The Bernoulli numbers $B_n$. where all numbers $B_n$ are zero with odd index $n>1$. first values are given by $B_{0} = 1$ , $B_{1} = -1/2$, $B_{2} = 1/6$, $B_{3} = -1/30$.

Agoh conjecture: let $n$ be a positive integer with $n \ge 2$, then

$$nB_{n-1} \equiv -1 \pmod{n}\iff n\text{ is prime} $$

The idea of ​​$\Upsilon$ number comes from my power sum formula and negative values of zeta function

Definition let $m$ be a non-negative integer $$\zeta(-m)=(-1)^{m}\frac{B_{m+1}}{m+1}=\sum_{b=1}^{m+1} \Upsilon_b\sum_{i=0}^{b-1} (-1)^{i}(b-i)^{m}\binom{b-1}i$$

so we can calculate values of $\Upsilon_b$ , if we substitute value $\zeta (0)=-1/2$ gives $\Upsilon_1=-1/2$.

Again we can calculate next value using or substituting previous values of $\Upsilon_b$.

first values are given by $$\begin{align*} \Upsilon_1&=-\frac{ 1}{2},\\ \Upsilon_2&=\frac{ 5}{12},\\ \Upsilon_3&=-\frac{ 3}{8},\\ \Upsilon_4&=\frac{ 251}{720},\\ \Upsilon_5&=-\frac{ 95}{288},\\ \Upsilon_6&=\frac{ 19087}{60480},\\ \Upsilon_7&=-\frac {5257}{17280}. \end{align*}$$

hypothesis: let $n$ be a positive integer with $n \ge 2$, then

$$n\Upsilon_{n-1} \equiv -1 \pmod{n} \iff n\text{ is prime} $$

Else show that hypothesis is equivalence to agoh's conjectures

Else show that if n is prime then $n\Upsilon_{n-1} \equiv -1 \pmod{n} $

Steven Clark had verified conjecture $n\Upsilon_{n-1} \equiv -1 \pmod{n}\iff n\in \mathbb{P}$ for $2 \leq n \leq 500 $

More simply

Let $D_{m,b}=\sum_{i=0}^{b-1} (-1)^{i}(b-i)^{m}\binom{b-1}i$

So $D_{m,m+1}=m!$

$$\Upsilon_{m+1}=\frac{(-1)^{m}B_{m+1}-(m+1)(\Upsilon_1 D_{m,1}+\Upsilon_2 D_{m,2}+\cdots+\Upsilon_m D_{m,m})}{(m+1)!}$$

Please see my post on MO

  • $\begingroup$ Are you familiar with Giuga's conjecture? It's in the same territory. en.wikipedia.org/wiki/Agoh–Giuga_conjecture $\endgroup$ Aug 10, 2019 at 6:48
  • $\begingroup$ @Gerry_Myerson yes I am giuga's and agoh's conjectures are equivalence $\endgroup$
    – Pruthviraj
    Aug 10, 2019 at 7:02
  • 1
    $\begingroup$ So, what evidence do you have for your hypothesis? $\endgroup$ Aug 10, 2019 at 7:10
  • $\begingroup$ @Gerry_Myerson Based on given $\Upsilon_n$ values I have assumed my hypothesis is true but I am working on a program for further observation $\endgroup$
    – Pruthviraj
    Aug 10, 2019 at 10:41
  • 1
    $\begingroup$ I deleted one of my earlier comments after I found an error in my code. Assuming no more errors in my code, I've now verified your conjecture $n\,\Upsilon_{n-1}\equiv -1\,(mod\ n)\,\iff\,n\in \mathbb{P}$ for $2\le n\le 500$. My interest in this question is partially motivated by my conjectured formulas for $\zeta(-j)$ in my question math.stackexchange.com/q/3262873. $\endgroup$ Aug 26, 2019 at 19:41

1 Answer 1


This post is not intended as an answer but rather a way to share results and code as originally requested by the OP.

Assuming no errors in my code, I have now verified the conjecture $n\,\Upsilon_{n-1}\equiv -1\,(mod\ n)\,\iff\,n\in \mathbb{P}$ for $2\le n\le 1000$.

I attached an image of my Mathematica code to provide others the opportunity to verify correctness of my code and use it to explore the conjecture for a larger range of $n$ values if so desired.

Screenshot of code


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.