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If I have a large known prime, is the number 2 greater than that or 2 less than that more likely than chance to be prime?

It seems that since finding new large primes is something of a newsworthy event (especially so if they are the new largest known prime of course), if a twin is more likely than a randomly selected odd number of similar size, once a large prime is found, one could effectively piggyback on that to find other nearby large primes. Even if they are the lower twin, "I found the second largest known prime" would still be of interest to some people, I would think.

Ze'ev misses Monica
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  • If a prime $p_1$ is not a twin prime, then $p_1 + 2$ and $p_1 - 2$ are both not twin primes. However, if a prime $p_2$ is the smaller of the twin prime pair, then $p_2 - 2$ is a prime, but for the prime $p_2 - 2$, $p_2$ is a prime. – Toby Mak Jun 12 '17 at 23:05
  • Since for each prime pair there is always one other number that is larger by 2 and one that is smaller by 2, considering that there are no other cases where this happens, the chance is $\frac{1}{2}$ for both cases. – Toby Mak Jun 12 '17 at 23:06
  • If you're asking if twin primes are likely to happen then that isn't really true for bigger numbers. – kingW3 Jun 12 '17 at 23:08
  • I don't think the OP is asking this - but since the OP hasn't clarified our comments it's hard to tell – Toby Mak Jun 12 '17 at 23:12
  • I am saying if there is a large prime, is the number two up or two down going to be more likely than some other number of that size to be prime? – Ze'ev misses Monica Jun 12 '17 at 23:24
  • The record-holders for largest known prime tend to be Mersenne primes, partly because the Lucas-Lehmer test makes primality testing for Mersenne primes faster than testing other numbers of similar magnitude, partly because more effort is being spent in looking for Mersenne primes than other very large primes. – Robert Israel Jun 13 '17 at 00:20
  • https://arxiv.org/abs/1603.03720 – Count Iblis Jun 13 '17 at 00:21
  • Of the first 30 Mersenne primes, the only ones that are twin primes are $3$ and $31$. I strongly suspect there are no others. – Robert Israel Jun 13 '17 at 00:34
  • (I deleted my previous comment because it badly messed up the formatting of this page). I'll try again... – Robert Israel Jun 13 '17 at 01:21
  • Finding new large primes, if they aren't special in some way (e.g. the largest known, or nearly so) is not particularly newsworthy, it is rather easy to do. For example, 71625597731433658276942269193081257843715495465728567 07266115016758537514777274853137059069235571383349602 33931623274911577848696419381310061542448638926324535 30002781073498992861174400146608608724415498029662547 47124836402432418849924548958686092674858660928225620 21118819370164088635110651339568313 is almost certainly "new" – Robert Israel Jun 13 '17 at 01:22
  • And, BTW, it is not a twin. – Robert Israel Jun 13 '17 at 01:26

1 Answers1

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It is not known whether there are infinite pairs of twin primes. So, it is possible (although very unlikely) that there is only a finite number of twin primes, in which case for large enough primes $p$ it would be impossible for $p+2$ to be prime (same for $p-2$).

Things get far more complicated in the more plausible scenario that there are infinite pairs of twin prime numbers. We can't say much if we don't know something about the way the twin primes are asymptotically distributed (only that the probability of $p+2$ being prime is positive). And unfortunately we know basically nothing about this.

One famous conjecture of Hardy an d Littlewood -- often regarded as true, but unproven -- establishes that the number of twin number pairs less than $x$ has order
$$ \pi_2(x)\sim C\frac x {(\log x)^2}. $$ for some constant $C$.On the other hand, from the prime number theorem we know that the number of primes less than $x$ is $$ \pi(x)=\frac x {\log x} $$ which is of course of larger order. Hence, if the conjecture is true, then for $p$ odd the ratio of probablilities $$ \frac {P(p+2\ {\rm prime,\ given\ } p\ {\rm prime})} {P(p+2\ {\rm prime})} $$ is asymptotically equal to $$ \left(\frac C {\log x}\right)/\left(\frac 1 {\log x}\right)=C. $$

And, by the way, the constant $C$ has been computed to be approximately $1.32$. So, for very large values of $p$ it will be slightly more likely for $p+2$ to be prime if we know that $p$ is prime. The same story holds for $p-2$.

http://mathworld.wolfram.com/k-TupleConjecture.html

Federico
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    Are you sure that means that? I think of it saying the the primes have a density $1/\log(x)$ and the twin primes $1/\log^2(x)$ which is what you'd expect if the primes were just independently placed with their density $1/\log(x).$ Not sure though. ( And I'd still expect there to be 2nd order effects in this direction since $p-2$ and $p+2$ can't both be prime if $p$ is prime and can if it isn't.) – spaceisdarkgreen Jun 12 '17 at 23:57
  • "only that the probability of p+2p+2 being prime is positive" It's consistent, so far as we know, that there are infinitely many twin primes but they have asymptotic density zero. – Noah Schweber Jun 13 '17 at 00:19
  • @spaceisdarkgreen you're right, I made a mistake. In the definition of conditional probability $P(A|B)=P(A\cap B)/P(B)$ I didn't divide by $P(B)$ if you see what I mean. I'll make the correction in a moment. – Federico Jun 14 '17 at 01:09