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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?

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    $\begingroup$ WLOG, we can assume $n\le m$ implying $p_n\le p_m$ $\endgroup$ – Peter May 25 '18 at 8:29
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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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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 $p_{10}=29>25$.

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