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

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

Problem

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 $$a>3$$.