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\begin{center}
\vskip 1cm{\LARGE\bf On Arithmetic Progressions of Integers \\
\vskip .1in
with a Distinct Sum of Digits}
\vskip 1cm
\large
Carlo Sanna \\
Italy\\
\href{mailto:carlo.sanna.dev@gmail.com}{\tt carlo.sanna.dev@gmail.com} \\
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\vskip .2in

\begin{abstract}
Let $b \geq 2$ be a fixed integer. Let $s_b(n)$ denote the sum of digits of 
the nonnegative integer $n$ in the base-$b$ representation. 
Further let $q$ be a positive integer.
In this paper we study the length $k$ of arithmetic progressions 
$n, {n + q}, \ldots, {n + q(k-1)}$ such that $s_b(n), {s_b(n + q)}, \ldots, s_b(n + q(k-1))$ 
are (pairwise) distinct. More specifically, let $L_{b,q}$ denote the supremum of $k$ 
as $n$ varies in the set of nonnegative integers $\mathbb{N}$.
We show that $L_{b,q}$ is bounded from above and hence finite.
Then it makes sense to define $\mu_{b,q}$ as the smallest $n \in \mathbb{N}$ 
such that one can take $k = L_{b,q}$.
We provide upper and lower bounds for $\mu_{b,q}$.
Furthermore, we derive explicit formulas for $L_{b,1}$ and $\mu_{b,1}$.
Lastly, we give a constructive proof that $L_{b,q}$ is unbounded with respect to $q$.
\end{abstract}

\section{Introduction}

Let $b \geq 2$ be a fixed integer and let $s_b(n)$ denote the sum of digits of 
the nonnegative integer $n$ in the base-$b$ representation.
Further let $q$ a positive integer, we are interested in the length $k$ of 
arithmetic progressions $n, {n + q}, \ldots, {n + q(k-1)}$
such that the integers $s_b(n), {s_b(n + q)}, \ldots, {s_b(n + q(k-1))}$ are (pairwise) distinct.

There are known results on the asymptotic behavior of the sum of digits function \cite{ref_Bush,ref_Fang}, 
and about its distribution along arithmetic progressions \cite{ref_Drmota,ref_Gelfond}.
But, to our knowledge, this particular problem has not been studied before.

More specifically, let $L_{b,q}$ denote
the supremum of $k$ as $n$ varies in the set of nonnegative integers $\mathbb{N}$. 
We show that $L_{b,q}$ is bounded from above and hence finite.
As a consequence, it makes sense to define $\mu_{b,q}$ as the smallest $n \in \mathbb{N}$ 
such that one can take $k = L_{b,q}$.
Then, we provide upper and lower bounds for $\mu_{b,q}$. 
Everything allows for an effective computation of $L_{b,q}$ and $\mu_{b,q}$ by checking 
a finite number of candidates, though this is feasible in a short amount 
of time only for small values of $b$ and $q$.
Furthermore, we derive explicit formulas for $L_{b,1}$ and $\mu_{b,1}$.
Lastly, we give a constructive proof that $L_{b,q}$ is unbounded with respect to $q$, 
in the sense that $\sup_{q \in \mathbb N^+} L_{b,q} = +\infty$.

\section{Bounds for $L_{b,q}$ and $\mu_{b,q}$}

\begin{theorem}\label{first_upper_bound}
Let $m$ be the least positive integer such that 
\begin{equation}
m(b-1) + 1 \leq \left\lfloor\frac{b^m}{q}\right\rfloor .
\end{equation}
Then $L_{b,q} \leq 2\!\left\lfloor b^m / q \right\rfloor$.
\end{theorem}
\begin{proof}
Let $n \in \mathbb{N}$, $k \in \mathbb{N}^+$ and $A := \{n + qi : i=0,1,\ldots,k-1\}$
such that $s_b(n)$, $s_b(n+q)$, $\ldots$, $s_b(n+q(k-1))$ are distinct. 
For any $t \in \mathbb{N}$ we define $A_t := A \cap [tb^m, (t+1)b^m-1]$.
For convenience, take $M := \left\lfloor b^m / q\right\rfloor$. 
Then for all nonnegative integers $t &lt; u$ the following statements are true:
\begin{enumerate}[(i).]
\item $|A_t| \leq M$.
\item $A_t, A_u \neq \varnothing \;\Rightarrow\; \forall v \in \mathbb{N}, t &lt; v &lt; u \quad |A_v| = M$.
\item $|A_t| = M, \; t \not\equiv -1 \pmod b \;\Rightarrow\; \forall u \in \mathbb{N}, u &gt; t \quad A_u = \varnothing$.
\item $|A_t| = M, \; t \not\equiv 0 \pmod b \;\Rightarrow\; \forall u \in \mathbb{N}, u &lt; t \quad A_u = \varnothing$.
\end{enumerate}

(i). For all $a \in A_t$ we have $s_b(a) = s_b(t) + s_b(a \bmod b^m)$ and therefore
\begin{equation}
s_b(A_t) := \{ s_b(a) : a \in A_t \} \subseteq \{s_b(t),s_b(t)+1,\ldots,s_b(t)+m(b-1)\} ,
\end{equation}
so by the hypotheses $|A_t| = |s_b(A_t)| \leq m(b-1) + 1 \leq M$.

(ii). Since $A_t$, $A_u$ are nonempty we have $n &lt; vb^m$ and $(v+1)b^m - 1 &lt; n + q(k-1)$. Then
\begin{align}
|A_v| &amp;= \big|\{qi : i = 0, \ldots, k - 1\} \cap \left[vb^m - n, (v+1)b^m - n - 1\right]\big| \nonumber \\ 
    &amp;= \big|q\mathbb{N} \cap \left[vb^m - n, (v+1)b^m - n - 1\right[\big| \geq \left\lfloor\frac{b^m}{q}\right\rfloor = M ,
\end{align}
because $\big|q\mathbb{N} \cap [x,y]\big|\geq \lfloor (y - x + 1) / q \rfloor$ for any integers $y \geq x \geq 0$.
 From point (i) it follows that $|A_v| = M$.

(iii). We have $s_b(t + 1) = s_b(t) + 1$. Suppose by contradiction that $A_{t+1}$ is nonempty, 
so there exists $a := \min(A_{t+1})$. 
Now $s_b(a) = s_b(t) + 1 + s_b(a \bmod b^m)$
and, since $|A_t| = M$ implies $s_b(A_t) = \{s_b(t),s_b(t)+1,\ldots,s_b(t)+m(b-1)\}$, then necessarily
$s_b(a \bmod b^m) = m(b-1)$, so that $a = (t+2)b^m-1$.
In fact, $s_b(a \bmod b^m) \leq m(b-1)$ and if we suppose $s_b(a \bmod b^m) &lt; m(b-1)$
then $s_b(a) \in s_b(A_t)$, in contradiction to our standing hypotheses.
But $q \leq \frac1{M}b^m \leq b^{m-1}$, so $a - q \geq (t + 1)b^m$ and $a - q \in A_{t+1}$, a contradiction.
In conclusion $A_{t+1} = \varnothing$ and, since $q &lt; b^m$, this implies $A_u = \varnothing$.

(iv). Note that $t \geq 1$, we have $s_b(t - 1) = s_b(t) - 1$. 
Suppose that $A_{t-1}$ is nonempty, so there exists $a := \max(A_{t-1})$. 
Then $s_b(a) = s_b(t) - 1 + s_b(a \bmod b^m)$ 
and, since $|A_t| = M$ implies $s_b(A_t) = \{s_b(t),s_b(t)+1,\ldots,s_b(t)+m(b-1)\}$, then it must be
$s_b(a \bmod b^m) = 0$ that is $a = (t-1)b^m$.
But $a + q \leq tb^m - 1$ so $a + q \in A_{t-1}$, a contradiction.
Thus $A_{t-1} = \varnothing$ and, since $q &lt; b^m$, it follows that $A_u = \varnothing$.

The sets $\{A_t\}_{t=0}^\infty$ form a partition of $A$,
hence $A = \bigcup_{t \in \mathbb{N}} A_t$. On the other hand, for the statements proved, we have that
at most two of the sets $\{A_t\}_{t=0}^\infty$ are nonempty and their
cardinality is less than or equal to $M$. In conclusion ${k = |A| \leq 2M}$.
\end{proof}

\begin{corollary}\label{case_q_equal_1}
$L_{b,1} = 2b$, $\mu_{2,1} = 14$ and $\mu_{b,1} = b^3 - b$ if $b \geq 3$.
\end{corollary}
\begin{proof}
 From Theorem \ref{first_upper_bound} we know that $L_{b,1} \leq 2b$. 
It is easy to verify that $L_{2,1} = 4$ and $\mu_{2,1} = 14$ (OEIS \seqnum{A000120}).
If $b \geq 3$ then $L_{b,1} = 2b$ and $\mu_{b,1} \leq b^3 - b$ because 
\begin{equation}
s_b(b^3 - b + i) = \begin{cases}
2(b-1) + i, &amp; \text{ if }\quad i=0,1,\ldots,b-1; \\
i - b + 1, &amp; \text{ if }\quad i = b,b+1,\ldots,2b-1.
\end{cases}
\end{equation}
Now let $n &lt; b^3 - b$ be a nonnegative integer. 
Write $n = d_2 b^2 + d_1 b + d_0$ with $d_0,d_1,d_2 \in \{0,1,\ldots,b-1\}$.
If $d_1 \neq b - 1$ then let $m$ be the least integer greater than or equal to $n$
and not divisible by $b$. We have $m \leq n + 1$ and $s_b(m) = s_b(m+b-1)$.
If $d_1 = b - 1$ and $d_0 = 0$ then $d_2 \leq b-2$, since $n &lt; b^3- b$, and so $s_b(n) = s_b (n + 2b - 2)$.
If $d_1 = b - 1$ and $d_0 \neq 0$ then let $h := (d_2 + 1)b^2$. 
We have $h \leq n + b - 1$ and $s_b(h + 1) = s_b(h + b)$.
In any case we have found two integers $u$, $v$ such that $n \leq u &lt; v \leq n + 2b - 1$
and $s_b(u) = s_b(v)$. Therefore $\mu_{b,1} \geq b^3 - b$ 
and actually $\mu_{b,1} = b^3 - b$.
\end{proof}

\begin{theorem}
$L_{b,bq} = L_{b,q}$ and $\mu_{b,bq} = b\;\;\!\!\! \mu_{b,q}$.
\end{theorem}
\begin{proof}
For all $n,i \in \mathbb{N}$ and $d \in \{0,1,\ldots,b-1\}$
we have $s_b(bn + d + bqi) = s_b(n + qi) + d$. Then for any $k \in \mathbb{N}$ we have that 
${s_b(bn + d)}, {s_b(bn + d + bq)}, \ldots, {s_b(bn + d + (k-1)bq)}$
are distinct if and only if ${s_b(bn)}, {s_b(bn + bq)}, \ldots, {s_b(bn + (k-1)bq)}$
are distinct, which in turn holds if and only if ${s_b(n)}, {s_b(n + q)}, \ldots, {s_b(n + (k-1)q)}$
are distinct. In conclusion ${L_{b,bq} = L_{b,q}}$ and $\mu_{b,bq} = b\;\;\!\!\! \mu_{b,q}$.
\end{proof}

\begin{theorem}\label{minimum_lower_bound}
$\mu_{b,q} &gt; b^{\frac{L_{b,q} - b}{b - 1}} - q(L_{b,q} - 1)$.
\end{theorem}
\begin{proof}
Let $r := \left\lfloor\log_b(\mu_{b,q} + q(L_{b,q} - 1))\right\rfloor + 1$.
Since $s_b(\mu_{b,q}), s_b(\mu_{b,q} + q), \ldots s_b(\mu_{b,q} + q(k-1))$ are distinct 
and less than or equal to $r(b-1)$, then
\begin{equation}\label{ineq1}
L_{b,q} \leq r(b-1) + 1 &lt; (b-1)\log_b(\mu_{b,q} + q(L_{b,q} - 1)) + b .
\end{equation}
Solving (\ref{ineq1}) for $\mu_{b,q}$ we get $\mu_{b,q} &gt; b^{\frac{L_{b,q} - b}{b - 1}} - q(L_{b,q} - 1)$.
\end{proof}

\begin{theorem}\label{minimum_upper_bound}
$\mu_{b,q} \leq b^3[q(L_{b,q} - 1)]^2 - 1$.
\end{theorem}
\begin{proof}
Take $r := \left\lfloor \log_b(q(L_{b,q} - 1))\right\rfloor + 1$, $\mu := (\mu_{b,q} \bmod b^r)$,
 $m := \left\lfloor \mu_{b,q} / b^r\right\rfloor$, so that $\mu_{b,q} = mb^r + \mu$, 
and let $i \in \{0,1,\ldots,L_{b,q}-1\}$. If $\mu + qi &lt; b^r$ then
\begin{equation}
s_b(\mu_{b,q} + qi) = s_b(m) + s_b(\mu + qi) ,
\end{equation}
else if $\mu + qi \geq b^r$ then
\begin{equation}
s_b(\mu_{b,q} + qi) = s_b(m + 1) + s_b(\mu + qi - b^r) ,
\end{equation}
for the fact that $\mu + qi \leq 2b^r - 2$.
Define $n := (b^{r+1} - 1)b^r + \mu$, if $\mu + qi &lt; b^r$ then
\begin{equation}
s_b(n + qi) = s_b(b^{r+1} - 1) + s_b(\mu + qi) = (r + 1)(b - 1) + s_b(\mu + qi) &gt; r(b-1) .
\end{equation}
On the other hand, if $\mu + qi \geq b^r$ then
\begin{equation}
s_b(n + qi) = s_b(b^{r+1}) + s_b(\mu + qi - b^r) = 1 + s_b(\mu + qi - b^r) \leq r(b - 1) ,
\end{equation}
because $\mu + qi - b^r \leq b^r - 2$ and so $s_b(\mu + qi - b^r) \leq r(b - 1) - 1$.
In conclusion $s_b(n), {s_b(n+q)}, \ldots, s_b(n + q(L_{b,q} -1))$ are distinct and
\begin{equation}
\mu_{b,q} \leq n \leq b^{2r+1} - 1 \leq b^3 [q(L_{b,q}-1)]^2 - 1 ,
\end{equation}
this completes our proof. 
\end{proof}

\section{Arbitrarily large values of $L_{b,q}$}

For the next theorem we need a lemma about the sum of digits of multiples of $b^r - 1$, $r \in \mathbb{N}^+$.
Similar results has been considered by Stolarsky \cite[Lemma 2.2]{ref_Stolarsky} in the case $b=2$, and by
Balog and Dartyge \cite[Lemma 1]{ref_Balog} for a generic base $b$.

\begin{lemma}\label{const_digit_sum}
If $r \in \mathbb{N}^+$ then $s_b((b^r - 1)i) = r(b-1)$ 
for all $i=1,2,\ldots,b^r - 1$.
\end{lemma}
\begin{proof}
Let $t$ be the greatest nonnegative integer such that $b^t \mid i$.
If $i_0 := b^{-t}\,i$ then there exist $d_0,d_1,\ldots,d_{r-1} \in \{0,1,\ldots,b-1\}$ 
with $d_0 \neq 0$ such that $i_0 = \sum_{j=0}^{r-1} d_j b^j$ is 
the base-$b$ representation of $i_0$. We have
\begin{align}
(b^r-1)i_0 &amp;= \sum_{j=r}^{2r-1} d_{j-r} b^j - \sum_{j=0}^{r-1} d_j b^j \nonumber \\
 &amp;= \sum_{j=r+1}^{2r-1} d_{j-r} b^j + (d_0 - 1) b^r + b^r - \sum_{j=0}^{r-1} d_j b^j \nonumber \\
 &amp;= \sum_{j=r+1}^{2r-1} d_{j-r} b^j + (d_0 - 1) b^r + \sum_{j=1}^{r-1} (b - 1 - d_j) b^j + (b - d_0) .
\end{align}
Then it is straightforward that
\begin{equation}
s_b((b^r-1)i_0) = \sum_{j=r+1}^{2r-1} d_{j-r} + (d_0 - 1) + \sum_{j=1}^{r-1} (b - 1 - d_j) + (b - d_0) = r(b-1) .
\end{equation}
The claim follows from $s_b((b^r-1)i) = s_b((b^r-1)i_0)$.
\end{proof}

\begin{theorem}
$\sup_{q \in \mathbb{N}^+} L_{b,q} = +\infty$.
\end{theorem}
\begin{proof}
Let $n \in \mathbb{N}$ and $q,k \in \mathbb{N}^+$ such that 
$s_b(n),s_b(n+q), \ldots, s_b(n+q(k-1))$ are distinct.
If $t, r$ are the least positive integers such that $n + qk &lt; b^t$ and $(b-1)b^{r-1} \geq k$
then we define
\begin{align}
n^\prime &amp;= \big((b^t - 1)b^{2r}+ b^{2r-1} + (b^r - 1)((b-1)b^{r-1} - k + 1)\big)b^t + n \nonumber \\
q^\prime &amp;= (b^r - 1)b^t + q .
\end{align}
For any $i=0,1,\ldots,k$ we have
\begin{equation}
n^\prime + q^\prime i = \big((b^t - 1)b^{2r}+ b^{2r-1} + (b^r - 1)((b-1)b^{r-1} + i - k + 1)\big)b^t + n + qi ,
\end{equation}
and then
\begin{equation}
s_b(n^\prime + q^\prime i) = s_b\big((b^t - 1)b^{2r} + b^{2r-1} + (b^r - 1)((b-1)b^{r-1} + i - k + 1)\big) + s_b(n + qi) .
\end{equation}
If $i \leq k - 1$ then $(b^r - 1)((b-1)b^{r-1} + i - k + 1) &lt; (b-1)b^{2r-1}$ and
\begin{equation}
s_b(n^\prime + q^\prime i) = t(b-1) + 1 + s_b\big((b^r - 1)((b-1)b^{r-1} + i - k + 1)\big) + s_b(n + qi) ,
\end{equation}
so Lemma \ref{const_digit_sum} implies that
\begin{equation}
s_b(n^\prime + q^\prime i) = (t + r)(b-1) + 1 + s_b(n + qi) &gt; (t + r)(b-1) .
\end{equation}
On the other hand, if $i = k$ then
\begin{align}
s_b(n^\prime + q^\prime k) &amp;= s_b\big(b^{2r+t} + b^{r-1} - 1\big) + s_b(n + qk) = 1 + (r-1)(b-1) + s_b(n + qk) \nonumber \\
&amp;\leq 1 + (t + r - 1)(b-1) \leq (t + r)(b-1) .
\end{align}
Therefore $s_b(n^\prime), s_b(n^\prime + q), \ldots, s_b(n^\prime + qk)$ are distinct, it follows
that $L_{b,q^\prime} &gt; L_{b,q}$ and hence $\sup_{q \in \mathbb{N}^+} L_{b,q} = +\infty$.
\end{proof}

\section{Further developments}

Thanks to Theorem \ref{first_upper_bound} we know that $L_{b,q}$ is not too large when $q$ 
is small compared to $b$, e.g., if $q \leq \frac1{2}b$ then $L_{b,q} \leq 2b^2$.
Actually, it is likely that for small $q$ there exist explicit formulas for $L_{b,q}$ and $\mu_{b,q}$ 
analogous to those of Corollary \ref{case_q_equal_1}. However, when $q$ is much larger than $b$
the question becomes more difficult. 
Theorem \ref{first_upper_bound} and \ref{minimum_upper_bound} 
allowed us, with the aid of a personal computer, to calculate some values of $\mu_{b,q}$ and $L_{b,q}$.

\begin{center}
\renewcommand{\arraystretch}{1.2}
\begin{tabular}{|c|*{9}{c|}}
\hline
$\mu_{b,q} ,  L_{b,q}$  
         &amp; $q=1$   &amp; $q=2$   &amp; $q=3$   &amp; $q=4$    &amp; $q=5$     &amp; $q=6$    &amp; $q=7$     &amp; $q=8$    &amp; $q=9$     \\
\hline
$b = 2$  &amp; $14, 4$ &amp; $28, 4$ &amp; $58, 4$ &amp; $56, 4$  &amp; $242, 6$  &amp; $116, 4$ &amp; $109, 5$  &amp; $112, 4$ &amp; $994, 6$  \\
\hline
$b = 3$  &amp; $24, 6$ &amp; $24, 3$ &amp; $72, 6$ &amp; $234, 5$ &amp; $705, 9$  &amp; $72, 3$  &amp; $697, 10$ &amp; $18, 3$  &amp; $216, 6$  \\
\hline
$b = 4$  &amp; $60, 8$ &amp; $56, 8$ &amp; $60, 3$ &amp; $240, 8$ &amp; $1004, 8$ &amp; $244, 4$ &amp; $977, 13$ &amp; $224, 8$ &amp; $239, 4$  \\
\hline
\end{tabular}
\end{center}

\begin{center}
Table 1: Values of $\mu_{b,q}$ and $L_{b,q}$ for $b=2,3,4$ and $q=1,2,\ldots,9$.
\end{center}

Through these series of numerical experiments we have reason to believe 
that the upper bound given by Theorem \ref{first_upper_bound} is in some 
sense ``astronomical'' and surely can be improved. Similarly, also the upper and lower bounds for $\mu_{b,q}$ can be improved.

Many other questions remain unsolved.
For instance, let $\kappa_{b,q}$ be the function
sending the nonnegative integer $n$ to the maximum $k \in \mathbb{N}$ such that
$s_b(n), {s_b(n + q)}, \ldots, {s_b(n + q(k - 1))}$ are distinct.
We proved that the function $\kappa_{b,q}$ is bounded.
Is $\kappa_{b,q}$ definitely periodic? 
Does it present a fractal behavior?
What is its Fourier expansion? 
What is its mean, variance, etc.? 
 On the other side, for which $n \in \mathbb{N}$ is $\kappa_{b,q}(n)$ particularly small?
These and other questions will be the subject of future investigations.

\section{Acknowledgements}
I am grateful to Salvatore Tringali
(LJLL, Universit\'e Pierre et Marie Curie) for his careful proofreading
and, more generally, his valuable support in the field of Mathematics.
I am also thankful to the anonymous referee for his constructive comments.

\vspace{10pt}

\begin{thebibliography}{10}

\bibitem{ref_Balog}
A. Balog and C. Dartyge, 
On the sum of the digits of multiples,
{\em Moscow J. Comb. Number Th.} {\bf 2} (2012), 3--15.

\bibitem{ref_Bush} 
L. E. Bush, 
An asymptotic formula for the average sum of the digits of integers, 
{\em Amer. Math. Monthly} {\bf 47} (1940), 154--156.

\bibitem{ref_Drmota}
M. Drmota, C. Mauduit, and J. Rivat, 
The sum of digits function of polynomial sequences, 
{\em J. Lond. Math. Soc.} {\bf 84} (2011), 81--102.

\bibitem{ref_Fang}
Y. Fang, 
A theorem on the k-adic representation of positive integers,
{\em Proc. Amer. Math. Soc.} {\bf 130} (2002), 1619--1622.

\bibitem{ref_Gelfond}
A. O. Gelfond, 
Sur les nombres qui ont des propri\'et\'e  additives et multiplicative donn\'ees, 
{\em Acta Arith.} {\bf 13} (1967/1968), 259--265.

\bibitem{ref_Stolarsky}
K. B. Stolarsky, 
Integers whose multiples have anomalous digital frequencies,
{\em Acta Arith.} {\bf 38} (1980), 117--128.

\end{thebibliography}

\bigskip
\hrule
\bigskip

\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11A63;
Secondary 11A25, 11B25.

\noindent \emph{Keywords: } 
Elementary number theory, radix representation, sum of digits, arithmetic progressions.

\bigskip
\hrule
\bigskip

\noindent (Concerned with sequence
\seqnum{A000120}.)

\bigskip
\hrule
\bigskip

\vspace*{+.1in}
\noindent
Received  August 5 2012;
revised version received September 23 2012.  
Published in {\it Journal of Integer Sequences}, October 2 2012.

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