4 ms·

I think of it like this. If a number n has a factor f, f cannot be a factor of n+1 or n-1 (unless f is 1, but we can ignore that for primeness). The next number

I think of it like this. If a number n has a factor f, f cannot be a factor of n+1 or n-1 (unless f is 1, but we can ignore that for primeness). The next numbers to have f as a factor are n+f or n-f. If a number n has loads of factors, all of those factors are excluded from n+1 and n-1, so there are not many numbers left to be factors of them and they are likely to be twin primes.

sweezyjeezy 5y ago
This article about the converse though - if you have twin primes is the central value more likely to be highly composite? Your statement can be true and this second statement not be true.
- feoren 5y ago
  Assume (1) A -> B, (2) A is sometimes true, and (3) B is sometimes false. Then isn't it provably true that P(A | B) > P(B)? In other words, if highly composite numbers essentially "cause" twin primes, then given a twin prime, they must be more likely to contain a highly composite number between them. Major edit: Clearly, P(twin prime | highly composite) > P(twin prime) -- this is GP's statement. You're (and the article's) question is: ok, what about P(highly composite | twin prime)? Well, Bayes rule says: P(highly composite | twin prime) * P(twin prime) = P(twin prime | highly composite) * P(highly composite). We're comparing P(highly composite | twin prime) vs. P(highly composite). But we know P(twin prime | highly composite) > P(twin prime) based on GP's reasoning. So P(highly composite | twin prime) * P(twin prime) = P(twin prime) * C * P(highly composite) for some C > 1. Then P(highly composite | twin prime) = C * P(highly composite) for some C > 1, and we know P(highly composite | twin prime) > P(highly composite) Doesn't this more or less prove the question posed in the article?
  - sweezyjeezy 5y ago
    Yes you're right, my bad - it doesn't answer the question in the article though. What's the expected number of divisors? That doesn't follow from this kind of analysis, because the ones that aren't highly composite could be 'very' uncomposite for some reason.
  - thaumasiotes 5y ago
    > Assume (1) A -> B, (2) A is sometimes true, and (3) B is sometimes false. Then isn't it provably true that P(A | B) > P(B)? No, this is not true. Consider a population of 200 people, with 20 in group AB, 0 in group A, 80 in group B, and 100 in group Nil. We see that A -> B, since group A has 0 members. We see that A is sometimes true, since group AB has 20 members. We see that B is sometimes false, since group Nil has 100 members. P(A | B) = 20 / (20 + 80) = 0.2 P(B) = (20 + 80) / (20 + 80 + 0 + 100) = 0.5
    - feoren 5y ago
      Thank you, you're absolutely right: I meant to say P(A | B) > P(A). By excluding !B, you're necessarily excluding only the Nil group. I think I got it right in my edit, but thank you for correcting my thinking here.
lkrubner 5y ago
"f cannot be a factor of n+1 or n-1" You are mostly just restating Euclid's proof that there are an infinite number of primes.
- thaumasiotes 5y ago
  This is an observation on which the proof depends, but, as is logically required, it is not the proof.

sweezyjeezy 5y ago

This article about the converse though - if you have twin primes is the central value more likely to be highly composite? Your statement can be true and this second statement not be true.

feoren 5y ago
Assume (1) A -> B, (2) A is sometimes true, and (3) B is sometimes false. Then isn't it provably true that P(A | B) > P(B)? In other words, if highly composite numbers essentially "cause" twin primes, then given a twin prime, they must be more likely to contain a highly composite number between them. Major edit: Clearly, P(twin prime | highly composite) > P(twin prime) -- this is GP's statement. You're (and the article's) question is: ok, what about P(highly composite | twin prime)? Well, Bayes rule says: P(highly composite | twin prime) * P(twin prime) = P(twin prime | highly composite) * P(highly composite). We're comparing P(highly composite | twin prime) vs. P(highly composite). But we know P(twin prime | highly composite) > P(twin prime) based on GP's reasoning. So P(highly composite | twin prime) * P(twin prime) = P(twin prime) * C * P(highly composite) for some C > 1. Then P(highly composite | twin prime) = C * P(highly composite) for some C > 1, and we know P(highly composite | twin prime) > P(highly composite) Doesn't this more or less prove the question posed in the article?
- sweezyjeezy 5y ago
  Yes you're right, my bad - it doesn't answer the question in the article though. What's the expected number of divisors? That doesn't follow from this kind of analysis, because the ones that aren't highly composite could be 'very' uncomposite for some reason.
- thaumasiotes 5y ago
  > Assume (1) A -> B, (2) A is sometimes true, and (3) B is sometimes false. Then isn't it provably true that P(A | B) > P(B)? No, this is not true. Consider a population of 200 people, with 20 in group AB, 0 in group A, 80 in group B, and 100 in group Nil. We see that A -> B, since group A has 0 members. We see that A is sometimes true, since group AB has 20 members. We see that B is sometimes false, since group Nil has 100 members. P(A | B) = 20 / (20 + 80) = 0.2 P(B) = (20 + 80) / (20 + 80 + 0 + 100) = 0.5
  - feoren 5y ago
    Thank you, you're absolutely right: I meant to say P(A | B) > P(A). By excluding !B, you're necessarily excluding only the Nil group. I think I got it right in my edit, but thank you for correcting my thinking here.

lkrubner 5y ago

"f cannot be a factor of n+1 or n-1" You are mostly just restating Euclid's proof that there are an infinite number of primes.

thaumasiotes 5y ago
This is an observation on which the proof depends, but, as is logically required, it is not the proof.