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This is motivated by a recent question by Wadim.

The negative answer should be known, since t is very natural, in this case I would be happy to see any reference.

May Pafnuty Lvovich Chebyshev's approach to distribution of primes lead to PNT itself, if we replace $\frac{(30 n)! n!}{(15 n)! (10 n)! (6 n)!}$ to other integer ratios of factorials? If not, what are the best constants in asymptotic relation $$ c_1 \frac{n}{\log n}< \pi(n)< c_2 \frac{n}{\log n} $$ which may be obtained on this way?

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    $\begingroup$ I just put my NT lecture with a simplified version of Chebyshev's theorem as answer to an earlier question. The factorial ratio used by Chebyshev gives best possible estimates on this way. There is a related extension of the method, computation of the $\mathbb Z$-transfinite diameter well discussed in H. Montgomery's CBMS(?) lectures. It is known that this cannot give $c=1$ in the asymptotics $\pi(n)\sim cn/\log(n)$. $\endgroup$
    – Wadim Zudilin
    Commented May 29, 2010 at 9:00

2 Answers 2

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Erdős and Diamond proved in [1] that Chebyshev could have achieved sharper bounds for the asymptotic behavior of the prime counting function. Nevertheless, their proof does not shed any light on the first question that you posed because they took the PNT for granted throughout their note.

References:

[1] H. G. Diamond; P. Erdős. On sharp elementary prime number estimates, Enseign. Math. (2) 26 (1980) 313-321.

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  • $\begingroup$ Just the opposite: Chebyshev was extremely sharp in his approach, and no other factorial ratio can produces better estimates. $\endgroup$
    – Wadim Zudilin
    Commented May 29, 2010 at 9:03
  • $\begingroup$ There is no denying his estimates were sharp. Though, what E & D proved is that pushing the attack a litle bit one can get even sharper estimates. $\endgroup$
    – José Hdz. Stgo.
    Commented May 29, 2010 at 9:34
  • $\begingroup$ There is also a summary of that paper of Diamond and Erdos in Diamond's "Elementary methods in the study of the distribution of prime numbers" in the Bulletin AMS (sections 3 and 9). It exposes the ideas in several steps: first $$\frac{(2n)!}{n!n!}$$, then $$\frac{(30 n)! n!}{(15 n)! (10 n)! (6 n)!}$$ (Chebyshev's approach) and finally a (related) general form, due to Erdos-Diamond, which uses a function $\mu_T$ close to the Moebius $\mu$ function. The closest analogue of Chebyshev's is then $$ \frac{(30 n)!^5(5n)!}{(15 n)!^5(10 n)!^5(6 n)!^5}. $$ $\endgroup$
    – Paul-Olivier Dehaye
    Commented Oct 1, 2010 at 13:23
  • $\begingroup$ There is also the work of Sylvester, but I haven't seen that presented anywhere in detail. $\endgroup$
    – Paul-Olivier Dehaye
    Commented Oct 1, 2010 at 13:26
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According to the notes in fifth edition of Niven, Zuckerman and Montgomery's An Introduction to the Theory of Numbers for each $\epsilon\in(0,1)$ there is a series of parameters in Chebyshev's method that proves $$(1-\epsilon)\frac{\log x}{x} < \pi(x) < (1+\epsilon)\frac{\log x}{x}$$ for all large enough $x$ but that the proof of this uses PNT so that it doesn't provide an alternative proof of PNT.

They cite a paper of H. G. Diamond and P. Erdos: "On sharp elementary prime estimates", L'Enseignment Math. 26 (1980), 313-321.

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  • $\begingroup$ Ah so: these are different parameters, not the ones from factorial ratios. I now understand JHS's answer. $\endgroup$
    – Wadim Zudilin
    Commented May 29, 2010 at 9:07
  • $\begingroup$ The account in NZM's book doesn't mention factorial ratios. The rough idea is to take a sequence of reals $(v(d))$ which is eventually zero and with $\sum_d v(d)/d=0$ and estimate $\sum_d v(d)T(x/d)$ where $T(x)=\sum_{n\le x} \log n$. This sum will equal $\sum_{r\le x} N(r,x)\Lambda(r)$ for suitable $N(r,x)$ and the idea is to choose the $v(d)$ so that most of the $N(r,x)$ are close to $1$. Then $\sum_d v(d)T(x/d)$ will be close to $\psi(x)$. See the book for more details (elementary but a little fiddly). $\endgroup$
    – Robin Chapman
    Commented May 29, 2010 at 9:18
  • $\begingroup$ Thanks, Robin. Yes, this is a familiar approach, the choice of $\nu(d)$ (its existence for each $\epsilon$) is guaranteed by the PNT. These are related with factorial ratios: take $\nu(d)=1$ for $n=30$ and 1 and $-1$ for $n=15$, 10, and 6. The resulting "step" function will correspond to Chebyshev's original choice of factorial ratio. $\endgroup$
    – Wadim Zudilin
    Commented May 29, 2010 at 9:29
  • $\begingroup$ If I were you, I would skip the NZM account and proceed directly to the perusal of the relevant note by Chebyshev: Mémoire sur les nombres premiers (it is somewhere on the first volume of his Œuvres). The NZM account does not do any justice to the clarity with which Chebyshev exposes his work. $\endgroup$
    – José Hdz. Stgo.
    Commented May 29, 2010 at 9:31
  • $\begingroup$ @JHS: As I mentioned above, I give a simplified version of Chebyshev's 1850 original theorem in mathoverflow.net/questions/26272/…. In that case I use $T(x)-2T(x/2)$ but it makes clear on how to use other weighted combinations. $\endgroup$
    – Wadim Zudilin
    Commented May 29, 2010 at 9:32

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