1. 8
  1. 5

    This seems to make intuitive sense to me. This is not a criticism of the article.

    For every span of n numbers, exactly one number there has n as a divisor. If you know nothing more about these numbers, you would expect each number to have a 1/n chance of having n as a divisor.

    So for such a span containing two twin primes, you would expect each number other than the twin primes to have a greater chance – that is, 1/(n-2) – of having n as a divisor. So it would have a 1/3 chance of having 5 as a divisor, when a randomly selected number has a 1/5 chance.

    I would also be interested if the numbers directly before and after the twin primes also have more factors on average. They wouldn’t have the advantage of having 3 as a divisor.