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 $a$.

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 $X_a$


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 $x\in X_{a+1}$


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

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

  2. 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

Recent same posts on M.S.E.

Proof for $W(10,9x+1)=9$ iff $x\in X_{10}$

  • $\begingroup$ Simulposted to m.se, math.stackexchange.com/questions/3465331/… without notice to this site. That's an abuse of the system. Please don't do that. $\endgroup$ Dec 7 '19 at 5:39
  • $\begingroup$ @GerryMyerson actually I didn't know that we can't posted at same time on both site without notice. I'm sorry, I'll take care $\endgroup$
    – Pruthviraj
    Dec 7 '19 at 6:43
  • $\begingroup$ You not supposed to post to both sites at the same time, whether you give notice or not, but it's much worse when you do it without giving notice. You are supposed to post to one site; then, if you don't get a satisfactory response after, say, a week, you can post to the other, remembering to leave a link at each site to the post at the other. $\endgroup$ Dec 7 '19 at 10:34

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 $$f(a,2m) = \frac{D(a,a^{2m+1} - S(a,2m))}{a-1} \leq 2m+1-q.$$

We have $f(a,2m)<2m+1<a-1$, and thus $(f(a,2m))_a$ forms a single digit.

So, to prove $(f(a,2m))_a\in X_a$ it remains to show that $f(a,2m)\ne 1$.

Suppose that $f(a,2m)=1$. Then $$a-1 = D(a,a^{2m+1} - S(a,2m)) = D(a,(a-1)a^{2m} - S(a-1,2m)).$$

Let $S(a-1,m) = ua^{2m}+v$, where $v<a^{2m}$. Noticing that $S(a-1,2m) < \frac{a^{2m+1}}{2m+1}$, we conclude that $u\leq \lfloor \frac{a}{2m+1} \rfloor$.

Now, $(a-1)a^{2m} - S(a-1,2m) = (a-2-u)a^{2m} + (a^{2m}-v)$, and thus $D(a,(a-1)a^{2m} - S(a-1,2m)) = a-2-u + D(a,a^{2m}-v)$, implying that $D(a,a^{2m}-v) = u+1$, i.e. $$v=a^{2m} - a^{k_1} - \dots - a^{k_{u+1}},$$ which heuristically is unlikely for small $u$. (I do not have rigorous proof.)

  • $\begingroup$ Actually I am not fully understand your proof,How do you show that If $(a-1)\mid S(a-1,2m)$ Then $D(a,a^{2m+1}-S(a,2m))+D(a,S(a-1,2m))=(a-1)(2m+1)$? $\endgroup$
    – Pruthviraj
    Jan 3 '20 at 16:38
  • $\begingroup$ Also ask on MSE math.stackexchange.com/q/3496119/647719 $\endgroup$
    – Pruthviraj
    Jan 3 '20 at 19:45
  • $\begingroup$ Professor Can you please elaborate, how you conclude $f(a,2m)=2m+1-q$? $\endgroup$
    – Pruthviraj
    Jan 4 '20 at 20:23
  • $\begingroup$ @Pruthviraj: It was a typo. It should be $f(a,2m) \leq 2m+1-q$, now corrected. Thanks for noticing. $\endgroup$ Jan 9 '20 at 3:40
  • $\begingroup$ Thanks professor, last problem, please can you also elaborate, how $2m+1-q\ge 2$? $\endgroup$
    – Pruthviraj
    Jan 9 '20 at 7:42

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