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\begin{center}
\vskip 1cm{\LARGE\bf Complementary Bell Numbers and \\
\vskip .11in
$p$-adic Series 
}
\vskip 1cm
\large
Deepak Subedi  \\
Faculty of Science and Technology\\
ICFAI University Himachal Pradesh \\
Kalujhinda, Solan, HP-174103\\
India\\
\href{mailto:deepak12321@gmail.com}{\tt deepak12321@gmail.com} \\
\end{center}

\vskip .2 in

\newcommand{\Q}{\mathbb{Q}}
\newcommand{\Z}{\mathbb{Z}}

\newtheorem {Congruence}{Congruence}

\begin{abstract}
In this article, we generalize a result of Murty  on  the non-vanishing of
complementary Bell numbers and irrationality of a $p$-adic series. This
generalization leads to a sequence of polynomials. We partially answer
the question of existence of an integral zero of those polynomials.
\end{abstract}

\section{Introduction}
 Murty and Sumner \cite{mu} have shown that there is a sequence of integers    $a_k, b_k$ such that the following equality
 \begin{equation}\label{base}
 \sum _{n=0}^{\infty}n^k n!=a_k \sum _{n=0}^{\infty} n! +b_k
 \end{equation}
 holds in $\Q_{p}$.
Alexander \cite{alexa} has shown that $a_k$ vanishes at most twice. In Proposition \ref{portion}, we generalize Eq.~\eqref{base}  to show that for  non-negative integers $k,j$ there exist two sequences of integers $a_k(j), b_k(j)$ such that the following equality
\begin{equation}\label{akj}
\sum_{n=0}^{\infty}n^k(n+j)!=a_k(j)\alpha+b_k(j)\end{equation}
holds in ${\Q}_p$,  $\alpha$ being the $p$-adic sum $\sum _{n=0}^{\infty} n!$.

It is obvious that we would like to identify non-negative integers $k$ and $j$ such that 
\[a_k(j)=0.\] This would mean that the infinite  sum on left-hand side of 
Eq.~\eqref{akj}
 is just an integer $b_k(j)$. In fact by Proposition \ref{portion} it would follow that
\[a_1(0)=0,a_2(1)=0\]and\[b_1(0)=-1,b_2(1)=2.\]
Again by Proposition  \ref{portion}, it follows that
\[ \sum_{n=0}^{\infty} n \cdot n! =-1 \]
and
\[ \sum_{n=0}^{\infty} n^2 (n+1)!=2.\]
On the other hand, Dragovich \cite{dr1} has shown that if the series
 \[\sum_{n=0}^{\infty}n!\] converges to a rational number in $\Q_{p}$ for every prime $p$, then the series cannot converge to the same rational number.
 Furthermore, the fact $a_2(j)&gt;0$ for every integer $j\geq 2$ leads us to conclude that for a fixed integer $j\geq 2$ if the  series 
\[\sum_{n=0}^{\infty}n^2(n+j)!\] converges to a rational number in $\Q_{p}$,
then it  cannot converge to a  fixed rational number in $\Q_{p}$ for every prime $p$. In order to show that $a_k(j)$ is non-zero for selected values of $k,j$ we need certain identities for $a_k(j)$. We derive a few of those identities in the next section.


\section{ Recurrence for the polynomial  }\label{a123}


We begin this section  by considering the series $\sum_{n=0}^{\infty}(n+j)!$ for a fixed non-negative integer $j$. Observe that

\[\sum _{n=0} ^{\infty} (n+j)!= \sum _{n=0} ^{\infty} n!-(0!+1!+2!+\cdots +(j-1)!).\]
 Kurepa's left factorial, $K(m)$ for a non-negative integer $m$  is given by


\begin{displaymath}
K(m) = \begin{cases}
	0,                               &amp; \text{if $m=0$;} \\
    \text{$0!+1!+2!+ \cdots +(m-1)! $}, &amp; \text{if $ m$ is a positive integer.}
	\end{cases}
\end{displaymath}




It then follows  that
\begin{align}
\sum_{n=0}^{\infty}(n+j)!&amp; = \alpha-K(j).\notag
\end{align}
For reasons which will be clear in a moment, we define
\begin{align}
a_0(x) &amp; =1 \notag\\
b_0(j) &amp; =-K(j).\notag
\end{align}
Therefore, it follows that
\begin{equation}
\sum_{n=0}^{\infty}(n+j)!=a_0(j)\alpha +b_0(j).
\end{equation}
We hereby present the main result of this section.
\begin{proposition} \label{portion}
Let $ k\geq 0$ and $j\geq 0$ be fixed integers. Then there is a polynomial $a_{k}(x)$ and an integer $b_{k}(j)$ such that
\[\sum_{n=0}^{\infty}n^k(n+j)!=a_k(j)\alpha+b_k(j)\]
where $a_k(x)$ and $b_k(j)$ are defined inductively on $k$ as follows: 
\[a_k(x)=a_{k-1}(x+1)-(x+1)a_{k-1}(x) \text{, } k\geq 1\]
\[b_k(j)=b_{k-1}(j+1)-(j+1)b_{k-1}(j)\text{, }k\geq 1\]
and
\begin{align}
a_0(x) &amp;=1 \notag \\
b_0(j) &amp; = -K(j). \notag
\end{align}

\end{proposition}
\begin{proof}
The proof is by induction on $k$. The case $k=0$ has already been worked out. So we may assume $k\geq 0$ and that the proposition holds for $k$.

Observe that
\[\sum_{n=0}^{\infty}n^{k}(n+j+1)! =\sum_{n=0}^{\infty}n^{k+1}(n+j)! +(j+1)\sum_{n=0}^{\infty}n^k(n+j)!.\]

Thus using $\sum_{n=0}^{\infty}n^{k}(n+j)!=a_k(j)\alpha+b_k(j)$ and then comparing the coefficient of $\alpha$, term without $\alpha$, we have
\begin{equation}\label{recurrenceforakj}
a_{k}(j+1)-(j+1)a_k(j)=a_{k+1}(j)
\end{equation}
and
\[b_{k}(j+1)-(j+1)b_k(j)=b_{k+1}(j).\]


\end{proof}
\begin{corollary}
The series $\sum_{n=0}^{\infty}n^k(n+j)!$ converges to an integer whenever $a_k(j)$ vanishes.
\end{corollary}
\begin{corollary}
\[a_{k}(0)=a_{k+1}(-1).\]
\end{corollary}



The next proposition may  remind us about a similar kind of property exhibited by Bernoulli polynomials.




\begin{proposition}\label{deri}
The derivative of $a_{k}(x)$ is given by
\[\dfrac{d }{dx}a_k(x)=-ka_{k-1}(x), \quad k\geq 1.\]
\end{proposition}
\begin{proof}
The proposition is easily seen to be true for $k=1 $ and  we prove the proposition by induction on $k$. We differentiate the expression
\[a_{K}(x+1)-(x+1)a_K(x)=a_{K+1}(x)\] given in Proposition \ref{portion} to obtain
\[a'_{K}(x+1)-a_K(x)-a'_K(x)(x+1)=a'_{K+1}(x).\]
We assume the proposition  holds for $k= K$ to obtain
\[-Ka_{K-1}(x+1)-a_K(x)+Ka_{K-1}(x)(x+1)=a'_{K+1}(x).\]
Again we consider Proposition \ref{portion} to obtain  the desired result.

\end{proof}




\begin{proposition}
If  $c_{i,k}$ denotes the coefficients of $x^i $ in   $a_{k}(x)$
then
\[-kc_{i,k-1}=(i+1)c_{i+1,k}\] for  non-negative integers $i,k$ and $i\leq  k-1$.
\end{proposition}

\begin{proof} The proof  follows by comparing constant term and the coefficient of powers of $x$ in  Proposition \ref{deri}.
\end{proof}

We note that if $k$ is a  prime then for $i \leq k-2$ \[\gcd(i+1,k)= 1.\] Hence $i+1$ must divide $c_{i,k-1}$ and $k$ must divide $c_{i+1,k}$ but $a_k(0)=a_k$ and so we can write \[a_p(x)\equiv a_p-x^p\pmod p.\]


We proceed for a few more congruences for $a_k(x)$. We start with a proposition  that states $a_k(x)$ can be determined using $a_k$ and the binomial coefficients. 
\begin{proposition}\label{akina}
The polynomial $a_k(x)$ is given by

\[a_{k}(x)=\sum_{i=0}^{k} {}^kC_{i}a_{i} (-1)^{k-i}x^{k-i}.\]
\end{proposition}
\begin{proof}
 Applying induction on Proposition \ref{deri}, it follows that 
 \[\dfrac{d^i}{dx^i}a_k(x)=(-1)^i k(k-1)(k-2)\cdots (k-i+1)a_{k-i}(x).\]
We write $a_k(x)$ as 
 \[a_{k}(x)=\sum_{i=0}^{k} b_{i}x^{i}.\] Then $\dfrac{d^i}{dx^i}a_k(x)$ at $x$ equal to 0 must be $b_i i!$.
 Hence $b_i$ must be 
 \[(-1)^{i} \,\,{}^kC_{i} a_{k-i}(0).\]
Using the fact that $a_{k-i}(0)=a_{k-i}$ the result follows.
 
\end{proof}

Now, we include a table containing first few polynomials $a_k(x)$.
\begin{center}
\begin{tabular}{|c|c|}
\hline
$ k $ &amp; $a_{k} (x)$ \\ \hline
 0 &amp; $1$ \\ \hline
 1 &amp; $-x$ \\ \hline
 2 &amp;   $-1+x^{2}$ \\ \hline
3 &amp; $1+3x-x^{3}$ \\ \hline
 4 &amp;  $+2-4x-6x^2+x^4$ \\ \hline
 5 &amp; $-9-10x+10x^2+10x^3-x^5$ \\ \hline
 6 &amp; $9 + 54\ x + 30\ x^2 - 20\ x^3 - 15\ x^4 + x^6$ \\ \hline
 7 &amp;  $50 - 63\ x - 189\ x^2 - 70\ x^3 + 35\ x^4 + 21\ x^5 - x^7$    \\ \hline
8 &amp;   $-267 - 400 x + 252 x^2 + 504\ x^3 + $
$140\ x^4 - 56 x^5 -
    28 x^6 + x^8$ \\ \hline
9 &amp;   $413 + 2403\ x + 1800\ x^2 - 756\ x^3- 1134\ x^4 - 252\
    x^5+ 84\ x^6$ \\ &amp; $  + 36\ x^7 - x^9$ \\ \hline

 10 &amp; $2180 - 4130\ x - 12015\ x^2 - 6000\ x^3 + 1890\ x^4 +2268\ x^5 + $
 \\ &amp; $ 420\ x^6 - 120\ x^7 - 45\ x^8 + x^{10}$ \\ \hline
11 &amp;    $-17731 - 23980\ x + 22715\ x^2 + 44055\ x^3 + 16500\ x^4 - 4158\ x^5  $
\\ &amp;
 $- 4158 x^6- 660x^7 +165 x^8 + 55 x^9 - x^{11} $ \\ \hline

12 &amp; $50533 + 212772 x + 143880 x^2 - 90860 x^3 - 132165 x^4 -39600 x^5$
\\ &amp; $ + 8316 x^6 + 7128 x^7 $
 $+ 990 x^8 - 220\ x^9 - 66 x^{10} + x^{12}$    \\ \hline
  13 &amp; $110176 - 656929 x -
    1383018 x^2 - 623480x^3 + 295295 x^4 + 343629\
    x^5$ \\ &amp; $+ 85800 x^6 - 15444 x^7 - 11583 x^8 - 1430 x^9 +
    286 x^{10} + 78 x^{11} - x^{13}$ \\ \hline
14 &amp; $ -1966797-1542464 x+4598503x^2+6454084 x^3+2182180x^4$
\\ &amp; $-826826x^5 -801801x^6-171600x^7+ 27027x^8+18018x^9 + 2002x^{10}  $\\ 
 &amp; $-364x^{11} - 91x^{12}+ x^{14}  $  \\ \hline


% 15 &amp; 9938669 + 29501955 x +  11568480 x^2 - 22992515 x^3 - 24202815 x^4 - 6546540 x^5 + 2067065x^6 + 1718145    x^7 + 321750x^8 - 45045 x^9 - 27027 x^{10} - 2730 x^{11} + 455 x^{12} + 105 x^{13} - x^{15}\)
   
\end{tabular}
\end{center}
It is easy to see from the table that \[a_1(0)=0, a_2(1)=0\] and so we would like to identify integers  $k$ and $j$ such that $a_k(j)$ is zero/nonzero. We partially identify such $k$ and $j$ in the next section.

\section{On non-vanishing of the polynomials}
The next proposition helps us in concluding  non-vanishing of $a_k(x)$  whenever $k$ is a prime. 


\begin{proposition}If  $p$ is a prime then
$a_p(j)$ does not vanish for every $j$ in ${{\Z} }$ with $j$ incongruent to $1$ modulo $p$.
\end{proposition}
\begin{proof}
The proposition can be easily verified for $p=2$. For $p\geq 3$, Proposition \ref{akina}  and the  congruence   \[\binom{p}{i}\equiv 0\pmod p  \text { for } 1\leq i\leq p-1  \] 
leads  us to
\[a_p(j)\equiv a_p-j^p \pmod p.\]
Murty \cite{mu} has shown that 
\[a_p \equiv 1\pmod p.\]
Considering  Fermat's theorem the proposition follows.

\end{proof}
Observe that $a_1(x)=-x&gt;0$ for $x&lt;0$. More generally, we have the following proposition.

\begin{proposition}
$a_k(x)&gt;(k-1)!$  for $k\geq 1$ and $x\leq -k$.
\end{proposition}
\begin{proof}
We prove this proposition by induction on $k$.

Assume $a_k(x)&gt;(k-1)!$  for   $x\leq -k$ holds for some fixed $k\geq 1$ then  the recurrence relation

\[a_{k+1}(x)=a_{k}(x+1)-(x+1)a_{k}(x)\] 
for $a_k(x)$  gives us 
\[a_{k+1}(x)&gt;(k-1)!+(k-1)(k-1)!\]  for   $x\leq -k-1$.

Hence the proposition follows.

\end{proof}
With this proposition it is clear that $a_k(x)$ does not vanish   for $k\geq 1$ and $x\leq -k$.

To analyse $a_k(x)$ further, we start with the following result.

\begin{proposition}\label{cona}
For a non-negative integer $m$ and an integer $j$  
\[a_{p+m}(j)\equiv a_{m+1}(j)+a_m(j) \pmod p .\]
\end{proposition}

\begin{proof}For an integer $j$,  by  Proposition \ref{akina} it follows  that 
\[a_p(j) \equiv 1-j \pmod p. \] 
Applying  the recurrence given in Proposition \ref{portion} and  the fact \[a_2(j)+a_1(j)=j^2-j-1,\]it follows that
\[a_{p+1}(j)\equiv a_2(j) +a_{1}(j)\pmod p.\]
Again, applying the recurrence in Proposition \ref{portion} repeatedly we obtain the desired result.

\end{proof}



\begin{proposition}  For a prime $p$ such that
\[p\equiv 2,3 \pmod {5}, \]
 $a_{p+1}(j)$ does not vanish for any integer $j$. 
\end{proposition}
\begin{proof}   

By previous proposition  
\[a_{p+1}(j)\equiv j^2-j -1 \pmod p .\]
However, for a prime $p \neq 2,5$ considering
\[4(j^2-j -1)=(2j-1)^2-5,\] it is clear that $a_{p+1}(j) $ is not congruent to $0$ modulo $p$ whenever 
the Legendre  symbol
\[\binom{5}{p} =-1 .\]

Now, $\binom{5}{p} =-1$ if and only if  
\[p\equiv 2,3 \pmod 5. \]
Hence the proposition follows.
\end{proof}

\begin{corollary}
$a_{8}(j)$, $a_{14}(j)$, $a_{18}(j)$ and $a_{24}(j)$ does not vanish for any integer $j$.
\end{corollary}


\begin{proposition}  For  a prime $p\equiv 2,5,6,7,8,11 \pmod {13}$  and an integer $j$  not divisible by $p$, $a_{p+2}(j)$ does not vanish.
\end{proposition}
\begin{proof}   

By  Proposition \ref{cona}, for $m=2$, one has 
\[a_{p+2}(j)\equiv -j(j^2-j-3) \pmod p .\]
Hence the proposition follows.
\end{proof}

\begin{proposition}\label{irred}
If  $a_p(1)$ is not divisible by $p^2$,
then  $a_p(x)$ is an irreducible polynomial over $\Q$. 
\end{proposition}
\begin{proof}
By  Proposition~\ref{akina} we have 
\[a_p(x+1)\equiv a_p-(x+1)^p \pmod p .\]
Considering the congruence for $a_p$ given by  Murty \cite{mu}  again it follows that
 \[a_p(x+1)\equiv  -x^p\pmod p.\]
Hence by Eisenstein's criterion,  the result follows.

\end{proof}

As a consequence of Proposition~\ref{irred} it is clear that if   $a_p(1)$ is not divisible by $p^2$, then  there does not exist an integer $j$ 
such that $a_p(j) = 0$. The next proposition gives us a conditional statement for deciding whether  $a_p(j)$ is different from $1$.
\begin{proposition}\label{ir}
For an odd prime $p$, if  $a_p-1$ is not divisible by  $p^2 $, then  $a_p(x)-1$ is an irreducible polynomial. 
\end{proposition}
\begin{proof}
Following  the steps of Proposition~\ref{irred} we have 
 \[a_p(x)-1\equiv  -x^p\pmod p.\]
Hence by Eisenstein's criterion for the irreducibility of a polynomial,
the result follows.

\end{proof}

 \begin{proposition}
For non-negative integers $m,t$ and an integer $j$ the following congruence holds
\begin{equation}\label{cong} 
a_{tp+m}(j)\equiv \sum_{i=o}^{t} {}^{t}C_{i} a_{m+i}(j)\pmod p .\end{equation}
\end{proposition}
\begin{proof}
The case $t=0$ is obviously true. As our induction hypothesis we assume that the congruence in Eq.~\eqref{cong} is true for some $t\geq 0$ and by Proposition \ref{cona}
it follows that 
\begin{equation}\label{} 
a_{(t+1)p+m}(j)\equiv a_{tp+m+1}(j)+a_{tp+m}(j)\pmod p .\end{equation}
Hence by our induction hypothesis 
\begin{align}
a_{(t+1)p+m}(j) &amp;\equiv \sum_{i=o}^{t} {}^{t}C_{i} a_{m+1+i}(j)  +\sum_{i=o}^{t} {}^{t}C_{i} a_{m+i}(j)\pmod p \\
&amp; \equiv\sum_{i=o}^{t+1} {}^{t+1}C_{i} a_{m+i}(j)\pmod p .
\end{align}
Hence the result follows by induction.
\end{proof}

\begin{proposition}\label{pi}
For non-negative integers $m,i$ and an integer $j$ the following congruence holds
\[ a_{p^i+m}(j)\equiv a_{m+1}(j)+ia_m(j)\pmod p .\]
\end{proposition}
\begin{proof}
The case $i=0$ is a trivial case and the case $i=1$ follows from Proposition \ref{cona}. So we assume $i\geq 2$.

For $1\leq j\leq p^{i-1}-1$, considering the congruence
\[\binom{p^{i-1}}{j} \equiv 0 \pmod p\]and
 $t=p^{i-1}$ in the above proposition  it follows that
\[ a_{p^i+m}(j)\equiv a_{m}(j)+a_{p^{i-1}+m}(j)\pmod p .\]
Repeating the previous step $r$  number of times where $r\leq i-1$ we have
\[ a_{p^i+m}(j)\equiv r a_{m}(j)+a_{p^{i-r}+m}(j)\pmod p .\]
Choosing $r=i-1$ we have
\[ a_{p^i+m}(j)\equiv (i-1)a_{m}(j)+a_{p+m}(j)\pmod p .\] The result follows from the above  congruence and Proposition \ref{cona}.
\end{proof}

  
\begin{corollary}For a positive integer $i$,
\[ a_{p^i}\equiv i \pmod p.\]
\end{corollary}
\begin{proof}
Choosing $m,j$ equal to $0$ the corollary follows.
\end{proof}

\begin{proposition}
$a_{p^{zp}}(j)$ does not vanish for any integer $j$ not divisible by $p$.
\end{proposition}
\begin{proof}
 
We consider $i=zp$ for some non-negative integer $z$ in the previous proposition to obtain 
\[a_{p^{zp}+m}(j) \equiv a_{m+1}(j)\pmod p .\] 
 Choosing $m=0$, the result follows.
\end{proof}


\begin{proposition}
For a non-negative integer $t$ and an integer $j$
\[a_{3t}(j)\neq 0.\]

\end{proposition}
\begin{proof}
Considering $i=2,p=2$ in Proposition \ref{pi}, it follows that 
\[a_{4+m}(j)\equiv a_{1+m}(j)\pmod 2.\] 
Choosing $m=2,5,8,\cdots$ it is  easy to see that for a positive integer $t$
\[ a_{3t}\equiv  a_{3}(j)\pmod 2 .\]  The fact  $$ a_{3}(j)\not \equiv 0 \pmod 2$$ leads to the desired result.
 
\end{proof}
\begin{proposition}
For a non-negative integer $t$ 
\[ a_{pt}\equiv a_{t-1} \pmod p .\]
\end{proposition}
\begin{proof}
Through Proposition \ref{akina} it is easy to see that
\[ a_{k}(-1)=\sum _{i=0}^{k} {}^{k}C_{i}a_i\]
However, by the recurrence \ref{portion} 
\[a_k(-1)=a_{k-1}(0).\]
By congruence \eqref{cong} 
\[a_{pt+m}(j)\equiv \sum _{i=0}^{t} {}^{t}C_{i}a_{m+i}(j) \pmod p.\]
For $m=0,j=0$ above congruence reduces to
\[a_{pt}(0)\equiv \sum _{i=0}^{t} {}^{t}C_{i}a_{i}(0) \pmod p\]

and so
\[a_{pt}(0)\equiv  a_{t-1} \pmod p.\]

\end{proof}



\begin{proposition}
For a positive integer $i$ and an integer $j$ if 
\[j\not \equiv i \pmod p\] then 
\[a_{p^i}(j)\neq 0.\]
\end{proposition}
\begin{proof}
 Choosing $m=0$ in Proposition \ref{pi} we have 
 \[a_{p^i}(j) \equiv a_{1}(j)+ia_{0}(j) \pmod p.\]
 The result follows.
 
\end{proof}



The next result gives us a much stronger congruence of $a_k(j)$. 



\begin{theorem}For non-negative integers $t,m$, a positive integer $n$, an odd prime $p$ and an  integer $j$ such that  
\[j\equiv 0,1,2 \pmod p\] the following congruence
\[a_{_{ \frac{p^{p}-1} {p-1}\cdot p^{n-1}t+m}}(j) \equiv a_{m}(j) \pmod {p^n}\] holds.
\end{theorem}
\begin{proof}
We consider three cases: $j\equiv 0 \pmod p$, $j \equiv 1 \pmod p$ and $j\equiv 2 \pmod p$
\\
{\bf{Case 1.} }
For an  integer $j\equiv 0 \pmod p$ and a positive integer $r$, it is easy to see that
\[\binom{ \frac{p^{p}-1} {p-1}\cdot p^{n-1}}{r}(-j)^r =\dfrac{\frac{p^{p}-1} {p-1}(-j)^r\cdot p^{n-1}}{r}\binom{\frac{p^{p}-1} {p-1}\cdot p^{n-1} -1}{r-1} \equiv 0 \pmod {p^n}.\]
For an odd prime $p$,  following a slightly different notation, Alexander \cite{alexa} has proved that 
\[a_{_{\frac{p^{p}-1} {p-1}\cdot p^{n-1}t+m}} \equiv a_m \pmod {p^n},\text{ }  t, m \text{  being non-negative integers}.\]
Hence by Proposition \eqref{akina}, it  follows that  for an   integer $j\equiv 0 \pmod p$

\begin{equation}
{a_{{\frac{p^{p}-1} {p-1}} }}(j) \equiv 1 \pmod {p^n} .
\end{equation}

For simplicity we denote $ \frac{p^{p}-1} {p-1}\cdot p^{n-1}$ by $k$.
\\
{\bf{Case 2.} } In this case we are expressing $a_k(j)$ in terms of $a_{k+1}(j-1) $ and $a_{k}(j-1)$ and then deriving the required congruence.

 Replacing  $j $ by $j-1$, Eq.~\eqref{recurrenceforakj} can be written as 
\begin{equation}\label{a}
a_{k}(j)=a_{k+1}(j-1)+ja_{k}(j-1).
\end{equation}
For an integer $j\equiv 1 \pmod {p}$ and
$k=\frac{p^{p}-1} {p-1}\cdot p^{n-1}$, $n\geq 1$ 

$$a_{k+1}(j-1)\equiv  a_{k+1} -(j-1)(k+1)a_{k}\pmod {p^n}.$$
 Again using the result of Alexander \cite{alexa}, it follows that
$$a_{k+1}(j-1)\equiv  -(j-1)\equiv  a_1(j-1)\pmod {p^n}.$$
Hence, it follows that
\begin{equation}
{a_{k}(j) } \equiv 1 \pmod {p^n} \text{ for } j \equiv 1 \pmod p
\end{equation}
\\
{\bf{Case 3.} } In this case, we express $a_{k}(j)$ in term of $a_{k+2}(j-2)$ and other similar terms.
\\
Replacing $k$ by $k+1$ and $j$ by $j-1$, Eq.~\eqref{a} can be written as
      
\begin{equation}\label{b}
 a{_{k+1}}(j-1)=a{_{k+2}}(j-2)+(j-1)  a{_{k+1}}(j-2) \text{ and}
\end{equation}
replacing $j$ by $j-1$, Eq.~\eqref{a} can be written as



\begin{equation}\label{c}
a_{k}(j-1)=a_{k+1}(j-2)+(j-1)a_{k}(j-2)
\end{equation}
Eliminating $a_{k}(j-1),a_{k+1}(j-1)$ from Eqs.~\eqref{a}, \eqref{b}, and \eqref{c}, it follows that
\begin{equation}\label{delta1}
a_{k}(j)=a_{k+2}(j-2)+(j-1)a_{k+1}(j-2)+j\{a_{k+1}(j-2)+(j-1)a_{k}(j-2)\}.
\end{equation}

But for an integer $j\equiv 2 \pmod {p}$, by Proposition \ref{akina} it is easy to see that
\[ {a_{k+2} }(j-2)\equiv a_{k+2}+(k+2)(2-j)a_{k+1}+\frac{(k+2)(k+1)}{2} (j-2)^ {2} a_{k}\pmod {p^n} \]

Again, considering the result of Alexander it would follow that

\begin{equation}\label{alpha}
a_{k+2} (j-2)\equiv -1 +(j-2)^{2} \equiv {a_{2}(j-2) }\pmod {p^n} .  
\end{equation} 
Also it is easy to obtain that
\begin{equation}\label{beta}
a_{k+1} (j-2)\equiv  {a_{0}(j-2) }\pmod {p^n} . 
\end{equation} 
Applying  Eqs.~\eqref{delta1},\eqref{alpha},\eqref{beta}  it would follow that for an integer $j\equiv 2 \pmod p$,
\begin{equation}
a_{k} (j)\equiv  1\pmod {p^n}.  
\end{equation} 
Therefore, using Eq.~\eqref{recurrenceforakj} the result follows.
\end{proof}


\begin{corollary}
For non-negative integers $t,m$ and  a positive integer $n$,  the following congruence
\[a_{_{ 13\cdot 3^{n-1}t+m}}(j) \equiv a_{m}(j) \pmod {3^n}\] holds.
\end{corollary}

\begin{proof}
The proof follows by considering $p=3$ in previous proposition.
\end{proof}





Remark: The polynomials $a_k(x)$ were previously analyzed by  Wannemacker  \cite{wan}. He  has  verified numerically that $a_k(x)$ is irreducible over $\Z$ for all $6 \leq k \leq 200$. He conjectured  that $a_k(x)$ is irreducible over $\Z$ for all $k \geq 6$.

\begin{thebibliography}{5}
\bibitem{dr0}
B. Dragovich,  On some $p$-adic series with factorials, 
in {\it p-adic Functional Analysis}, Lect. Notes Pure Appl. Math., Vol.\ 192,
Dekker, 1997, pp.\ 95--105.

\bibitem{dr1}
B. Dragovich, {\em On $p$-adic power series}, preprint,   \newline
\url{http://arxiv.org/abs/math-ph/0402051}.

\bibitem{d1}
B. Dragovich, {On some finite sums with factorials}, 
{\em Facta Universitasis (Nis) Ser. Math. Inform.} {\bf 14} (1999), 1--10.

\bibitem{bell}
L. Comtet, {\it Advanced Combinatorics}, D. Reidel Publishing Company,
1974.

\bibitem{mu}
M. Ram Murty  and S. Sumner, {\em On the $p$-adic series
$\sum_{n=1}^{\infty}n^{k} \cdot n!$},
in {\it Number Theory}, CRM Proc.\ Lecture Notes,
{\bf 36}, Amer. Math. Soc., 2004,  pp.\ 219--227.

\bibitem{alexa}
N. C. Alexander,  Non-vanishing of Uppuluri-Carpenter numbers,
preprint, \url{http://tinyurl.com/oo36das}.

\bibitem{slo} N. J. A. Sloane, {The On-line  Encyclopedia of Integer
Sequences,} \url{http://oeis.org}.

 \bibitem{wan}
S. D. Wannemacker, {\em Annihilating polynomials and Stirling numbers
of the second kind}, Ph.\ D.\ thesis, University College Dublin, Ireland,
2006.

 \bibitem{yang}
Y. Yang,  
\href{http://www.combinatorics.org/ojs/index.php/eljc/article/view/v8i1r19}{On a  multiplicative partition function},
{\em Electron. J.
Combin.} {\bf 8} (2001) \#R19.  

\end{thebibliography}
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\noindent 2001 {\it Mathematics Subject Classification}:
Primary 11A07; Secondary 40A30.

\noindent \emph{Keywords: } 
 $p$-adic series, complementary Bell number.

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\noindent (Concerned with sequence \seqnum{A000587}.)

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\vspace*{+.1in}
\noindent
Received June 24 2013;
revised version received  February 3 2014.
Published in {\it Journal of Integer Sequences}, February 15 2014.

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\noindent
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