Define $X_a$ be the set as, namely$\{ x=(\ \underbrace{ 1\ 1\cdots\ 1\ 1}_{\text{$n$ terms}}\ \ 0 \ \ \underbrace{ \alpha_t\ \alpha_{t-1} \cdots \alpha_1 \ \alpha_0}_{\text{$k$ terms, k=t+1}})_a \mid\ n,k\ge 0\ and \ a-1 \ge \alpha_j\ge \alpha_{j-1} \ge 1 \ for \ t\ge j \ge 1 \} $

and $x\notin \{1,11,111,...\}$

*Note*: $x$ have at most only one '0' digit.

*Example* for set$X_{10}$

$x= \begin{align} 5 \\ 932 \\ 1108552 \\ 1111097322 \\110111 \\ 11103221 \\ 11110 \\ \vdots \end{align}$

For positive integers $n,m$, let $$S(n,m)=\sum_{i=1}^{n}i^m$$ and for positive integers $m,b$, with $b>1$, let $D(b,m)$ be the sum of the base-$b$ digits of $m$.

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

ProblemGiven $a\in \mathbb{Z}_{\ge 4}$ and $m\in \mathbb{Z}_{\ge 1}$

Show that, If $a-1\mid S(a-1,2m)$ and $a-1>2m+1$ then $(f(a,2m))_a\in X_a$

$(f(a,2m))_a$ is representing the value of $f(a,2m)$ in base $a$.

proof for $m=1$

suppose $a$ is a positive integer such that $a \mid S(a,2)$, and let $b=a+1$.

Identically, we have $$ S(n,2) = \sum_{i=1}^n i^2 = \frac{n(n+1)(2n+1)}{6} $$ hence \begin{align*} &a \mid S(a,2)\\[4pt] \implies\;&a{\;|}\left( \frac{a(a+1)(2a+1)}{6} \right)\\[4pt] \implies\;&6 \mid \left((a+1)(2a+1)\right)\\[4pt] \implies\;&6 \mid \left(b(2b-1)\right)\\[4pt] \implies\;&6 \mid b\;\;\text{or}\;\;\Bigl(2 \mid b\;\;\text{and}\;\;3 \mid (2b-1)\Bigr)\\[4pt] \end{align*} If $6 \mid b$, then \begin{align*} S(b,2)&=\frac{b(b+1)(2b+1)}{6}\\[4pt] &=\frac{b^3}{3}+\frac{b^2}{2}+\frac{b}{6}\\[4pt] &= \left({\small{\frac{b}{3}}}\right)\!{\cdot}\,b^2 + \left({\small{\frac{b}{2}}}\right)\!{\cdot}\,b^1 + \left({\small{\frac{b}{6}}}\right)\!{\cdot}\,b^0 \end{align*} hence $$ D(b,S(b,2)) = \left({\small{\frac{b}{3}}}\right) + \left({\small{\frac{b}{2}}}\right) + \left({\small{\frac{b}{6}}}\right) = b $$ If $2 \mid b$ and $3 \mid (2b-1)$, then $b\equiv 2 \pmod3$, so \begin{align*} S(b,2)&=\frac{b(b+1)(2b+1)}{6}\\[4pt] &=\frac{b^3}{3}+\frac{b^2}{2}+\frac{b}{6}\\[4pt] &= \left({\small{\frac{b+1}{3}}}\right)\!{\cdot}\,b^2 + \left({\small{\frac{b-2}{6}}}\right)\!{\cdot}\,b^1 + \left({\small{\frac{b}{2}}}\right)\!{\cdot}\,b^0 \end{align*} hence $$ D(b,S(b,2)) = \left({\small{\frac{b+1}{3}}}\right) + \left({\small{\frac{b-2}{6}}}\right) + \left({\small{\frac{b}{6}}}\right) = b. $$ Thus, for all subcases, we have $D(b,S(b,2))=b$

$\implies D(b,b^3-S(b,2))$

$= 3a+1-D(b,S(b,2))= 2a$

and $2\in X_b$

and also note $a\in \{6t\pm 1\}$ then $a|S(a,2)$ and $a>3$.

Motivation and application of $X_a$

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

Theorem $1$: $W(a+1,ax+1)=a$ iff $x\in X_{a+1}$

*Proof*:

First, some definitions of sets that are crucial to this problem.

Let $ S(k)$ be the set of $k$ digit numbers with digit sum of a.

Let $D(k)$ be the set of $k$ digit numbers whose digits are non-increasing, namely $ \{x=(\ \underbrace{ \alpha_t\ \alpha_{t-1} \cdots \alpha_1 \ \alpha_0}_{\text{$u$ terms, u=t+1}})_{a+1} \mid\ u\ge 0\ and \ a \ge \alpha_j\ge \alpha_{j-1} \ge 1 \ for \ t\ge j \ge 1 \}$.

These are the tail end of the set $X_a$.

**Lemma:** For $k\geq 2$, given $s_k \in S(k)$, $(a+1)^k - s_k - 1 = a d_{k-1}$ iff $d_{k-1} \in D(k-1)$.

**Proof:** Given $d_{k-1} \in D(k-1)$

$a d_{k-1} = ((a+1)-1) d_{k-1} = \underbrace{ (\alpha_t -1 )\ (a -\alpha_{t-1}+\alpha_{t-2}) \cdots (a - \alpha_1+\alpha_0) \ ((a+1) - \alpha_0})$.

Observe that each place value is nonnegative, so this is indeed the base a+1 representation (possibly ignoring leading 0's).

$(a+1)^k -1 - ad_{k-1} = \underbrace{ ((a+1)-\alpha_t )\ (\alpha_{t}-\alpha_{t-1}) \cdots ( \alpha_1-\alpha_0) \ (\alpha_0} - 1)$ The sum of the digits is $a+1-\alpha_t +\alpha_{t}-\alpha_{t-1} + \ldots + \alpha_0 -1 = a$.

For the converse, just reverse these steps.

**Corollary:** Given $s_k \in S(k)$, $(a+1)^{k+n} - s_k - 1 = a x_{k-1}$ iff $x_{k-1} \in X$.

**Proof:** $ \frac{ (a+1)^{k+n} - (a+1)^k } { a} = (a+1)^k \frac{ {\underbrace {a\ a \ a }_\text{$n$ terms}}} {a} = \underbrace {1\ 1 \ 1 }_\text{$n$ terms} \times (a+1)^k$ as desired.

**Corollary** $W(a+1, ax+1) = a $ iff $ (x)_{a+1} \in X_{a+1}$.

**Proof:** This is a restatement of the previous corollary.

Above theorem $1$ help to show

$W(a+1,W(a+1,s(a+1,2m)))=W(a+1,a f(a+1,2m)+1) = a $

iff $(f(a+1,2m))_{a+1}\in X_{a+1}$ for $a>2m+1$

Here $a\mid S(a,2m)$

*this question is equivalent to my unsolved question check*

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