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


Given $a\in \mathbb{Z}_{\ge 4}$ 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 $a$.

this question is equivalent to my unsolved question check

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 $a>3$.


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