Sum of the digits in base $p+1$

Question

Check link for M.O. post https://mathoverflow.net/q/346862/149083

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 it's 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 $W(6,W(6,S(6,2)))=W(6,W(6,91))=W(6,10)=6$

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

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

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

Special observation on prime number and π(n)

Answer 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 $(f(a,2m))_a\in X_a$

half 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 $\implies 2\le r_2 < p$

Hence $W(z,pr_2)=p+1=z$

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Reference

https://mathoverflow.net/q/347742/149083