Let's recall: Von Mangoldt function $\Lambda$ is a following function: $$\Lambda: \mathbb{N}\rightarrow \mathbb{R}$$ $$ \Lambda(n) = \left\{ \begin{array}{ll} \log p & \textrm{if $n = p^k$ for some prime $p$}\\ 0 & \textrm{otherwise}\\ \end{array} \right. $$ Ternary Goldbach Conjecture states that: $$W(N):=\sum _{k_1+k_2+k_3 = N}\chi(k_1)\chi(k_2)\chi(k_3)>0$$ For odd numbers greater than $5$. Here $\chi$ is characteristic function for prime numbers.

But here is the problem. In proof of this concjecture we see a quite different expression. $$G(N) = \sum_{k_1+k_2+k_3 = N}\Lambda(k_1)\Lambda(k_2)\Lambda(k_3)$$ This sum is proved to be sufficiently big.

My question is:

How are $W(N)$ and $G(N)$ connected? How to estimate from below the first expression ($W(N)$) using just $G(N)$?

Thank you in advance.


1 Answer 1


In analytic number theory, when one studies primes, it is customary to study sums involving $\Lambda$ instead of $\chi$. Note that $\Lambda$ is essentially supported on primes. For any function $f$, we have $$ \sum_{n\leq x} f(n)\Lambda(n) = \sum_{p\leq x} f(p)\log p + \sum_{\substack{p^k\leq x\\ k\geq 2}} f\left(p^k\right)\log p. $$ If $f$ is small, say bounded by 1, the second sum is $O(\sqrt{x}\log x)$, which is quite small if the main term is expected to be of size $x$. This is a general theme that occurs when studying sums over primes, namely that the sum over prime powers (with the power at least 2) doesn't give a significant contribution to the estimate. Thus summing $f$ against $\Lambda$ is essentially summing $f$ over primes, with the additional weight $\log p$ (indeed, sums such as the one above are often regarded in the literature as simply "sums over primes"). To address your question specifically, we have $$ G(N) = \sum_{p_1+p_2+p_3=N} (\log p_1)(\log p_2)(\log p_3) + O\Big(\sum_{\substack{p_1^k+p_2^k+p_3^k=N\\ k\geq 2}} (\log p_1)(\log p_2)(\log p_3)\Big) $$ By a trivial estimation, the error term is $O\left(N^{3/2}(\log N)^3\right)$. Since the main term is known to be of size $N^2$, it follows that for $N$ sufficiently large, the sign of $G(N)$ is not affected by the error term, and so $G(N) > 0$ for $N$ sufficiently large. Note that if $G(N) > 0$, we certainly must have $W(N) > 0$ as well. That is, one does not bound $W(N)$ using $G(N)$ directly, but rather infers that $W(N) > 0$ from the fact that $G(N) > 0$, as both are non-negative sums over primes.

  • $\begingroup$ Thank you for the answer .What is $T(N)$? Do you mean $T(N) = W(N)$? $\endgroup$
    – mkultra
    May 28, 2021 at 13:12
  • $\begingroup$ Oh, yes. I was working on something else using that notation. I've edited my answer to reflect this. $\endgroup$ May 28, 2021 at 16:03

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