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\begin{center}
\vskip 1cm{\LARGE\bf Infinite Sets of $b$-Additive and\\
\vskip .1in
$b$-Multiplicative Ramanujan-Hardy Numbers}
\vskip 1cm
\large
Viorel Ni\c tic\u a\\
Department of Mathematics\\
West Chester University of Pennsylvania\\
West Chester, PA 19383\\
USA\\
\href{mailto:vnitica@wcupa.edu}{\tt vnitica@wcupa.edu} \\
\end{center}
\vskip .2 in
\begin{abstract} Let $b$ a numeration base. A $b$-additive Ramanujan-Hardy
number $N$ is an integer for which there exists at least one integer $M$,
called the additive multiplier, such that the product of $M$ and the sum
of base-$b$ digits of $N$, added to the reversal of the product, gives
$N$. We show that for any $b$ there exist infinitely many $b$-additive
Ramanujan-Hardy numbers and infinitely many additive multipliers. A
$b$-multiplicative Ramanujan-Hardy number $N$ is an integer for which
there exists at least an integer $M$, called the multiplicative multiplier,
such that the product of $M$ and the sum of base-$b$ digits of $N$,
multiplied by the reversal of the product, gives $N$. We show that
for $b\equiv 4 \pmod {6}$, and for $b=2$, there exist infinitely many
$b$-multiplicative Ramanujan-Hardy numbers and infinitely many
multiplicative multipliers. If $b$ even, $b\equiv 0 \pmod {3}$ or
$b\equiv 2 \pmod {3},$ we show there exist
infinitely many numeration bases for which
there exist infinitely many $b$-multiplicative Ramanujan-Hardy numbers
and infinitely many multiplicative multipliers.
These results completely answer two questions and partially answer
two other questions asked in a previous paper of the author.
\end{abstract}
\section{Introduction}\label{sec:1}
Let $b\ge 2$ be a numeration base. In Ni\c tic\u a \cite{N}, motivated by some properties of the taxicab number, 1729, we introduce the classes of {\it $b$-additive Ramanujan-Hardy (or $b$-ARH) numbers} and {\it $b$-multiplicative Ramanujan-Hardy (or $b$-MRH) numbers}. The first class consists of numbers $N$ for which there exists at least an integer $M$, called the
\emph{additive multiplier}, such that the product of $M$ and the sum of base-$b$ digits of $N$, added to the reversal of the product, gives $N$. The second class consists of numbers $N$ for which there exists at least an integer $M$,
called the \emph{multiplicative multiplier}, such that the product of $M$ and the sum of base-$b$ digits of $N$, multiplied by the reversal of the product, gives $N$.
It is asked \cite[Question 6]{N} if the set of $b$-ARH numbers is infinite and it is asked \cite[Question 8]{N} if the set of additive multipliers is infinite. It is shown \cite[Theorems 12 and 15]{N} that the answer is positive if $b$
is even. The case $b$ odd is left open. It is asked \cite[Question 7]{N} if the set of $b$-MRH numbers is infinite and it is asked \cite[Question 9]{N} if the set of multiplicative multipliers is infinite. It is shown \cite[Theorem 30]{N} that the answer is positive if $b$ is odd. The case $b$ even is left open.
We recall that \emph{Niven (or Harshad) numbers} are numbers divisible by the sum of their decimal digits. Niven numbers have been extensively studied. See, for instance, Cai \cite{C}, Cooper and Kennedy \cite{CK}, De Koninck and Doyon \cite{KD}, and Grundman \cite{G}. Of interest are also $b$-Niven numbers, which are numbers divisible by the sum of their base-$b$ digits. See, for example, Fredricksen, Iona\c scu, Luca, and St\u anic\u a \cite{FILS}. A $b$-MRH-number is a $b$-Niven number. High degree $b$-Niven numbers are introduced in \cite{N2}.
The goal of this paper is to show that, for any numeration base, there
exist infinitely many $b$-ARH numbers and infinitely many distinct
additive multipliers. We also show that, for $b\equiv 4 \pmod {6}$,
and for $b=2$, there exist infinitely many $b$-MRH numbers, and
infinitely many distinct multiplicative multipliers. If $b$ even,
$b\equiv 0 \pmod {3}$ or $b\equiv 2 \pmod {3},$ we show there are
infinitely many numeration bases for which there exist infinitely
many $b$-multiplicative Ramanujan-Hardy numbers and infinitely many
multiplicative multipliers. These results completely answer the first
two questions from \cite{N} revisited above, and partially answer the
other two. We observe that a trivial example of infinitely many $b$-MRH
numbers is given by the powers of $10$. Our examples have at least two
digits different from zero. Finding infinitely many $b$-MRH numbers with
all digits different from zero remains an open question.
Our results about $b$-ARH numbers also give solutions to the Diophantine
equation $N\cdot M=\text{reversal}(N\cdot M)$. Motivated by this link,
we show that the Diophantine equation has solution for all integers $N$
not divisible by the numeration base $b$. We do not know how to answer
the following related question:
\begin{question} Does there exist, for any integer $N$, an integer $M$ such that $N\cdot M$ is a $b$-ARH number (or a $b$-MRH number, or a $b$-Niven number)?
\end{question}
Our final result shows that for any string of digits $I$ there exist
infinitely many $b$-Niven numbers that contain $I$ in their base-$b$ representation. We do not know a similar result for the classes of $b$-ARH and $b$-MRH numbers.
\section{Statements of the main results}\label{sec:1-main}
Let $s_b(N)$ denote the sum of base-$b$ digits of integer $N$. If $x$ is a string of digits, let $(x)^{\land k}$ denote the base-$10$ integer obtained by repeating $x$ $k$-times. Let $[x]_b$ denote the value of the string $x$ in base $b$.If $N$ is an integer, let $N^R$ denote the reversal of $N$, that is, the number obtained from $N$ writing its digits in reverse order. The operation of taking the reversal is dependent on the base. In the definition of a $b$-ARH-number/$b$-MRH number $N$ we take the reversal of the base-$b$ representation of $s_b(N)M$.
\begin{theorem}\label{thm:1} Let $\alpha\ge 1$ integer, $b\ge \alpha+1$ integer, and $k=(1+\alpha)^\ell, \ell\ge 0$.
Assume $b\equiv 2+\alpha \pmod{2+2\alpha}$. Define
\begin{equation*}
N_k=[(1\alpha)^{\land k}]_b.
\end{equation*}
Then there exists $M\ge 0$ integer such that
\begin{equation*}
s_b(N_k)\cdot M=(s_b(N_k)\cdot M)^R=\frac{N_k}{2}.
\end{equation*}
In particular, the numbers $N_k, k\ge 1,$ are $b$-ARH numbers and $b$-Niven numbers.
\end{theorem}
The proof of Theorem \ref{thm:1} is done in Section \ref{sec:2}.
\begin{remark} The particular case $b=10, \alpha=2,$ of Theorem \ref{thm:1}, which gives $N_k=(12)^{3^\ell}$, is covered by \cite[Example 10]{N}.
Theorem \ref{thm:1} does not give any information if $b=2$.
\end{remark}
The following proposition gives positive answers to \cite[Questions 5 and 6]{N}.
\begin{proposition}\label{prop:1} For any $b\ge 2$, there exist infinitely
many $b$-ARH numbers and infinitely many additive multipliers. The $b$-ARH numbers are also $b$-Niven numbers.
\end{proposition}
The proof of Proposition \ref{prop:1} is done in Section \ref{sec:3-pre}.
\begin{remark} Note that \cite[Theorems 12 and 15]{N} show, for all even bases, infinitely many $b$-ARH numbers that are not $b$-Niven numbers. The case of odd
base is open. The question of finding infinitely many $b$-Niven numbers that are not $b$-ARH numbers is also open. It is shown in \cite[Theorem 28]{N} that for any base there exist infinitely many numbers that are not $b$-ARH numbers.
\end{remark}
The result in Theorem \ref{thm:1} gives many base-$10$ solutions for the equation:
\begin{equation}\label{eq:star10}
N\cdot M=(N\cdot M)^R.
\end{equation}
One can try to solve the equation \eqref{eq:star10} ,where $(N\cdot M)^R$ is the reversal of $N\cdot M$ written in base $b$, for any numeration base $b$.
Observe that if $N$ is divisible by $b$, then $(N\cdot M)^R$ has less digits then $N\cdot M$, therefore $N$ is not a solution of \eqref{eq:star10}. Note also that if $N=N^R$ and $N$ has $k$ digits then \eqref{eq:star10} always has an infinite set of solutions with
\begin{equation*}
M=[(1(0)^{\land\ell})^{\land p}1]_b, \ell \ge k-1, p\ge 0.
\end{equation*}
Consequently, if $(N_0,M_0)$ is a solution of \eqref{eq:star10}, then \eqref{eq:star10} has infinite sets of solutions of types $(N_0,M)$ and $(N,M_0)$.
\begin{theorem}\label{thm:new} Let $b\ge 2$ and $N\ge 1$ integer such that $b\not \vert N$. Then $N$ is a solution of \eqref{eq:star10}.
\end{theorem}
The proof of Theorem \ref{thm:new} is done in Section \ref{sec:3}. For
base 10, a proof belonging to David Radcliffe can be found at \cite{Rat}. We learned about this reference from J. Shallit. We generalize the proof for an arbitrary numeration base. After our paper was written, we learned from J. Shallit \cite{S} that he also has a proof of Theorem \ref{thm:new}.
A \emph{$b$-numeric palindrome} is a base-$b$ integer $N$ such that $N=N^R$.
\begin{corollary} All integers, not divisible by $b$, are factors of $b$-numeric palindromes.
\end{corollary}
\begin{definition} The \emph{multiplicity} of a multiplicative multiplier $M$ is the number of $(N,M)$ solutions of \eqref{eq:star10}.
\end{definition}
It was observed above that for any solution $(N,M)$ of \eqref{eq:star10}, $M$ has infinite multiplicity. The following theorem shows infinitely many solutions of \eqref{eq:star10} independent of above.
\begin{theorem}\label{thm:33} Let $b\ge 2$ a numeration base. Then, for all $k\ge 0$, we have
\begin{equation*}
[1(b-1)]_b\cdot [(b-1)^{\land k}]_b=[1(b-2)(b-1)^{\land k-2}(b-2)1]_b.
\end{equation*}
\end{theorem}
The proof of Theorem \ref{thm:33} is done in Section \ref{sec:44}.
Our next results show, for $b$ even, more examples of infinite sets of of $b$-ARH.
\begin{theorem}\label{thm:final} Let $b\ge 2$ even.
Let $a\in\{1,2,\dots,b-1\}$ and let $k\ge 0$ be an integer.
\noindent (a) Let
\begin{equation*}
N_k=[a(0)^{\land k}a]_b.
\end{equation*}
Then $N_k$ is a $b$-ARH number, but not a $b$-Niven number.
\bigskip
\noindent (b) Let
\begin{equation*}
N_k=[\left ( 1(0)^{\land k} \right )^{\land b}0 \left ( (0)^{\land k} 1\right )^{\land b}]_b.
\end{equation*}
Then $N_k$ is a $b$-ARH number, but not a $b$-Niven number.
\bigskip
\noindent (c) Let
\begin{equation*}
N_k=[\left ( (0)^{\land k}1 \right )^{\land b}0 \left (1 (0)^{\land k} \right )^{\land b}]_b.
\end{equation*}
Then $N_k$ is a $b$-ARH number and a $b$-Niven number.
\end{theorem}
The proof of Theorem \ref{thm:final} is done in Section \ref{sec:final}.
The following theorem gives partial answers to \cite[Questions 7 and 8]{N}.
\begin{theorem}\label{thm:final2}
\leavevmode
\vphantom{x} \\
\noindent (a) Let $b\equiv 4\pmod{6}$. Let $k\ge 1$ integer such that $k\equiv 1\pmod{3}$. Define
\begin{equation*}
\alpha_k=[1(0)^{\land k}(b-2)]_b.
\end{equation*}
Then $N_k=\alpha_k\cdot(\alpha_k)^R$ is a $b$-MRH number.
\medskip
\noindent (b) Let $b=2$ and let $k\ge 1$ be an even integer. Define
\begin{equation*}
\alpha_k=[1(0)^{\land k}1]_2.
\end{equation*}
Then $N_k=\alpha_k\cdot(\alpha_k)^R$ is a $b$-MRH number.
In particular, for any numeration base $b$, $b\equiv 4\pmod{6},$ and for $b=2,$ there exist infinitely many $b$-MRH numbers and infinitely many multipliers.
\end{theorem}
The proof of Theorem \ref{thm:final2} is done in Section \ref{sec:final2}.
Our next result lists several infinite sequences of $10$-MRH-numbers.
\begin{proposition}\label{prop:examplesMRH} Assume $k\ge 1$ integer and define $N_k=\alpha_k\cdot (\alpha_k)^R$, where $\alpha_k$ is one of the following numbers:
\begin{itemize}
\item $[1(0)^{\land k} 8]_{10}, k\equiv 1\pmod{3}$,
\item $[7(0)^{\land k} 2]_{10},$
\item $[5(0)^{\land k} 4]_{10},$
\item $[4(0)^{\land k} 5]_{10}$
\end{itemize}
Then $N_k$ is a 10-MRH number.
\end{proposition}
The first item in Proposition \ref{prop:examplesMRH} follows as a corollary of Theorem \ref{thm:final2}. The other items can be proved using the same approach as in the proof of Theorem \ref{thm:final2}.
If $b$ even, $b\equiv 0 \pmod {3}$ or $b\equiv 2 \pmod {3}$, the
next theorem shows there are infinitely many numeration bases for which there exist infinitely many $b$-MRH numbers and infinitely many
multipliers.
\begin{theorem}\label{thm:3k0,2}
\leavevmode
\vphantom{x} \\
\noindent (a) Let $b\ge 18$, $b=6a$, and $ a\equiv 1\pmod {25}$. Let $\alpha_k=[1 (0)^{\land k}4]_b$ with $k\equiv 4\pmod 5$. Then $N_k=\alpha_k\cdot \alpha_k^R$ is a $b$-MRH number. The corresponding multipliers are distinct.
\bigskip
\noindent (b) Let $b\ge 18$, $b=8a$, $a\equiv 1\pmod {25}$, and $a\equiv 1 \pmod{3}$. Let $\alpha_k=[1 (0)^{\land k}4]_b$ with $k\equiv 4\pmod {20}$. Then $N_k=\alpha_k\cdot \alpha_k^R$ is a $b$-MRH number. The corresponding multipliers are distinct.
\end{theorem}
The proof of Theorem \ref{thm:3k0,2} is done in section \ref{sec:thm3k02}.
\begin{theorem}\label{thm:last-thm} For any base $b$ and for any string of base $b$ digits $I$ there exist infinitely many $b$-Niven numbers that contain the string $I$ in their base-$b$ representation.
\end{theorem}
\begin{proof} Let $I$ be a string of base-$b$ digits. There exist
infinitely many base-$b$ strings $J$ such that $s_b([IJ]_b)$ is a power of $b$, say $b^k, k\ge 1$. Then the number $N_J=[IJ(0)^{\land k}]_b$ is a $b$-Niven number.
\end{proof}
\section{Proof of Theorem \ref{thm:1}}\label{sec:2}
\begin{proof} The condition $b\equiv 2+\alpha \pmod{2+2\alpha}$ implies that $b+\alpha$ is even. The base-$b$ representation for $N_k/2$ is $N_k/2=\left [ \left ( 0\frac{b+\alpha}{2}\right )^{\land k}\right ]_b$. One has that:
\begin{equation}\label{eq:55}
s_b(N_k)=k\cdot(1+\alpha)=(1+\alpha)^{\ell+1}.
\end{equation}
The value of $N_k/2$ in base 10 is obtained summing a geometric series.
\begin{align}\label{eq:66}
\frac{N_k}{2}&=\frac{b+\alpha}{2}\cdot b^{2k-2}+\frac{b+\alpha}{2}\cdot b^{2k-4}+\cdots
+ \frac{b+\alpha}{2}\cdot b^{2}+\frac{b+\alpha}{2}=\frac{b+\alpha}{2}\cdot \frac{b^{2k}-1}{b^2-1} \nonumber \\
&=\frac{b+\alpha}{2}\cdot \frac{b^{2(1+\alpha)^{\ell}}-1}{b^2-1}.
\end{align}
Note that $N_k/2=(N_k/2)^R$. We finish the proof of the theorem if we show that:
\begin{equation}\label{eq:1}
(1+\alpha)^{\ell+1} \Big \vert \frac{b+\alpha}{2}\cdot \frac{b^{2(1+\alpha)^{\ell}}-1}{b^2-1}.
\end{equation}
We prove \eqref{eq:1} by induction on $\ell$. For $\ell=0$ equation \eqref{eq:1} becomes $1+\alpha \Big \vert \frac{b+\alpha}{2},$
which is true because $b\equiv 2+\alpha \pmod{2+2\alpha}$.
Now we assume that \eqref{eq:1} is true for $\ell$ and show that it is true for $\ell+1$.
\begin{equation}\label{eq:account}
\begin{gathered}
\frac{b+\alpha}{2}\cdot \frac{b^{2(1+\alpha)^{\ell+1}}-1}{b^2-1}=\frac{b+\alpha}{2}\cdot \frac{\left ( b^{2(1+\alpha)^{\ell}}\right )^{1+\alpha}-1}{b^2-1}\\
=\frac{b+\alpha}{2}\cdot \frac{b^{2(1+\alpha)^{\ell}}-1}{b^2-1}\left ( B^{\alpha}+B^{\alpha-1}+\cdots +B^2+B+1\right ),
\end{gathered}
\end{equation}
where
\begin{equation}\label{eq:B}
B=b^{2(1+\alpha)^{\ell}}.
\end{equation}
The congruence $b\equiv 2+\alpha \pmod{2+2\alpha}$ implies that
\begin{equation*}
b^2\equiv (2+\alpha)^2\equiv \alpha^2+4\alpha+4\equiv \alpha^2\equiv 1 \pmod{1+\alpha},
\end{equation*}
which implies that
\begin{equation}\label{eq:bsquare}
b^m\equiv 1 \pmod{1+\alpha}, \text{ $m$ even}.
\end{equation}
From \eqref{eq:B} and \eqref{eq:bsquare} follows that $B^p\equiv 1 \pmod{1+\alpha}, 1\le p\le \alpha$, so
\begin{equation}\label{eq:bigB}
1+\alpha\vert B^{\alpha}+B^{\alpha-1}+\cdots +B^2+B+1.
\end{equation}
Combining \eqref{eq:1} (for $\ell$) and \eqref{eq:bigB}, and using \eqref{eq:account}, it follows that \eqref{eq:1} is true for $\ell+1$.
\end{proof}
\section{Proof of Proposition \ref{prop:1}} \label{sec:3-pre}
\begin{proof} The case $b=2$ is covered by \cite[Theorem 12]{N}. If $b\ge 3$, choose $\alpha=b-2$ and apply Theorem \ref{thm:1}. We show that, for a fixed $b$, the multipliers appearing in the proof of Theorem \ref{thm:1} are all distinct. It follows from \eqref{eq:55} and \eqref{eq:66} that the multiplier for $N_k$ is given by:
\begin{equation}\label{eq:comp}
M=\frac{\frac{N_k}{2}}{s_b(N_k)}=\frac{\frac{b+\alpha}{2}\cdot\frac{b^{2(1+\alpha)^\ell}-1}{b^2-1}}{(1+\alpha)^{\ell+1}}.
\end{equation}
Note that $\alpha=b-2$. After algebraic manipulations, equation \eqref{eq:comp} becomes
\begin{equation*}
M=\frac{b^{2(1+\alpha)^{\ell}}-1}{(b-1)^{\ell}(b^2-1)}.
\end{equation*}
In order to show that the multipliers are distinct it is enough to show that the sequence of multipliers is strictly increasing as a function of $\ell$ That is, we need to show that:
\begin{equation}\label{eq:22}
\frac{b^{2(1+\alpha)^{\ell}}-1}{(b-1)^{\ell}(b^2-1)}<\frac{b^{2(1+\alpha)^{\ell+1}}-1}{(b-1)^{\ell+1}(b^2-1)}.
\end{equation}
After algebraic manipulations \eqref{eq:22} becomes
\begin{equation}\label{eq:33}
(b-1)(b^{2(1+\alpha)^{\ell}}-1)< b^{2(1+\alpha)^{\ell+1}}-1.
\end{equation}
After denoting
\begin{equation*}
B=b^{2(1+\alpha)^{\ell}}=b^{2(b-1)^{\ell}},
\end{equation*}
right hand side of \eqref{eq:33} factors as follows:
\begin{equation}\label{eq:131}
b^{2(1+\alpha)^{\ell+1}}-1=(b^{2(1+\alpha)^{\ell}}-1)(B^{\alpha}+B^{\alpha-1}+\cdots +B+1).
\end{equation}
Now \eqref{eq:33} follows from \eqref{eq:131} and the following inequality:
\begin{equation*}
b-1< b^{2(b-1)^{\ell}}, \ell\ge 0, \ell\ge 0, b\ge 3.
\end{equation*}
\end{proof}
\section{Proof of Theorem \ref{thm:new}}\label{sec:3}
\begin{proof} Let $b=p_1^{\alpha_1}p_2^{\alpha_2}\cdots p_k^{\alpha_k}, \alpha_i\ge 1$, $p_i$ prime, $1\le i\le k$. We recall that a base-$b$ integer $N$ is divisible by $p_i^{\gamma}$ if the last $\gamma$ digits of $N$ form a base-$b$ integer divisible by $p_i^{\gamma}$.
Let $N=p_1^{\beta_1}p_2^{\beta_2}\cdots p_k^{\beta_k}w$, where $\gcd(w,b)=1$. Let $m=\max(\beta_1, \beta_2, \cdots , \beta_k)$. Let $L$ be the base-$b$ integer equal to $p_1^{\beta_1}p_2^{\beta_2}\cdots p_k^{\beta_k}$. As $b\not\vert N$, the last digit of $L$ is not $0$. Let $\ell$ be the length of $L$. Consider the
base-$b$ palindrome $P=[L^R(0)^{\land {m-\ell}}L]_b$, where $L^R$ is the reversal of base-$b$ representation of $L$. As $P$ is divisible by $p_1^{\beta_1}p_2^{\beta_2}\cdots p_k^{\beta_k}$, this is the end of the proof if $w=1$.
Assume $w>1$. Let $\phi$ be Euler's totient function which counts the positive integers up to a given integer $n$ that are relatively prime to $n$. As $\gcd(w,b)=1$ Euler's theorem implies that $b^{\phi(w)} - 1\equiv 0 \pmod w$.
Let $r$ be an multiple of $\phi(w)$ which is greater than $l+m$, the length of $P$. Let $q\ge 1$ a multiple of $b^{\phi(w)}-1$.
Consider
the infinite family of integers given by
\begin{align}
\label{eq:newalpha}
Q_{r,q}&=\big[1\big((0)^{\land r-1}1\big)^{\land q}\big]_b=1+b^{r}+b^{2r}+\cdots+b^{qr}\nonumber\\
&=1+b^{r}+b^{2r}+\cdots+b^{qr}+q-q\\
&=\big (b^{r}-1\big )+\big (b^{2r}-1\big )+\big (b^{3r}-1\big )+\cdots + \big (b^{qr}-1\big )+q. \nonumber
\end{align}
All terms in the last part of \eqref{eq:newalpha} are divisible by $b^{\phi(w)}-1$, so $Q_{r,q}$ is divisible by $b^{\phi(w)}-1$ and by $w$.
We finish the proof observing that $P\cdot Q_{r,q}$ is a base-$b$ palindrome divisible by $N$.
\end{proof}
\section{Proof of Theorem \ref{thm:33}}\label{sec:44}
\begin{proof} Observe that:
\begin{equation}\label{eq:77}
\begin{gathered}
(b-1)\cdot (b-1)=b(b-2)+1=[(b-2)1]_b\\
(b-1)b^{k}+(b-1)b^{k}=b^k+(b-2)b^{k-1}=[1(b-2)0^{\land k}]_b.
\end{gathered}
\end{equation}
Using \eqref{eq:77} we get
\begin{align*}
\ [1(b-1)]_b\cdot[(b-1)^{\land k}]_b &=(b+b-1)\cdot \left ( \sum_{i=0}^{k-1}(b-1)b^i\right )\\
&=\sum_{i=0}^{k-1} \Big ( (b-1)b^{i+1}+ \left ( b(b-2)+1\right )b^i\Big )\\
&=\sum_{i=1}^{k}(b-1)b^i+\sum_{i=0}^{k-1} \left ( b(b-2)+1\right )b^i \\
&=(b-1)b^{k}+\sum_{i=1}^{k-1}\Big ((b-1)+b(b-2)+1\Big ) b^i+b(b-2)+1\\
&=(b-1)b^{k}+\sum_{i=1}^{k-1} (b-1)b^{i+1}+b(b-2)+1\\
&=(b-1)b^{k}+(b-1)b^{k}+\sum_{i=1}^{k-2} (b-1)b^{i+1}+b(b-2)+1
\end{align*}
\begin{align*}
&=b^k+(b-2)b^{k-1}+\sum_{i=1}^{k-2} (b-1)b^{i+1}+b(b-2)+1\\
&=[1(b-2)(b-1)^{\land k-2}(b-2)1]_b.
\end{align*}
\end{proof}
\section{Proof of Theorem \ref{thm:final}}\label{sec:final}
\begin{proof}
\noindent (a) Note that $s_b(N_k)=2a$. As $b$ is even, there exists an integer $M$ such that:
\begin{equation*}
2a\cdot M=[a(0)^{\land k+1}]_b.
\end{equation*}
The following computation shows that $N_k$ is a $b$-ARH number:
$$
s_b(N_k)\cdot M+(s_b(N_k)\cdot M)^R\\
=[a(0)^{\land k+1}]_b+[a]_b=[a(0)^{\land k}a]_b=N_k.
$$
To show that $N_k$ is not $b$-Niven observe that $N_k/a=[1(0)^{\land k}1]_b$ is odd.
\bigskip
\noindent (b) Note that $s_b(N_k)=2b$. As $b$ is even, the multiplier $M=[(1(0)^{\land k})^{\land b}(0)^{\land kb+b-1}]_b/2$ is an integer.
The following computation shows that $N_k$ is a $b$-ARH number:
\begin{equation*}
\begin{gathered}
s_b(N_k)\cdot M+(s_b(N_k)\cdot M)^R\\
=[(1(0)^{\land k})^{\land b}(0)^{\land kb+b}]_b+[((0)^{\land k}1)^{\land b}]_b=[(1(0)^{\land k})^{\land b}0 ((0)^{\land k}1)^{\land b}]_b=N_k.
\end{gathered}
\end{equation*}
To show that $N_k$ is not $b$-Niven observe that $N_k$ is not divisible by $b$.
\bigskip
\noindent (c) The proof is similar to that of b).
\end{proof}
\section{Proof of Theorem \ref{thm:final2}}\label{sec:final2}
\begin{proof}
\noindent (a) Using the fact that
\begin{equation*}
(b-2)^2=b^2-4b+4=b(b-4)+4=[(b-4)4]_b,
\end{equation*}
an equivalent base-$b$ representation for $N_k$ is given by
\begin{equation}
N_k =
\begin{cases}
[(b-2)(0)^{\land k-1}(b-4)5(0)^{\land k}(b-2)]_b, & \text {if }b\not = 4; \\
[2(0)^{\land k-1}11(0)^{\land k}2]_4, & \text {if }b = 4.
\end{cases}
\end{equation}
If $b\not = 4$ one has $s_b(N_k)=3(b-1)$ and if $b=4$ one has $s_4(N_k)=6$. To finish the proof of case a) it is enough to show that $\alpha_k$ is divisible by $s_b(N_k)$.
If $b\not = 4$ we get
$$
\alpha_k=b^{k+1}+b-2=b^{k+1}-1+b-1\\
=(b-1)\big (b^k+b^{k-1}+\dots +b^2+b+2 \big)
$$
and
$$b^k+b^{k-1}+\dots +b^2+b+2\\
\equiv k+2 \equiv 0\pmod{3}.
$$
For the first congruence we used $b\equiv 1\pmod{3}$ and for the second we used $k\equiv 1 \pmod{3}$.
If $b=4$, then clearly $\alpha_k$ is divisible by 2. Moreover
\begin{equation*}
\alpha_k=4^{k+1}+2=(3+1)^{k+1}+2\equiv 0\pmod{3},
\end{equation*}
which shows that $\alpha_k$ is divisible by 6.
\bigskip
\noindent (b) Now assume that $b=2$. Then an equivalent base-2 representation for $N_k$ is given by
\begin{equation*}
N_k=[1(0)^{\land k-1}10(0)^{\land k}1]_2,
\end{equation*}
so $s_2(N_k)=3$. To finish the proof, we use the fact that $k$ is even to show that $\alpha_k$ is divisible by 3:
\begin{equation*}
\alpha_k=2^{k+1}+1=(3-1)^{k+1}+1\equiv 0\pmod{3}.
\end{equation*}
To prove the last claim in the theorem, we show that the multipliers corresponding to various values of $k$ are distinct. This follows from the explicit formulas below. All sequences of multipliers are strictly increasing as functions of $k$.
If $b=2$ the sequence of multipliers is given by $M_k=\frac{2^{k+1}+1}{3}$.
If $b=4$ the sequence of multipliers is given by $M_k=\frac{4^{k+1}+2}{6}$.
If $b>4$ the sequence of multipliers is given by $M_k=\frac{b^{k+1}+b-2}{3(b-1)}$.
\end{proof}
\section{Proof of Theorem \ref{thm:3k0,2}}\label{sec:thm3k02}
\begin{proof}
\noindent (a) The base-$b$ representation for $N_k$ is
\begin{equation*}
N_k=[4(0)^{\land k-1}(17)(0)^{\land k}4]_b.
\end{equation*}
Therefore $s_b(N_k)=25$. If $k=5\ell+4$. one has that:
\begin{equation*}
\alpha_k=6^ka^k+4\equiv (6^5)^\ell6^4+4\equiv (7776)^\ell\cdot 296+4\equiv 0 \pmod {25}.
\end{equation*}
Hence $N_k$ is a $b$-MRH number with multiplier $\frac{\alpha_k}{25}=\frac{(6a)^k+4}{25}$.
\bigskip
\noindent (b) As above, $s_b(N_k)=25$. If $k=20\ell+4$, one has that:
\begin{equation*}
\alpha_k=8^ka^k+4\equiv (8^{20})^\ell8^4+4\equiv (76)^\ell\cdot 96+4\equiv 0 \pmod {25}.
\end{equation*}
Hence $N_k$ is a $b$-MRH number with multiplier $\frac{\alpha_k}{25}=\frac{(8a)^k+4}{25}$.
\end{proof}
\section{Acknowledgments} The author would like to thank the editor and the referee for valuable comments that helped him write a better paper.
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\end{thebibliography}
\bigskip
\hrule
\bigskip
\noindent 2010 {\it Mathematics Subject Classification}:
Primary 11B83; Secondary 11B99.
\noindent \emph{Keywords:} numeration base, $b$-Niven number, reversal, additive $b$-Ramanujan-Hardy number, multiplicative $b$-Ramanujan-Hardy number, palindrome.
\bigskip
\hrule
\bigskip
\noindent (Concerned with sequences
\seqnum{A005349},
\seqnum{A067030},
\seqnum{A305130}, and
\seqnum{A305131}.)
\bigskip
\hrule
\bigskip
\vspace*{+.1in}
\noindent
Received December 26 2018;
revised versions received March 14 2019; March 30 2019;
April 4 2019.
Published in {\it Journal of Integer Sequences}, May 24 2019.
\bigskip
\hrule
\bigskip
\noindent
Return to
\htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}.
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