% EJC papers *must* begin with the following two lines. \documentclass[12pt]{article} \usepackage{e-jc} \specs{P9}{20(1)}{2013} % Please remove all other commands that change parameters such as % margins or pagesizes. % only use standard LaTeX packages % only include packages that you actually need % we recommend these ams packages \usepackage{amsthm,amsmath,amssymb} % we recommend the graphicx package for importing figures \usepackage{graphicx} % use this command to create hyperlinks (optional and recommended) \usepackage[colorlinks=true,citecolor=black,linkcolor=black,urlcolor=blue]{hyperref} % use these commands for typesetting doi and arXiv references in the bibliography \newcommand{\doi}[1]{\href{http://dx.doi.org/#1}{\texttt{doi:#1}}} \newcommand{\arxiv}[1]{\href{http://arxiv.org/abs/#1}{\texttt{arXiv:#1}}} % all overfull boxes must be fixed; % i.e. there must be no text protruding into the margins % declare theorem-like environments \theoremstyle{plain} \newtheorem{theorem}{Theorem} \newtheorem{lemma}[theorem]{Lemma} \newtheorem{corollary}[theorem]{Corollary} \newtheorem{proposition}[theorem]{Proposition} \newtheorem{fact}[theorem]{Fact} \newtheorem{observation}[theorem]{Observation} \newtheorem{claim}[theorem]{Claim} \theoremstyle{definition} \newtheorem{definition}[theorem]{Definition} \newtheorem{example}[theorem]{Example} \newtheorem{conjecture}[theorem]{Conjecture} \newtheorem{open}[theorem]{Open Problem} \newtheorem{problem}[theorem]{Problem} \newtheorem{question}[theorem]{Question} \theoremstyle{remark} \newtheorem{remark}[theorem]{Remark} \newtheorem{note}[theorem]{Note} \def\Z{\Bbb Z} \def\N{\Bbb N} \def\zp{\Z^+} \def\Q{\Bbb Q} \def\R{\Bbb R} \def\C{\Bbb C} %\def\l{\left} %\def\r{\right} \def\bg{\bigg} \def\({\bg(} \def\[{\bg\lfloor} \def\){\bg)} \def\]{\bg\rfloor} \def\t{\text} \def\f{\frac} \def\si{\sim_2} \def\pr{\prec_2} \def\pe{\preceq_2} \def\B{B_{k+1}} \def\E{E_k} \def\P{\Z_p} \def\p{\ (\roman{mod}\ p)} \def\pc{p\t{-adic}} %\def\em{\emptyset} \def\se {\subseteq} \def\sp {\supseteq} \def\sm{\setminus} \def\Ar{\Longrightarrow} \def\La{\Longleftrightarrow} \def\bi{\binom} \def\eq{\equiv} \def\cs{\cdots} \def\ls{\leqslant} \def\gs{\geqslant} \def\mo{{\rm mod}} \def\ord{\roman{ord}} \def\lcm{\roman{lcm}} \def\o{\t{ord}} \def\ve{\varepsilon} \def\al{\alpha} \def\da{\delta} \def\Da{\Delta} \def\dk{\da^{(k)}_{p,m}} \def\la{\lambda} \def\ta{\theta} \def\ch{\roman{ch}_p} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% % if needed include a line break (\\) at an appropriate place in the title \title{\bf Products and sums divisible by\\ central binomial coefficients} % input author, affilliation, address and support information as follows; % the address should include the country, and does not have to include % the street address \author{Zhi-Wei Sun\thanks{Supported by the National Natural Science Foundation (grant 11171140) of China and the PAPD of Jiangsu Higher Education Institutions.}\\ \small Department of Mathematics\\[-0.8ex] \small Nanjing University\\[-0.8ex] \small Nanjing 210093, People's Republic of China\\ \small\tt zwsun@nju.edu.cn } %\date{\dateline{May 20, 2011}{Oct. 18, 2012}\\ % \small Mathematics Subject Classifications: comma separated list of % MSC codes available from http://www.ams.org/mathscinet/freeTools.html} \date{\dateline{May 20, 2011}{}\\ \small Mathematics Subject Classifications: 05A10, 11B65, 11A07.} \begin{document} \maketitle % E-JC papers must include an abstract. The abstract should consist of a % succinct statement of background followed by a listing of the % principal new results that are to be found in the paper. The abstract % should be informative, clear, and as complete as possible. Phrases % like "we investigate..." or "we study..." should be kept to a minimum % in favor of "we prove that..." or "we show that...". Do not % include equation numbers, unexpanded citations (such as "[23]"), or % any other references to things in the paper that are not defined in % the abstract. The abstract will be distributed without the rest of the % paper so it must be entirely self-contained. \begin{abstract} In this paper we study products and sums divisible by central binomial coefficients. We show that $$2(2n+1)\bi{2n}n\ \bigg|\ \bi{6n}{3n}\bi{3n}n\ \ \t{for all}\ n=1,2,3,\ldots.$$ Also, for any nonnegative integers $k$ and $n$ we have $$\bi {2k}k\ \bigg|\ \bi{4n+2k+2}{2n+k+1}\bi{2n+k+1}{2k}\bi{2n-k+1}n$$ and $$\bi{2k}k\ \bigg|\ (2n+1)\bi{2n}nC_{n+k}\bi{n+k+1}{2k},$$ where $C_m$ denotes the Catalan number $\f1{m+1}\bi{2m}m=\bi{2m}m-\bi{2m}{m+1}$. On the basis of these results, we obtain certain sums divisible by central binomial coefficients. % keywords are optional \bigskip\noindent \textbf{Keywords:} central binomial coefficients; divisibility; congruences \end{abstract} \section{Introduction} Central binomial coefficients are given by $\bi{2n} n$ with $n\in\N=\{0,1,2,\ldots\}$. The Catalan numbers $$C_n=\f1{n+1}\bi{2n}n=\bi{2n}n-\bi{2n}{n+1}\ \ (n=0,1,2,\ldots)$$ play important roles in combinatorics (cf. R. P. Stanley~\cite[pp.\,219-229]{St}). There are many sophisticated congruences involving central binomial coefficients and Catalan numbers (see, e.g., \cite{S10a, ST1, ST2}). In 1998 N. J. Calkin \cite{C} proved that $\bi{2n}n\mid\sum_{k=-n}^n(-1)^k\bi{2n}{n+k}^m$ for any $m,n\in\Z^+$. See also V.J.W. Guo, F. Jouhet and J. Zeng \cite{GJZ}, and H. Q. Cao and H. Pan \cite{CP} for further extensions of Calkin's result. In this paper we investigate a new kind of divisibility problems involving central binomial coefficients. Our first theorem is as follows. \begin{theorem}\label{Th1.1} {\rm (i)} For any positive integer $n$ we have \begin{equation}\label{1.1}2(2n+1)\bi{2n}n\ \bigg|\ \bi{6n}{3n}\bi{3n}n.\end{equation} {\rm (ii)} Let $k$ and $n$ be nonnegative integers. Then \begin{equation}\label{1.2}\bi {2k}k\ \bigg|\ \bi{4n+2k+2}{2n+k+1}\bi{2n+k+1}{2k}\bi{2n-k+1}n\end{equation} and \begin{equation}\label{1.3}\bi{2k}k\ \bigg|\ (2n+1)\bi{2n}nC_{n+k}\bi{n+k+1}{2k}.\end{equation} \end{theorem} In view of (\ref{1.1}) it is worth introducing the sequence $$S_n=\f{\bi{6n}{3n}\bi{3n}n}{2(2n+1)\bi{2n}n}\ \ (n=1,2,3,\ldots).$$ Here we list the values of $S_1,\ldots,S_8$: \begin{gather*} 5,\ 231,\ 14568,\ 1062347,\ 84021990, \\7012604550,\ 607892634420,\ 54200780036595. \end{gather*} The author generated this sequence as A176898 at N.J.A Sloane's OEIS (cf. \cite{S10b}). By Stirling's formula, $S_n\sim 108^n/(8n\sqrt{n\pi})$ as $n\to+\infty$. Set $S_0=1/2$. Using {\tt Mathematica} we find that $$\sum_{k=0}^\infty S_kx^k=\f{\sin(\f23\arcsin(6\sqrt{3x}))}{8\sqrt{3x}}\ \ \left(03$ we have $$\sum_{k=1}^{p-1}\f{S_k}{108^k}\eq\begin{cases} 0\ (\mo\ p)&\t{if}\ p\eq\pm1\ (\mo\ 12), \\-1\ (\mo\ p)&\t{if}\ p\eq\pm5\ (\mo\ 12).\end{cases}$$ \end{conjecture} \begin{remark} Part (i) of Conjecture \ref{Conj1.1} might be shown by our method for proving Theorem \ref{Th1.1}(i). \end{remark} \medskip Our following conjecture is concerned with a companion sequence of $\{S_n\}_{n\gs0}$. \begin{conjecture}\label{Conj1.2} There are positive integers $T_1,T_2,T_3,\ldots$ such that $$\sum_{k=0}^\infty S_kx^{2k+1}+\f1{24}-\sum_{k=1}^\infty T_kx^{2k} =\f{\cos(\f23\arccos(6\sqrt 3x))}{12}$$ for all real $x$ with $|x|\ls 1/(6\sqrt3)$. Also, $T_p\eq-2\ (\mo\ p)$ for any prime $p$. \end{conjecture} Here we list the values of $T_1,\ldots,T_8$: \begin{gather*} 1,\ 32,\ 1792,\ 122880,\ 9371648, \\763363328,\ 65028489216,\ 5722507051008. \end{gather*} In 1914 Ramanujan \cite{R} obtained that $$\sum_{k=0}^\infty\f{4k+1}{(-64)^k}\bi{2k}k^3=\f2{\pi}$$ and $$\sum_{k=0}^\infty(20k+3)\f{\bi{2k}k^2\bi{4k}{2k}}{(-2^{10})^k}=\f 8{\pi}.$$ (See also \cite{BB,BBC,Be} for such series.) Actually the first identity was originally proved by G. Bauer in 1859. Both identities can be proved via the WZ (Wilf-Zeilberger) method (cf. M. Petkov\v sek, H. S. Wilf and D. Zeilberger \cite{PWZ}, and Zeilberger \cite{Z} for this method). For WZ proofs of the two identities, see S.~B. Ekhad and D. Zeilberger \cite{EZ} and Guillera \cite{G}. van Hamme \cite{vH} conjectured that the first identity has a $p$-adic analogue. This conjecture was first proved by E. Mortenson \cite{M}, and recently re-proved in \cite{Zu} via the WZ method. On the basis of Theorem \ref{Th1.1}, we deduce the following new result. \begin{theorem}\label{Th1.2} For any positive integer $n$ we have \begin{equation}\label{1.4}4(2n+1)\bi{2n}n\ \bigg|\ \sum_{k=0}^{n}(4k+1)\bi{2k}k^3(-64)^{n-k}\end{equation} and \begin{equation}\label{1.5}4(2n+1)\bi{2n}n\ \bigg|\ \sum_{k=0}^{n}(20k+3)\bi{2k}k^2\bi{4k}{2k}(-2^{10})^{n-k}.\end{equation} \end{theorem} \medskip Now we pose two more conjectures. \begin{conjecture}\label{Conj1.3} {\rm (i)} For any $n\in\Z^+$ we have $$a_n:=\f1{8n^2\bi{2n}n^2}\sum_{k=0}^{n-1}(205k^2+160k+32)(-1)^{n-1-k}\bi{2k}k^5\in\Z^+.$$ {\rm (ii)} Let $p$ be an odd prime. If $p\not=3$ then $$\sum_{k=0}^{(p-1)/2}(205k^2+160k+32)(-1)^k\bi{2k}k^5\eq 32p^2+\f{896}3p^5B_{p-3}\ (\mo\ p^6),$$ where $B_0,B_1,B_2,\ldots$ are Bernoulli numbers. If $p\not=5$ then $$\sum_{k=0}^{p-1}(205k^2+160k+32)(-1)^k\bi{2k}k^5\eq 32p^2+64p^3H_{p-1}\ (\mo\ p^7),$$ where $H_{p-1}=\sum_{k=1}^{p-1}1/k$. \end{conjecture} \begin{remark}. Note that $a_1=1$ and $$4(2n+1)^2a_{n+1}+n^2a_n=(205n^2+160n+32)\bi{2n-1}n^3\ \ \ \t{for}\ n=1,2,\ldots.$$ The author generated the sequence $\{a_n\}_{n>0}$ at OEIS as A176285 (cf. \cite{S10b}). In 1997 T. Amdeberhan and D. Zeilberger \cite{AZ} used the WZ method to obtain $$\sum_{k=1}^\infty\f{(-1)^k(205k^2-160k+32)}{k^5\bi{2k}k^5}=-2\zeta(3).$$ \end{remark} \begin{conjecture}\label{Conj1.4} {\rm (i)} For any odd prime $p$, we have $$\sum_{k=0}^{p-1}\f{28k^2+18k+3}{(-64)^k}\bi{2k}k^4\bi{3k}k\eq 3p^2-\f 72p^5B_{p-3}\ (\mo\ p^6),$$ and $$\sum_{k=0}^{(p-1)/2}\f{28k^2+18k+3}{(-64)^k}\bi{2k}k^4\bi{3k}k\eq 3p^2+6\left(\f{-1}p\right)p^4E_{p-3}\ (\mo\ p^5),$$ where $E_0,E_1,E_2,\ldots$ are Euler numbers. {\rm (ii)} For any integer $n>1$, we have $$\sum_{k=0}^{n-1}(28k^2+18k+3)\bi{2k}k^4\bi{3k}k(-64)^{n-1-k}\eq0\ \ \(\mo\ (2n+1)n^2\bi{2n}n^2\).$$ Also, $$\sum_{k=1}^\infty\f{(28k^2-18k+3)(-64)^k}{k^5\bi{2k}k^4\bi{3k}k}=-14\zeta(3).$$ \end{conjecture} \begin{remark} The conjectural series for $\zeta(3)=\sum_{n=1}^\infty1/n^3$ was first announced by the author in a message to Number Theory Mailing List (cf. \cite{S10c}) on April 4, 2010. \end{remark} \medskip For more conjectures similar to Conjectures \ref{Conj1.3} and \ref{Conj1.4} the reader may consult \cite{S09} and \cite{S10b}. In the next section we will establish three auxiliary inequalities involving the floor function. Sections 3 and 4 are devoted to the proofs of Theorem \ref{Th1.1} and Theorem \ref{Th1.2} respectively. %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Three auxiliary inequalities} In this section, for a rational number $x$ we let $\{x\}=x-\lfloor x\rfloor$ be the fractional part of $x$, and set $\{x\}_m=m\{x/m\}$ for any $m\in\Z^+$. \begin{theorem}\label{Th2.1} Let $m>1$ be an integer. Then for any $n\in\Z$ we have \begin{equation}\label{2.1}\left\lfloor\f nm\right\rfloor+\left\lfloor\f{6n}m\right\rfloor\gs\left\lfloor\f{2n}m\right\rfloor+\left\lfloor\f{2n+1}m\right\rfloor +\left\lfloor\f{3n}m\right\rfloor.\end{equation} \end{theorem} \begin{proof} Let $A_m(n)$ denote the left-hand side of (\ref{2.1}) minus the right-hand side. Then $$A_m(n)=\left\{\f{2n}m\right\}+\left\{\f{2n+1}m\right\}+\left\{\f{3n}m\right\}-\f1m-\left\{\f nm\right\}-\left\{\f{6n}m\right\},$$ which only depends on $n$ modulo $m$. So, without any loss of generality we may simply assume that $n\in\{0,\ldots,m-1\}$. Hence $A_m(n)\gs0$ if and only if \begin{equation}\label{2.2}\left\{\f{2n}m\right\}+\left\{\f{2n+1}m\right\}+\left\{\f{3n}m\right\}\gs \f{n+1}m.\end{equation} (Note that $2n+(2n+1)+3n-(n+1)=6n$.) (\ref{2.1}) is obvious when $n=0$. If $1\ls nn+1$, then $$n+2k+\{n-k+1\}_m=n+2k+(n-k+1+m)=2n+k+1+m\gs 2m$$ since $2n+k+1>k\gs m/2$ and $\{2n+k+1\}_mm/2$ then $3k-1\gs 3(m+1)/2-1>3m/2$. If $k\ls n$ then $2n+k+1>3k\gs 3m/2$. So, except the case $k=n+1=m/2$ we have $$n+2k+\{n-k+1\}_m=2n+k+1\gs 3m/2$$ and hence $n+2k+\{n-k+1\}_m=2n+k+1\gs 2m$ since $\{2n+k+1\}_mn+1$, then $$n+2k+\{n-k+1\}_m=n+2k+(n+1-k)+m=2n+k+1+m>m.$$ In view of the above, we have completed the proof of Theorem \ref{Th2.2}. \end{proof} \begin{theorem}\label{Th2.3} Let $m\in\Z^+$ and $k,n\in\Z$. Then we have \begin{equation}\label{2.5}\begin{aligned}&\left\lfloor\f{2n+2k}m\right\rfloor -\left\lfloor\f{n+k}m\right\rfloor+2\left\lfloor \f km\right\rfloor-2\left\lfloor\f{2k}m\right\rfloor \\&\ \ \gs2\left\lfloor\f nm\right\rfloor-\left\lfloor\f{2n+1}m\right\rfloor+\left\lfloor\f{n-k+1}m\right\rfloor, \end{aligned}\end{equation} unless $2\mid m$ and $k\eq n+1\eq m/2\ (\mo\ m)$ in which case the right-hand side of the inequality equals the left-hand side plus one. \end{theorem} \begin{proof} Since $$2n+2k-(n+k)+2k-2(2k)=2n-(2n+1)+(n-k+1),$$ (\ref{2.5}) is equivalent to the following inequality: \begin{equation}\label{2.6}\begin{aligned}&\left\{\f{2n+2k}m\right\}-\left\{\f{n+k}m\right\}+2\left\{\f km\right\}-2\left\{\f{2k}m\right\} \\ &\ \ \ls2\left\{\f nm\right\}-\left\{\f{2n+1}m\right\}+\left\{\f{n-k+1}m\right\}. \end{aligned}\end{equation} As (\ref{2.6}) only depends on $k$ and $n$ modulo $m$, without loss of generality we simply assume that $k,n\in\{0,\ldots,m-1\}$. {\it Case} 1. $k\left\{\f{2n+1}m\right\}-1.$$ {\it Case} 3. $k\gs m/2$ and $\{n+k\}_mn+1$, then the left-hand side of the inequality (\ref{2.7}) is $$\f{n+k}m+\f{n+1-k}m+1=\f{2n+1}m+1>1.$$ Now assume that $n+k2k\gs m$. If $k=n+1\not= m/2$, then $k=n+1\gs(m+1)/2$ and hence $2n+1=2(n+1)-1\gs m$. When $k=n+1=m/2$, the left-hand side of (\ref{2.6}) minus the right-hand side equals $$\f{m-2}m-\f{m-1}m+2\f{m/2}m-2\f{m/2-1}m+\f{m-1}m=1.$$ Combining the discussion of the four cases we obtain the desired result. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Proof of Theorem \ref{Th1.1}} For a prime $p$, the $p$-adic evaluation of an integer $m$ is given by $$\nu_p(m)=\sup\{a\in\N:\ p^a\mid m\}.$$ For a rational number $x=m/n$ with $m\in\Z$ and $n\in\Z^+$, we set $\nu_p(x)=\nu_p(m)-\nu_p(n)$ for any prime $p$. Note that a rational number $x$ is an integer if and only if $\nu_p(x)\gs0$ for all primes $p$. \medskip\noindent \begin{proof}[Proof of Theorem \ref{Th1.1}] (i) Fix $n\in\Z^+$, and define $A_m(n)$ for $m>1$ as in the proof of Theorem \ref{Th2.1}. Observe that $$Q:=\f{\bi{6n}{3n}\bi{3n}n}{(2n+1)\bi{2n}n}=\f{n!(6n)!}{(2n)!(2n+1)!(3n)!}.$$ So, for any prime $p$ we have $$\nu_p(Q)=\sum_{i=1}^\infty A_{p^i}(n)\gs0$$ by Theorem \ref{Th2.1}. Therefore $Q$ is an integer. Choose $j\in\Z^+$ such that $2^{j-1}\ls n<2^j$. As $2n+1\ls 2(2^j-1)+1<2^{j+1}$, we have \begin{align*} &\left\lfloor\f n{2^{j+1}}\right\rfloor+\left\lfloor\f{6n}{2^{j+1}}\right\rfloor -\left\lfloor\f{2n}{2^{j+1}}\right\rfloor-\left\lfloor\f{2n+1}{2^{j+1}}\right\rfloor -\left\lfloor\f{3n}{2^{j+1}}\right\rfloor \\=&\left\lfloor\f{3n}{2^j}\right\rfloor-\left\lfloor\f{3n}{2^{j+1}}\right\rfloor =\left\lfloor\f{3n+2^j}{2^{j+1}}\right\rfloor\gs\left\lfloor\f{2n+2^j}{2^{j+1}}\right\rfloor\gs1. \end{align*} Therefore $$\nu_2(Q)=\sum_{i=1}^\infty A_{2^i}(n)\gs A_{2^{j+1}}(n)\gs1.$$ and hence $Q$ is even. This proves (\ref{1.1}). (ii) (\ref{1.2}) and (\ref{1.3}) are obvious in the case $k=0$. If $k>n+1$, then $$\bi{2n+k+1}{2k}=\bi{n+k+1}{2k}=0$$ and hence (\ref{1.2}) and (\ref{1.3}) hold trivially. Below we assume that $1\ls k\ls n+1$. Recall that for any nonnegative integer $m$ and prime $p$ we have $$\nu_p(m!)=\sum_{i=1}^\infty\left\lfloor\f m{p^i}\right\rfloor.$$ Since $$\f{\bi{4n+2k+2}{2n+k+1}\bi{2n+k+1}{2k}\bi{2n+k+1}n}{\bi{2k}k} =\f{(4n+2k+2)!(k!)^2}{(2n+k+1)!((2k)!)^2n!(n-k+1)!}$$ and $$\f{(2n+1)\bi{2n}nC_{n+k}\bi{n+k+1}{2k}}{\bi{2k}k}=\f{(2n+1)!(2n+2k)!(k!)^2}{(n!)^2(n+k)!((2k)!)^2(n-k+1)!},$$ it suffices to show that for any prime $p$ we have $$\sum_{i=1}^\infty C_{p^i}(n,k)\gs0\ \ \t{and}\ \ \sum_{i=1}^\infty D_{p^i}(n,k)\gs0,$$ where \begin{align*} C_m(n,k)=&\left\lfloor\f{4n+2k+2}m\right\rfloor-\left\lfloor\f{2n+k+1}m\right\rfloor +2\left\lfloor\f km\right\rfloor-2\left\lfloor\f{2k}m\right\rfloor \\&-\left\lfloor\f nm\right\rfloor-\left\lfloor\f{n-k+1}m\right\rfloor \end{align*} and \begin{align*} D_m(n,k)=&\left\lfloor\f{2n+2k}m\right\rfloor-\left\lfloor\f{n+k}m\right\rfloor +2\left\lfloor\f km\right\rfloor-2\left\lfloor\f{2k}m\right\rfloor \\&-2\left\lfloor\f nm\right\rfloor+\left\lfloor\f{2n+1}m\right\rfloor-\left\lfloor\f{n-k+1}m\right\rfloor. \end{align*} (a) By Theorem \ref{Th2.2}, $C_{p^i}(n,k)\gs0$ unless $p=2$ and $k\eq n+1\eq 2^{i-1}\ (\mo\ 2^i)$ in which case $C_{2^i}(n,k)=-1$. Suppose that $k\eq n+1\eq 2^{i-1}\ (\mo\ 2^i)$, $k=2^{i-1}k_0$ and $n+1=2^{i-1}n_0$, where $1\ls k_0\ls n_0$ and $k_0$ and $n_0$ are odd. If $i\gs2$, then $$C_{2^{i-1}}(n,k)=4n_0+2k_0-1-(2n_0+k_0-1)+2k_0-4k_0-(n_0-1)-(n_0-k_0)=1$$ and hence $C_{2^{i-1}}(n,k)+C_{2^i}(n,k)=1+(-1)=0$. So it remains to consider the case $k\eq n+1\eq1\ (\mo\ 2)$. Assume that $k$ is odd and $n$ is even. Write $k+1=2^jk_1$ and $n=2n_1$ with $k_1,n_1\in\Z^+$ and $2\nmid k_1$. Then it is easy to see that \begin{align*} C_{2^{j+1}}(n,k) &=\left\lfloor\f{4n_1}{2^j}\right\rfloor+k_1-\left\lfloor\f{2n_1-2^{j-1} +2^{j-1}(k_1-1)}{2^j}\right\rfloor \\ &\quad+2\left\lfloor\f{k_1}2\right\rfloor-2\left\lfloor\f{2^jk_1-1}{2^j}\right\rfloor -\left\lfloor\f{n_1}{2^j}\right\rfloor-\left\lfloor\f{n_1+1-2^{j-1}k_1}{2^j}\right\rfloor \\ &=\left\lfloor\f{4n_1}{2^j}\right\rfloor+k_1-\left\lfloor\f{2n_1-2^{j-1}}{2^j}\right\rfloor-\f{k_1+1}2+k_1-1-2(k_1-1) \\ &\quad-\left\lfloor\f{n_1}{2^j}\right\rfloor-\left\lfloor\f{n_1+1+2^{j-1}}{2^j}\right\rfloor+\f{k_1+1}2 \\ &=1+\left\lfloor \f{n_1+(n_1+1+2^{j-1})+(2n_1-2^{j-1})}{2^j}\right\rfloor \\ &\quad-\left\lfloor\f {n_1}{2^j}\right\rfloor-\left\lfloor\f{n_1+1+2^{j-1}}{2^j}\right\rfloor-\left\lfloor\f{2n_1-2^{j-1}}{2^j}\right\rfloor \\ &\gs 1 \end{align*} and hence $C_2(n,k)+C_{2^{j+1}}(n,k)\gs0$. By the above, we do have $\sum_{i=1}^\infty C_{p^i}(n,k)\gs0$ for any prime $p$. So (\ref{1.2}) holds. (b) By Theorem \ref{Th2.2}, $D_{p^i}(n,k)\gs0$ unless $p=2$ and $k\eq n+1\eq 2^{i-1}\ (\mo\ 2^i)$ in which case $D_{2^i}(n,k)=-1$. So, to prove (\ref{1.2}) it suffices to find a positive integer $j$ such that $D_{2^j}(n,k)\gs1$. Clearly there is a unique positive integer $j$ such that $2^{j-1}\ls n+k<2^j$. Note that $k\ls(n+k)/2<2^{j-1}$ and $$D_{2^j}(n,k)=1+\left\lfloor\f{2n+1}{2^j}\right\rfloor\gs1.$$ This concludes the proof of (\ref{1.3}). The proof of Theorem \ref{Th1.1} is now complete. \end{proof} %%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% \section{Proof of Theorem \ref{Th1.2}} \begin{proof}[Proof of Theorem \ref{Th1.2}] (i) We first prove (\ref{1.4}). For $k,n\in\N$ define $$F(n,k)=\f{(-1)^{n+k}(4n+1)}{4^{3n-k}}\bi{2n}n^2\f{\bi{2n+2k}{n+k}\bi{n+k}{2k}}{\bi{2k}k}$$ and $$G(n,k)=\f{(-1)^{n+k}(2n-1)^2\bi{2n-2}{n-1}^2}{2(n-k)4^{3(n-1)-k}}\bi{2(n-1+k)}{n-1+k}\f{\bi{n-1+k}{2k}}{\bi{2k}k}.$$ Clearly $F(n,k)=G(n,k)=0$ if $n