\documentclass[12pt,reqno]{article} \usepackage[usenames]{color} \usepackage{amssymb} \usepackage{graphicx} \usepackage{amscd} \usepackage[colorlinks=true, linkcolor=webgreen, filecolor=webbrown, citecolor=webgreen]{hyperref} \definecolor{webgreen}{rgb}{0,.5,0} \definecolor{webbrown}{rgb}{.6,0,0} \usepackage{color} \usepackage{fullpage} \usepackage{float} \usepackage{psfig} \usepackage{graphics,amsmath,amssymb} \usepackage{amsthm} \usepackage{amsfonts} \usepackage{latexsym} \usepackage{epsf} \setlength{\textwidth}{6.5in} \setlength{\oddsidemargin}{.1in} \setlength{\evensidemargin}{.1in} \setlength{\topmargin}{-.1in} \setlength{\textheight}{8.4in} \newcommand{\seqnum}[1]{\href{http://oeis.org/#1}{\underline{#1}}} \begin{document} \begin{center} \epsfxsize=4in \leavevmode\epsffile{logo129.eps} \end{center} \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{proposition}[theorem]{Proposition} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \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 $a_2(j)>0$ 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, & \text{if $m=0$;} \\ \text{$0!+1!+2!+ \cdots +(m-1)! $}, & \text{if $ m$ is a positive integer.} \end{cases} \end{displaymath} It then follows that \begin{align} \sum_{n=0}^{\infty}(n+j)!& = \alpha-K(j).\notag \end{align} For reasons which will be clear in a moment, we define \begin{align} a_0(x) & =1 \notag\\ b_0(j) & =-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) &=1 \notag \\ b_0(j) & = -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 $ & $a_{k} (x)$ \\ \hline 0 & $1$ \\ \hline 1 & $-x$ \\ \hline 2 & $-1+x^{2}$ \\ \hline 3 & $1+3x-x^{3}$ \\ \hline 4 & $+2-4x-6x^2+x^4$ \\ \hline 5 & $-9-10x+10x^2+10x^3-x^5$ \\ \hline 6 & $9 + 54\ x + 30\ x^2 - 20\ x^3 - 15\ x^4 + x^6$ \\ \hline 7 & $50 - 63\ x - 189\ x^2 - 70\ x^3 + 35\ x^4 + 21\ x^5 - x^7$ \\ \hline 8 & $-267 - 400 x + 252 x^2 + 504\ x^3 + $ $140\ x^4 - 56 x^5 - 28 x^6 + x^8$ \\ \hline 9 & $413 + 2403\ x + 1800\ x^2 - 756\ x^3- 1134\ x^4 - 252\ x^5+ 84\ x^6$ \\ & $ + 36\ x^7 - x^9$ \\ \hline 10 & $2180 - 4130\ x - 12015\ x^2 - 6000\ x^3 + 1890\ x^4 +2268\ x^5 + $ \\ & $ 420\ x^6 - 120\ x^7 - 45\ x^8 + x^{10}$ \\ \hline 11 & $-17731 - 23980\ x + 22715\ x^2 + 44055\ x^3 + 16500\ x^4 - 4158\ x^5 $ \\ & $- 4158 x^6- 660x^7 +165 x^8 + 55 x^9 - x^{11} $ \\ \hline 12 & $50533 + 212772 x + 143880 x^2 - 90860 x^3 - 132165 x^4 -39600 x^5$ \\ & $ + 8316 x^6 + 7128 x^7 $ $+ 990 x^8 - 220\ x^9 - 66 x^{10} + x^{12}$ \\ \hline 13 & $110176 - 656929 x - 1383018 x^2 - 623480x^3 + 295295 x^4 + 343629\ x^5$ \\ & $+ 85800 x^6 - 15444 x^7 - 11583 x^8 - 1430 x^9 + 286 x^{10} + 78 x^{11} - x^{13}$ \\ \hline 14 & $ -1966797-1542464 x+4598503x^2+6454084 x^3+2182180x^4$ \\ & $-826826x^5 -801801x^6-171600x^7+ 27027x^8+18018x^9 + 2002x^{10} $\\ & $-364x^{11} - 91x^{12}+ x^{14} $ \\ \hline % 15 & 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 $a_1(x)=-x>0$ for $x<0$. More generally, we have the following proposition. \begin{proposition} $a_k(x)>(k-1)!$ for $k\geq 1$ and $x\leq -k$. \end{proposition} \begin{proof} We prove this proposition by induction on $k$. Assume $a_k(x)>(k-1)!$ 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 \[a_{k+1}(x)>(k-1)!+(k-1)(k-1)!\] 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) &\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 \\ & \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} \bigskip \hrule \bigskip \noindent 2001 {\it Mathematics Subject Classification}: Primary 11A07; Secondary 40A30. \noindent \emph{Keywords: } $p$-adic series, complementary Bell number. \bigskip \hrule \bigskip \noindent (Concerned with sequence \seqnum{A000587}.) \bigskip \hrule \bigskip \vspace*{+.1in} \noindent Received June 24 2013; revised version received February 3 2014. Published in {\it Journal of Integer Sequences}, February 15 2014. \bigskip \hrule \bigskip \noindent Return to \htmladdnormallink{Journal of Integer Sequences home page}{http://www.cs.uwaterloo.ca/journals/JIS/}. \vskip .1in \end{document}