I did some experiments in SAGE and it seems like the prime sequence $p_n$ satisfies:

$$p_n+p_m \le p_{n+m} < p_n p_m$$ for $(n,m) \neq (1,1)$.

For $n=1$ the last inequality is Bertrands postulate.

Here is some SAGE code to test this for some primes:

lp = list(primes(10000))
for n in range(len(lp)):
    for m in range(len(lp)):
        pn = lp[n]
        pm = lp[m]
        pnm = nth_prime(n+m+2)
        print pn*pm>=pnm,pnm>=pn+pm, n+1,m+1,n+m+2,pn,pm,pn*pm,pnm

Is this something known ( if it is true)? And if so, how does one prove or disprove it, are there heuristics?

  • 1
    $\begingroup$ WLOG, we can assume $n\le m$ implying $p_n\le p_m$ $\endgroup$ – Peter May 25 '18 at 8:29
  • 4
    $\begingroup$ IMO for larger $n,m$ it should be possible to use the inequaltities en.wikipedia.org/wiki/Prime-counting_function#Inequalities $n (\ln (n \ln n) - 1) < p_n < n \ln (n \ln n)$ for $n \ge 6$, and the remaining $n,m$ with direct computation. $\endgroup$ – gammatester May 25 '18 at 8:32
  • $\begingroup$ We have the inequality $p_{n+1}>2p_n$, but if your conjecture is true, then letting $m=n$ implies that $2p_{n+1}<p_{2n+1}$ for $n>1$ which is definitely a tighter upper bound. Doing some tests, however, I think a stronger inequality is that $$2p_{n+1}\le p_{2n+1}-1\tag{$n>1$}$$ $\endgroup$ – Mr Pie May 25 '18 at 8:52
  • 1
    $\begingroup$ The first part of the inequality holds for $1\le n\le m\le 2\cdot 10^4$ with the exception $m=n=1$. I am currently checking the second part in the same range, yet no counterexample. $\endgroup$ – Peter May 25 '18 at 9:03
  • 3
    $\begingroup$ Here a sketch for the first. Let $f(n)=\ln(n\ln n))$. This function is strictly increasing and $p_n < n f(n), \; n>6$. Therefore with $n>m$ we have $$p_n+p_m <nf(n)+mf(m) < nf(n)+mf(n) = (n+m)f(n) < (n+m)f(n+m) \le p_{n+m}$$ $\endgroup$ – gammatester May 25 '18 at 10:30

The left inequality is equivalent to the Second Hardy-Littlewood conjecture.

The k-tuple conjecutre, AKA First Hardy-Littlewood conjecture is the statement that the density of every Prime Constellation can be computed using a single general formula. If it is true, then there are infinitely many twin primes, also infinitely many prime tuples of form $(p, p+4)$, $(p, p+6)$, $(p, p+2, p+6, p+8)$, etc. Note that it does not imply the prime tuple $(p, p+2, p+4)$ is infinite, which is not true.

If k-tuple conjecture is true, then Second Hardy-Littlewood conjecture is not true. $\pi(3159)=446$, but there may be $447$-tuple primes spanning $3159$ integers. Such tuple is not yet discovered, but the formula in k-tuple conjecture suggests that the first such tuple is likely to be between $1.5\times10^{174}$ and $2.2\times10^{1198}$.

For right inequality, WLOG assume that $m \le n$. Then for $10 \le n$ and $3 \le m$, we can apply the inequality mentioned by @gammatester. $$p_{m+n}\le p_{2n}<2n\ln{(2n\ln{2n})}<4n{(\ln(n\ln n)-1)}<4p_n\le p_mp_n$$

For $n<10$, one can check all the cases manually. For $m=1$, it is Bertrand's Postulate. For $m=2$, it is proving $p_{n+2}<3p_n$. It is known that if $k\ge25$, then there exists a prime between $k$ and $1.2k$. It is trivial corollary that $p_{n+2}<1.2^3p_n<3p_n$ if $n>10$, especially since $p_{10}=29>25$.


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.