[seqfan] Re: A conjecture for double primes

[Vladimir Shevelev]

Thank you, Hans!

It is clear that, if k is odd, then for prime p>=5 one of the numbers
p+/- k is divisible by 4, and only in case k=p-4 both p+/- k could be semiprimes.
But for large p, it is unlikely that p-4 will be the smallest value of k. Michel
have not found any odd term in the sequence for n>=12 and up to 5*10^4.
Most likely that, for n>=12, all terms are even.
Sorry, I wrong understood Michel's comment in discussion fields. 
A dual sequence is A241539: "Smallest k>=1 such that numbers Semiprime(n) + or - k are both primes, or a(n)=0 if there is no such k".  If it has not zeros, at least, for sufficiently large n, then one can believe that, for a pair of sufficiently large semiprimes {r,s} there is a number
k=k(r,s), such that either {r-k, s+k} or {r+k, s-k} is a pair of primes. Then, if to consider  a
representations 2*n = r+s with, say, min{r,s}>log(n), then we can reduce such a representation to  Goldbach representation 2*n = p+q with primes p,q. Probably, Goldbach-like conjecture with semiprimes could be proven easier (cf. best known great Chen's result).

Best regards,
Vladimir



________________________________________
From: SeqFan [seqfan-bounces at list.seqfan.eu] on behalf of Hans Havermann [gladhobo at teksavvy.com]
Sent: 26 April 2014 18:40
To: Sequence Fanatics Discussion list
Subject: [seqfan] Re: A conjecture for double primes

A241536(78) = 6. A241536(79) = 6. It's likely that there are only three zeros: at indices 1, 2, and 5. I believe what Michel noted was that the number of *odd* terms was limited to indices 3, 4, 6, 8, and 11. This is a consequence (I think) of odd k being only ever the prime minus four and there is an increasing likelihood as we move along the sequence to infinity that another (even) k will be encountered before we reach it.


On Apr 26, 2014, at 6:59 AM, Vladimir Shevelev <shevelev at bgu.ac.il> wrote:

> Yesterday I have submitted a simple sequence A241536: "Smallest k>=1 such that numbers Prime(n) + or - k are both semiprimes, or a(n)=0 if there is no such k". The  behavior of the fist terms did not seem to bode special. Michel Marcus continued to calculate new terms, but starting with n=78 (prime(78)=397) and already up to n= 5*10^4 he obtained only zeros!! It is clear that large numbers have, as a rule,  more and more divisors, and one can expect that from a moment we will obtain sometimes zeros, but unexpectedly we obtained, maybe, ALL zeros.  If so, I think that it is a unique phenomenon of prime 397.


_______________________________________________

Seqfan Mailing list - http://list.seqfan.eu/