What did Yu Jianchun discover about Carmichael numbers? There's a news story going around (see for example [1]; other accounts are even more breathless) about an amateur mathematician, Yu Jianchun, finding an "alternative method to verify Carmichael numbers". I'm curious as to 
the method; does anyone know about this?
I don't think this is smoke and mirrors, as was the claimed proof of the Riemann Hypothesis by Opeyemi Enoch of the Federal University Oye-Ekiti, Nigeria. [2] Cai Tianxin of Zhejiang University has apparently read the proof and invited Yu Jianchun to give a talk at a graduate seminar, and CNN apparently ran the story by William Banks who apparently compared the discovery to the proof of the infinitude of Carmichael numbers (though from what appears in the article it seems he did not have access to the work itself).
But the article (appropriately, I suppose, for its nontechnical audience) doesn't give any real information on the discovery itself. The closest it comes is a picture of a letter (in Chinese) from Yu Jianchun with most of a formula:
$$
\frac{\left(\frac{N^{P_1}-N}{N} - \frac{N^{P_1-P_2+1}-N}{P_1-P_2}\right)(P_1-P_2)}{P_2}(N,P_1,P_2
$$
(note that the rightmost part is cut off). There is also what might be a worked example, starting
$$
\begin{align}
=&\frac183^{3n}(3^n+1)^3(3^n+2)^3(3^{2n-1}+2\cdot3^{n-1}+1)^3\\
=&\left[\frac123^n(3^n+1)(3^n+2)(3^{2n-1}+2\cdot3^{n-1}+1)\right]^3
\end{align}
$$
but it's hard to make sense of this without understanding the surrounding text (in Chinese).

[1] China's 'Good Will Hunting?' Migrant worker solves complex math problem
[2] Riemann Hypothesis not proved
 A: 
For example, the two simplest formulas which contain the quadratic term are ...

I found simpler formulas: $$(6n+1)(12n+1)(24n^2+6n+1)$$
$$(1008n-79)(2268n-179)(326592n^2-51372n+2021)$$
A: On page 136 of Cai Tianxin, The Book of Numbers, it says,  
In 2015 the Chinese amateur, Jianchun Yu, who was then a packer in a bookstore warehouse in Hangzhou found that all the numbers like $(6k+1)(12k+1)(24k^2+6k+1)$ are Carmichael numbers if these three factors are all primes. 
The proof is left as an exercise on page 142. I found no other mention of Jianchun's work in the book. 
A: Yu Jianchun's formulas: 
$$n=(6k+1)(18k+1)(54k^2+12k+1)=5832^4+2592k^3+450k^2+36k+1$$
$(n-1)/(6k)=972k^3+432k^2+75k+6$$
$$(n-1)/(18k)=324k^3+144k^2+25k+2$$
$$(n-1)/(54k^2+12k)=108k^2+24k+3$
Hope this helps.
A: Apparently it is an alternative proof of the infinitude of Carmichael numbers.
The other proof mentioned in the articles ("done by academics 20 years ago") is:


*

*W. R. Alford, Andrew Granville, Carl Pomerance (1994) "There are Infinitely Many Carmichael Numbers"


As for the method, unless someone took notes of the lecture at Zhejiang University, I guess it is unpublished at this point (except for the fragments shown in the news).
As a side note, notice two equations shown in one the photos:

$$1. \quad(6n+1)(18n+1)(54n^2+12n+1)$$ 
$$2. \quad(1764n-139)(2268n+179)(1000188n^2-157752n+6221)$$
This looks very similar to Chernick's result that $(6n+1)(12n+1)(18n+1)$ is a Carmichael number if each of the factors is prime. It is open whether this family of Charmichael numbers is infinite or not.
This is pure speculation, but the non-linear factors in Yu's numbers might make a big difference, since that is not a Dickinson's conjecture-type problem anymore.
Added data for very small values ($n<11$) for the first expression. $C_i$ indicates the $i$-th Carmichael number.
\begin{array}{|c|c|c|c|}
\hline
\mathrm{n}& \mathrm{eq. 1 } &   \\ \hline
1 & 7 \cdot 19\cdot 67 &8911=C_7\\ \hline
2 & 13 \cdot 37\cdot 241 &115921=C_{18}\\ \hline
3 & - &-\\ \hline
4 & - &-\\ \hline
5 & - &-\\ \hline
6 & 37 \cdot 109\cdot 2017& 8134561=C_{93}\\ \hline
7 & 43 \cdot 127\cdot 2731& 14913991=C_{125}\\ \hline
8 & - & -\\ \hline
9 & - & -\\ \hline
10 & 61 \cdot 181\cdot 5521& 60957361=C_{209}\\ \hline
\end{array}
Edit. See the comment of Zhiyun Cheng below for a translation of the text in chinese.
