DefinitionLet $W$ be the function , defined as $W(a,b)=r$

given $a,b\in \mathbb{Z_+}$ and $a>1$

Take $m$ to be the integer s.t. $a^{m+1} \ge b > a^{m}$, i.e. $m = \lceil \log{b}/\log{a} \rceil - 1$.

Convert number $a^{m+1} - b$ in base $a$ and add its digits

$$a^{m+1} - b = (r_{l} r_{l-1} ... r_{1} r_{0})_{a}$$

Where $r=\sum_{i=0}^{l}r_{i}$

**Example**

$W(5,77)=8$

**Identity$1$**

if $W(a,b)=r$ then $b+r\equiv 1($ mod $a-1)$

◆ $S$ is a function defined as

$$S(a,n)=\sum_{i=1}^{a}i^{n}$$

Where $a$ and $n$ are positive integer.

Let $p$ is prime and $p+1=z$

Questionshow that

If $ z>2n+2$ Then $W(z,W(z,S(z,2n)))=z$

**Example**

Let $n=1$ here, choose any $z>4$

Let $z=6$

So $W(6,W(6,S(6,2)))=W(6,W(6,91))=W(6,10)=6$

**Table For $W(t,W(t,S(t,2)))$.**

$$\begin{array}{c | c | c |c | } t & W(t,S(t,2)) & W(t,W(t,S(t,2))) \\ \hline 2 & 2 & 0 \\ \hline 3^{*} & 3 & 0 \\ \hline 4^{*} & 4 & 0 \\ \hline 5 & 6 & 7 \\ \hline 6^{*} & 10 & 6 \\ \hline 7 &5 & 2 \\ \hline 8^{*} &14& 8 \\ \hline 9 &12& 13 \\ \hline 10 &12& 16 \\ \hline 11 & 15 & 16 \\ \hline 12^{*} & 22 & 12 \\ \hline 13 & 10 & 3 \\ \hline 14^{*} & 26 & 14 \\ \hline 15 & 21 & 22 \\ \hline 16 &20 & 26 \\ \hline 17 &24& 25 \\ \hline 18^{*} &34& 18 \\ \hline 19 &15& 4 \\ \hline 20^{*} &38& 20 \\ \hline 21 &30& 31 \\ \hline \vdots &\vdots & \vdots \\ \hline \end{array}$$

$t^{*} = z $

Python programming for calculate $W$ function

```
n1=5
n2=77
rem_array = []
while n2 != 1:
mod = n2%n1
if mod != 0:
rem = n1-mod
n2 = n2 + rem
rem_array.append(round(rem))
n2=n2/n1
else:
n2 = n2/n1
rem_array.append(0)
print(rem_array[::-1])
print(sum(rem_array))
```

Proof for, if $p>n+1$ then $p|S(p,n)$

**Formula**

$$ S(a,n)= \sum_{i=1}^{a} i^{n}=\sum_{b=1}^{n+1} \binom{a}b\sum_{j=0}^{b-1} (-1)^{j}(b-j)^{n}\binom{b-1}j$$

**Proof**

Let $a=p(prime)>n+1$

We can see, $a$ can be common out from $\sum_{b=1}^{n+1}\binom{a}b\sum_{j=0}^{b-1} ...$

$\implies a|S(a,n)$

Proof for, If $ p|S(p,2n)$ Then $W(z,W(z,S(z,2n)))=(z-1)r+1=pr+1$

**Proof**

See $S(z,2n)=pr_1+1$

$\implies W(z,W(z,S(z,2n)))$ $\ \ \ by\ identity1$

$=W(z,W(z,pr_1+1))$

$=W(z,pr_2)$

$=pr+1=(z-1)r+1$

For some $r,r_1,r_2\in\mathbb{Z}$

I believe $r$ is always $1$ for all $z>2n+2$, that's my question.

*Related questions on* **MSE**

To count such $p$ which $p\nmid S(p,2n)$