
> 
> 
> > 
> > *Definition*
> > 
> > 
> > Let $W$ 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 $p+1=z$



> 
> **Question**
> 
> 
> 
> > 
> > show 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**


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


[for formula](https://math.stackexchange.com/q/3265732/647719)


**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 $p\nmid S(p,2n)$](https://math.stackexchange.com/q/3251653/647719)


[Special observation on prime number and π(n)](https://math.stackexchange.com/q/3271631/647719)


[Same question](https://math.stackexchange.com/q/3398042/647719)



Sum of the digits in base $p+1$
Define $X\_a$ be the set as, $\{2,3,...,a-1,a\}$


let $D(b,m)$ be the sum of the base-$b$ digits of $m$.


Define $f(a,k)=\frac{D(a,a^{k+1}-S(a,k))}{a-1}$


**Theorem**:


Given $a\in \mathbb{Z}\_{\ge 4}$ and $m\in \mathbb{Z}\_{\ge 1}$, If $a-1\mid S(a-1,2m)$ and $a-1>2m+1$ then $(f(a,2m))\_a\in X\_a$


**incomplete Proof**: This proof is incomplete to show $1\notin f(a,2m)$ but I assumed here it's true.


Clearly, we have $(a-1)|S(a-1,2m)$ iff $(a-1)|D(a,S(a-1,2m))$.


Let $q:=\frac{D(a,S(a-1,2m))}{a-1}$. Since for $a\geq 4$ and $m\geq 1$,
$S(a-1,2m) < (a-1)a^{2m}$ and $S(a,2m) = S(a-1,2m) + a^{2m}$, we have $D(a,S(a,2m)) = 1+q(a-1)$. Then
$$f(a,2m) = \frac{D(a,a^{2m+1} - a^{2m} - S(a-1,2m))}{a-1} \le 2m+1-q.$$


Since $2m+1-q<2m+1<a-1$, we conclude that $(f(a,2m))\_a$ forms a single digit $2m+1-q\geq 2$, and thus $(f(a,2m))\_a\in X\_a$.


**Corollary 1:** $W(a+1,ax+1)=a$ if $x\in X\_{a+1}$


**Corollary 2:** If $ p|S(p,2n)$ and $p\ge 2n+1$ Then $W(z,W(z,S(z,2n)))=z$


**Proof**


For $z>2n+2$


See $S(z,2n)=pr\_1+1$ For some $r\_1\in\mathbb{Z}$


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


$=W(z,W(z,pr\_1+1)) $


$=W(z,pr\_2)$


Here by theorem 2 $\implies 2\le r\_2 < p$


Hence $W(z,pr\_2)=p+1=z$




---


Reference


[Show that, If $a-1\mid S(a-1,2m)$ and $a-1>2m+1$ then $(f(a,2m))\_a\in X\_a$](https://mathoverflow.net/q/347742/149083)


