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 $\endgroup$ – Gerry Myerson Aug 10 '19 at 6:48
  • $\begingroup$ @Gerry_Myerson yes I am giuga's and agoh's conjectures are equivalence $\endgroup$ – Pruthviraj Aug 10 '19 at 7:02
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    $\begingroup$ So, what evidence do you have for your hypothesis? $\endgroup$ – Gerry Myerson Aug 10 '19 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 '19 at 10:41
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    $\begingroup$ 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. $\endgroup$ – Steven Clark Aug 26 '19 at 19:41

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


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