/-2n are all primes.


> Just submitted A112530:
> Numbers n such that prime(n) +/- n and prime(n) +/- 2n
> are all primes
> 720,1920,5580,14370,17160,21090,26040,28560,...
> Why are all n's  divisible by 30?
> Thanks, Zak
>
>
>
>
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