Any positive integer can be written as the difference between two coprime semiprimes.

Tested up to 1,000,000.

See also:

I'm not surprised that there are heuristics backing up the conjecture, but think it would be strange if the proof was so far away when the question above had an easy answer.

  • $\begingroup$ Do you count primes or 1 as semiprimes? Because actually every positive integer up to 100000 is the difference of a semiprime and a prime (or is a semiprime minus 1). $\endgroup$ – მამუკა ჯიბლაძე Dec 26 '17 at 8:36
  • $\begingroup$ @მამუკაჯიბლაძე: A semiprime (also called biprime or 2-almost prime, or pq number) is a natural number that is the product of two (not necessarily distinct) prime numbers. The semiprimes less than 100 are 4, 6, 9, 10, 14, 15, 21, 22, 25, 26, 33, 34, 35, 38, 39, 46, 49, 51, 55, 57, 58, 62, 65, 69, 74, 77, 82, 85, 86, 87, 91, 93, 94, and 95. (sequence A001358 in the OEIS) (Wikipedia). $\endgroup$ – Lehs Dec 26 '17 at 9:27

It seems to me most likely to be true that every integer can be written in infinitely many ways as the difference between a pair of co-prime semiprimes and that the number of pairs with both members under $x$ is a simpley described fiunction. I say this by analogy with twin primes and their generalizations as described below. For primes it seems that proofs using the current methods are unlikely. For semi-primes there might be proofs but I would bet against it (but not very much.)

It is highly expected, but unknown, that there are infinitely many pairs of primes $p,q$ with $p-q=2.$ There has been some progress in recent years but no proof as of yet.

The number of primes up to $x$ is $\pi(x) \sim \frac{x}{\log x}$ The number of primes $p \lt x$ such that $p+2$ is also prime is conjectured to be $\sim \frac{Cx}{(\log x)^2}$ for a certain constant $C.$ There are simple heuristic arguments for this and the computational data is very good.

It has long been conjectured that every even integer $k$ is the difference of two primes in infinitely many ways. It is now known that there is at least one $k \lt 248$ for which it is true.

For $k=4$ the density for pairs $p,p+4$ should be almost exactly the same as for $k=2.$ However for $k=6$ it should be higher, after all, knowing that $p$ is prime and not tiny makes $p+6$ not only odd, but also a non-multiple of $3.$ The same constant that works for $2$ should be correct for any power of $2$ and the one which works for $6$ should also work for $k=2^a3^b.$ In general the constant for $k$ should depend only on the distinct primes dividing $k.$ Again, the computation evidence looks very strong.

The number of semiprimes up to $x$ is $\pi_2(x) \sim \frac{x \log \log x}{\log x}$ So perhaps the number of semi-prime pairs $s_1,s_2$ with $s_1-s_2=k$ is $\sim \frac{C x( \log \log x)^2}{(\log x)^2}$ where $C$ depends only on the prime divisors of $k$ or maybe for semiprimes that is less of a consideration. I've kind of ignored the relatively prime condition. Without it we could look at the odd prime divisors of $k$ (for $k$ even) pick any one odd divisor $p*$ then find (if possible) pairs $p,q$ with $p-q=\frac{k}{p*}$ and use semiprimes $pp*,qp*.$ Using relatively prime semi-primes would allow many more solutions.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.