\input amstex \input amsppt.sty \nologo \magnification1200 \hsize13cm \vsize19cm %First page headline in AmS-TeX for S\'eminaire Lotharingien de Combinatoire \catcode`\@=11 \def\output@{\shipout\vbox{% \iffirstpage@ \global\firstpage@false %insert headline: \headline={\frheadline} \makeheadline \pagebody \makefootline% \else \ifrunheads@ \makeheadline \pagebody \else \pagebody \makefootline \fi \fi}% \advancepageno \ifnum\outputpenalty>-\@MM\else\dosupereject\fi} \catcode`\@=13 \font\rms=cmr8 \font\its=cmti8 \font\bfs=cmbx8 \def\frheadline{\its S\'eminaire Lotharingien de Combinatoire \bfs 47 \rms (2002), Article~B47g\hfill} \TagsOnRight \def\ZeilAV{14} \def\ZeilAM{13} \def\WiZeAC{12} \def\WegsAA{11} \def\StemAE{10} \def\SlatAC{9} \def\PeWZAA{8} \def\PeWiAA{7} \def\MeWaAA{6} \def\KratBN{5} \def\BailAA{4} \def\BacRAA{3} \def\BacRAZ{2} \def\AmZeAB{1} \def\sgn{\operatorname{sgn}} \def\si{\sigma} \def\om{\omega} \def\3{\ss} \def\({\left(} \def\){\right)} \def\coef#1{\langle#1\rangle} \def\po#1#2{(#1)_#2} \def\sgn{\operatorname{sgn}} \topmatter \title Evaluations of some determinants of matrices related to the Pascal triangle \endtitle \author C.~Krattenthaler$^\dagger$ \endauthor \affil Institut f\"ur Mathematik der Universit\"at Wien,\\ Strudlhofgasse 4, A-1090 Wien, Austria.\\ e-mail: KRATT\@Ap.Univie.Ac.At\\ WWW: \tt http://www.mat.univie.ac.at/\~{}kratt \endaffil \address Institut f\"ur Mathematik der Universit\"at Wien, Strudlhofgasse 4, A-1090 Wien, Austria. \endaddress %\email KRATT\@Ap.Univie.Ac.At\\ %WWW: \tt http://www.mat.univie.ac.at/\~{}kratt\endemail %\dedicatory \enddedicatory %\date \enddate \thanks{$^\dagger$Research partially supported by the Austrian Science Foundation FWF, grants P12094-MAT and P13190-MAT, and by EC's IHRP Programme, grant HPRN-CT-2001-00272}\endthanks \subjclass Primary 05A19; Secondary 05A10 11C20 15A15 33C45 \endsubjclass \keywords determinant, bivariate recurrent sequence, binomial coefficient,\linebreak Dodgson's condensation method, hypergeometric series\endkeywords \abstract We prove several evaluations of determinants of matrices, the entries of which are given by the recurrence $a_{i,j}=a_{i-1,j}+a_{i,j-1}$, or variations thereof. These evaluations were either conjectured or extend conjectures by Roland Bacher [J. Th\'eorie Nombres Bordeaux {\bf 14} (2002), to appear]. \endabstract \endtopmatter \document \leftheadtext{C. Krattenthaler} \rightheadtext{Evaluations of some determinants} \subhead 1. Introduction\endsubhead In the recent preprint \cite{\BacRAZ} (see \cite{\BacRAA} for the final version), Bacher considers determinants of matrices, the entries of which are given by the Pascal triangle recurrence $a_{i,j}=a_{i-1,j}+a_{i,j-1}$ or by similar recurrences, or are related in some other way to such bivariate recurrent sequences. He predicts (in form of precise conjectures) that such determinants also obey recurrence relations. This is most intriguing, because it points towards the possibility of {\it automatizing determinant evaluations}\footnote{The reader should recall that the successful automatization \cite{\PeWZAA, \WegsAA, \WiZeAC, \ZeilAM, \ZeilAV} of the evaluation of binomial and hypergeometric sums is fundamentally based on producing reccurrences by the computer.}, something that several authors (cf\. e.g\. \cite{\AmZeAB, \KratBN, \PeWiAA}) have been aiming at (albeit, with only limited success up to now). In addition, the preprint \cite{\BacRAZ} contains several conjectures on closed form evaluations of such determinants in some instances. These are also interesting since they do not fit into the known families of determinant evaluations (see the recent survey article \cite{\KratBN} for a long list of known determinant evaluations). While we are not able to address Bacher's conjectured recurrence relations, we are able to provide proofs for all the conjectured determinant evaluations in \cite{\BacRAZ}, and, in some cases, to prove in fact generalizations of conjectures in \cite{\BacRAZ}. (Some of these results are stated without proof in \cite{\BacRAA}.) In Section~2 we provide the proofs for the three conjectures in \cite{\BacRAZ} on determinants of matrices, the entries of which are given by the recurrence $a_{i,j}=a_{i-1,j}+a_{i,j-1}+x\,a_{i-1,j-1}$. Section~3 contains three theorems on binomial determinants which generalize (conjectural) determinant evaluations in Section~5 of \cite{\BacRAZ}. All of the methods that we use to prove our theorems are already contained in the review \cite{\KratBN} of the ``state of the art" of determinant evaluations, namely, on the one hand, the ``LU-factorization method" (see \cite{\KratBN, Sec.~2.6}) and, on the other hand, the ``identification of factors method" (see \cite{\KratBN, Sec.~2.4}). There is one case where elementary row and column operations suffice. \subhead 2. Some determinants of matrices with entries given by an extension of the Pascal triangle recurrence\endsubhead Our first theorem proves Conjecture~1.8 in \cite{\BacRAZ}. It is stated (without proof) as Theorem~1.5 in \cite{\BacRAA}. \proclaim{Theorem 1}Let $(a_{i,j})_{i,j\ge0}$ be the sequence given by the recurrence $$a_{i,j}=a_{i-1,j}+a_{i,j-1}+x\,a_{i-1,j-1},\quad \quad i,j\ge1 \tag2.1$$ and the initial conditions $a_{i,0}=\rho^i$ and $a_{0,i}=\si^i$, $i\ge0$. Then $$\det_{0\le i,j\le n-1}(a_{i,j})=(1+x)^{\binom {n-1}2}(x+\rho+\si-\rho\si)^{n-1}.$$ \endproclaim \demo{Proof}Let $F(u,v)=\sum _{i,j\ge0} ^{}a_{i,j}u^iv^j$ be the bivariate generating function of the sequence $(a_{i,j})$. By multiplying both sides of (2.1) by $u^iv^j$ and summing over all $i$ and $j$, we get an equation for $F(u,v)$ with solution $$F(u,v)=\frac {\frac {1-u} {1-\rho u}+\frac {1-v} {1-\si v}-1} {1-u-v-uvx}.\tag2.2$$ Now we use the {\it LU-factorization method\/} (see \cite{\KratBN, Sec.~2.6}). Let $M$ denote the matrix $(a_{i,j})_{0\le i,j\le n-1}$. I claim that $$M\cdot U=L,$$ where $U=(U_{i,j})_{0\le i,j\le n-1}$ with $$U_{i,j}=\cases 1&i=j\\ (-1)^{j-i}\(\binom {j-1}i\si+\binom {j-1}{i-1}\)&ij,\endcases$$ and $L=(L_{i,j})_{0\le i,j\le n-1}$ with $$L_{i,j}=\cases \rho^i&j=0,\\ (x+\rho+\si-\rho\si)(1+x)^{j-1}\sum _{\ell=0} ^{i-1}\binom {\ell}{j-1}\rho^{i-\ell-1}&j>0.\endcases$$ The matrix $U$ is an upper triangular matrix with 1's on the diagonal, whereas $L$ is a lower triangular matrix with diagonal entries $$1,\ (x+\rho+\si-\rho\si),\ (x+\rho+\si-\rho\si)(1+x),\ \dots,\ (x+\rho+\si-\rho\si)(1+x)^{n-2}.$$ It is obvious that the claimed factorization of $M$ immediately implies the validity of the theorem. For the proof of the claim we compute the $(i,j)$-entry of $M\cdot U$. By definition, it is $\sum _{k=0} ^{j}a_{i,k}U_{k,j}$. Let first $j=0$. Then the sum is equal to $a_{i,0}=\rho^i$, in accordance with the definition of $L_{i,0}$. If $j>0$, then (here, and in the following, $\coef{z^N}f(z)$ denotes the coefficient of $z^N$ in the formal power series $f(z)$) $$\align (M\cdot U)_{i,j}&=\sum _{k=0} ^{j}a_{i,k}U_{k,j}\\ &=M_{i,j}+ \sum _{k=0} ^{j-1}a_{i,k}U_{k,j}\\ &=\coef{u^iv^j}F(u,v)\(1+\sum _{k=0} ^{j-1}(-1)^{j-k}v^{j-k} \(\binom {j-1}k\si+\binom {j-1}{k-1}\)\)\\ &=\coef{u^iv^j}F(u,v)\((1-v)^{j-1}-\si v(1-v)^{j-1}\)\\ &=\coef{u^iv^j}F(u,v)(1-v)^{j-1}(1-\si v). \tag2.3 \endalign$$ Now we rewrite $F(u,v)$, as given by (2.2), slightly, $$\align F(u,v)&=\frac {1-u-v+uv(x+\rho+\si-\rho\si)-xuv} {(1-\rho u)(1-\si v)(1-u-v-uvx)}\\ &=\frac 1 {(1-\rho u)(1-\si v)}+ \frac {uv(x+\rho+\si-\rho\si)} {(1-\rho u)(1-\si v)(1-u-v-uvx)}\\ &=\frac 1 {(1-\rho u)(1-\si v)}+ \frac {uv(x+\rho+\si-\rho\si)} {(1-\rho u)(1-\si v)(1-u)(1-v)\(1-\frac {uv(x+1)}{(1-u)(1-v)}\)}. \endalign$$ This is substituted in (2.3): $$\multline (M\cdot U)_{i,j}\\= \coef{u^iv^j}(1-v)^{j-1} \(\frac 1 {(1-\rho u)}+ \frac {uv(x+\rho+\si-\rho\si)} {(1-\rho u)(1-u)(1-v)\(1-\frac {uv(x+1)}{(1-u)(1-v)}\)}\). \endmultline$$ Since the degree of $(1-v)^{j-1}$ in $v$ is smaller than $j$, it is obvious that the first expression in parentheses does not contribute anything to the coefficient of $u^iv^j$. Hence we obtain $$\align (M\cdot U)_{i,j}&= \coef{u^iv^j}(1-v)^{j-1} \frac {uv(x+\rho+\si-\rho\si)} {(1-u)(1-v)}\sum _{h=0} ^{\infty}\frac {u^hv^h(x+1)^h} {(1-u)^h(1-v)^h}\sum _{\ell=0} ^{\infty}\rho^\ell u^\ell\\ &=(x+\rho+\si-\rho\si)\sum _{h,\ell\ge0} ^{}\rho^\ell(1+x)^h \binom {i-\ell-1}h\binom {0}{h+1-j}. \endalign$$ Because of the second binomial coefficient, the summand is nonvanishing only for $h=j-1$. This yields exactly the claimed expression for $L_{i,j}$.\quad \quad \qed \enddemo The next theorem proves the conjecture about the evaluation of $\det(A(2n))$ on page~4 of \cite{\BacRAZ}. It is stated (without proof) in \cite{\BacRAA, equation for $\det(B(2n))$ after Theorem~1.5}. \proclaim{Theorem 2}Let $(a_{i,j})_{i,j\ge0}$ be the sequence given by the recurrence $$a_{i,j}=a_{i-1,j}+a_{i,j-1}+x\,a_{i-1,j-1},\quad \quad i,j\ge1 $$ and the initial conditions $a_{i,i}=0$, $i\ge0$, $a_{i,0}=\rho^{i-1}$ and $a_{0,i}=-\rho^{i-1}$, $i\ge1$. Then $$\det_{0\le i,j\le 2n-1}(a_{i,j})=(1+x)^{2(n-1)^2} (x+\rho)^{2n-2}.$$ \endproclaim \demo{Proof} We compute again the generating function $F(u,v)=\sum _{i,j\ge0} ^{}a_{i,j}u^iv^j$. This time we have $$F(u,v)=\frac {(1-u-v+\rho uv)} {(1-u-v-xuv)} \frac {(u-v)} {(1-\rho u)(1-\rho v)}.$$ Also here, we apply the LU-factorization method. However, we cannot apply LU-factorization directly to the original matrix since it is skew-symmetric and, hence, all odd-dimensional principal minors vanish. Instead, we apply LU-factorization to the matrix that results from the original matrix by first reversing the order of rows and then transposing the resulting matrix. This is the matrix $M=(a_{2n-1-j,i})_{0\le i,j\le 2n-1}$. I claim that $$M\cdot U=L,$$ where $U=(U_{i,j})_{0\le i,j\le n-1}$ with $$U_{i,j}=\cases 0&\kern-4.1cm i>j,\\ \frac {(-1)^{j-i}} {\rho}\(\binom {j-1}i+\binom {j-1}{i-1}\rho\)& \kern-4.1cm i\le j<2n-1,\\ 1&\kern-4.1cm i=j=2n-1,\\ -\frac {1} {(\rho+x)(1+x)^{n-2}}\Big(\sum _{s=0} ^{n-i/2-2}2^{2s+1} \binom {n-\frac {i} {2}-1+s}{2s+1} \sum _{t=0} ^{i/2-s-1}\binom {n-1}tx^{\frac {i} {2}-s-t-1}\\ \kern1cm +\rho \sum _{s=0} ^{n-i/2-1}2^{2s} \binom {n-\frac {i} {2}-1+s}{2s} \sum _{t=0} ^{i/2-s-1}\binom {n-1}tx^{\frac {i} {2}-s-t-1}\Big)\\ &\kern-4.1cm i\text { even, }i0$ by assumption. We obtain $$\align (M\cdot U)_{i,j}&= -\coef{u^{2n-1}}\frac {(u-1)^{j-1}} {\rho(1-\rho+\rho u+xu)} \bigg(-(1-\rho)(1-\rho u) \rho^{i-1}\\ &\kern1cm +{u(\rho+x)(u^2x+2u-1)} \frac {(1+xu)^{i-1}} {(1-u)^i}\bigg). \tag2.7\endalign$$ Let first $i0$. We split the expression on the right-hand side of (2.9) into $$\align (M&\cdot U)_{i,2n-1}=-\rho^{i-1} -\frac {1} {(\rho+x)(1+x)^{n-2}}\coef{v^i} \frac {(1-u-v+\rho uv)(u-v)} {(1-u-v-xuv)(1-\rho v)}\\ &\kern7cm \times \frac {(1+x)^{n-1}} {(1-2u-xu^2)}\\ &\kern.9cm +\frac {1} {(\rho+x)(1+x)^{n-2}}\coef{u^{2n-1}v^i} \frac {(1-u-v+\rho uv)(u-v)} {(1-u-v-xuv)(1-\rho v)} \frac {u(1-u)^{2n-2}} {(1-2u-xu^2)}\\ &=-\rho^{i-1}+\rho^{i-1}\frac {1+x} {\rho+x}\\ &\kern.9cm +\frac {1} {(\rho+x)(1+x)^{n-2}}\coef{u^{2n-1}v^i} \frac {(1-u-v+\rho uv)(u-v)} {(1-u-v-xuv)(1-\rho v)} \frac {u(1-u)^{2n-2}} {(1-2u-xu^2)}.\\ \tag2.10 \endalign$$ To the expression in the last line we apply again the partial fraction expansion (2.6). After having substituted this in (2.10), we may directly read off the coefficient of $v^i$. Once again, the first term on the right-hand side of (2.6) does not contribute to it since we have $i>0$ by assumption. We obtain $$\align (&M\cdot U)_{i,2n-1}\\ &=\rho^{i-1}\frac {1-\rho} {\rho+x} +\frac {1} {(\rho+x)(1+x)^{n-2}}\coef{u^{2n-1}} \frac {u(1-u)^{2n-2}} {(1-2u-xu^2)}\frac {1} {(1-\rho+\rho u+xu)}\\ &\kern2cm \times \bigg(-{(1-\rho)(1-\rho u)} \rho^{i-1} + {u(\rho+x)(u^2x+2u-1)} \frac {(1+xu)^{i-1}} {(1-u)^i}\bigg)\\ &=\rho^{i-1}\frac {1-\rho} {\rho+x} -\frac {1} {(\rho+x)(1+x)^{n-2}}\coef{u^{2n-1}} \frac {u(1-u)^{2n-2}(1-\rho)(1-\rho u)} {(1-2u-xu^2)(1-\rho+\rho u+xu)} \rho^{i-1}\\ &\kern1cm -\frac {1} {(\rho+x)(1+x)^{n-2}}\coef{u^{2n-1}} \frac {u^2(1-u)^{2n-2-i}(\rho+x)(1+xu)^{i-1}} {(1-\rho+\rho u+xu)}. \tag2.11 \endalign$$ Let us for the moment assume that $i<2n-1$. The coefficient of $u^{2n-1}$ in the last line can be easily determined by making use of the previously used fact that the coefficient of $u^N$ in a formal power series of the form $P(u)/(1-q u)$, where $P(u)$ is a polynomial in $u$ of degree $\le N$, is equal to $q^NP(1/q)$. Thus we obtain $$\align (&M\cdot U)_{i,2n-1}\\ &=\rho^{i-1}\frac {1-\rho} {\rho+x} -\frac {1} {(\rho+x)(1+x)^{n-2}}\coef{u^{2n-1}} \frac {u(1-u)^{2n-2}(1-\rho)(1-\rho u)} {(1-2u-xu^2)(1-\rho+\rho u+xu)} \rho^{i-1}\\ &\kern1cm +\frac {1} {(\rho+x)(1+x)^{n-2}}\frac {(\rho+x)(1+x)^{2n-3}} {(\rho-1)^{2n-2}}\rho^{i-1}. \tag2.12 \endalign$$ In order to determine the remaining coefficient of $u^{2n-1}$ on the right-hand side, we apply once again a partial fraction expansion: $$\multline \frac {(1-\rho u)} {(1-2u-xu^2)(1-\rho+\rho u+xu)}= \frac {\rho+x} {(1+x)(1-\rho+\rho u+xu)}\\ -\frac {1} {2(1+x)}\frac {1} {(u-\om)}-\frac {1} {2(1+x)}\frac {1} {(u-\overline\om)}, \endmultline$$ where $\om=-\frac {1} {x}(1-\sqrt{1+x})$ and $\overline\om=-\frac {1} {x}(1+\sqrt{1+x})$. This is now substituted in (2.12). By a routine calculation, which uses again the above fact of how to read off the coefficient of $u^N$ in a formal power series of the form $P(u)/(1-q u)$, it is seen that (2.12) vanishes for $i<2n-1$. On the other hand, if we have $i=2n-1$, then the expression in the last line of (2.11) is equal to $$-\frac {1} {(\rho+x)(1+x)^{n-2}}\coef{u^{2n-1}} \frac {u^2(\rho+x)(1+xu)^{2n-2}} {(1-u)(1-\rho+\rho u+xu)}.$$ In order to read off the coefficient of $u^{2n-1}$ in this expression, we must apply a partial fraction expansion another time: $$\frac {1} {(1-u)(1-\rho+\rho u+xu)}=\frac {1} {(1+x)} \frac {1} {(1-u)}+\frac {1} {(1+x)}\frac {\rho+x} {(1-\rho+\rho u+xu)}.$$ After having done that, a routine calculation has to be performed that shows that for $i=2n-1$ the expression (2.11) simplifies to $-(1+x)^{n-1}$, in accordance with (2.4). This completes the proof of the theorem.\quad \quad \qed \enddemo Finally we proof the conjectured evaluation of $\det(\tilde A(2n))$ on page~4 of \cite{\BacRAZ}. As it turns out, this is much simpler than the previous evaluations. \proclaim{Theorem 3}Let $(a_{i,j})_{i,j\ge0}$ be the sequence given by the recurrence $$a_{i,j}=a_{i-1,j}+a_{i,j-1}+x\,a_{i-1,j-1},\quad \quad i,j\ge1 $$ and the initial conditions $a_{i,0}=i$ and $a_{0,i}=-i$, $i\ge0$. Then $$\det_{0\le i,j\le 2n-1}(a_{i,j})=(1+x)^{2n(n-1)}.$$ \endproclaim \demo{Proof} Again, we compute the generating function $F(u,v)=\sum _{i,j\ge0} ^{}a_{i,j}u^iv^j$. Here we get $$F(u,v)=\frac {u-v} {(1-u-v-uvx)(1-u)(1-v)}.$$ This can be written in the following way: $$\align F(u,v)&=\frac {u} {(1-u)(1-u-v-uvx)}-\frac {v} {(1-v)(1-u-v-uvx)}\\ &=\frac {u} {(1-u)^2(1-v)\(1-\frac {uv(x+1)} {(1-u)(1-v)}\)}-\frac {v} {(1-u)(1-v)^2\(1-\frac {uv(x+1)} {(1-u)(1-v)}\)}\\ &=\sum _{\ell=0} ^{\infty}\frac {u^\ell v^\ell (x+1)^\ell} {(1-u)^{\ell+1}(1-v)^{\ell+1}}\(\frac {u} {1-u}-\frac {v} {1-v}\). \endalign$$ Therefore we have $$a_{i,j}=\sum _{\ell=0} ^{}\(\binom {i}{\ell+1}\binom {j}\ell-\binom {i}{\ell}\binom {j}{\ell+1}\)(x+1)^\ell.$$ Now we apply the following row and column operations: We subtract row $i$ from row $i+1$, $i=2n-2,2n-3,\dots,0$, and subsequently we subtract column $j$ from column $j+1$, $j=2n-2,2n-3,\dots,0$. This is repeated another $2n-2$ times. It is not too difficult to see (using the above expression for $a_{i,j}$) that, step by step, the rows and columns are ``emptied'' until finally the determinant $$\det\pmatrix 0&-1&0&0&\innerhdotsfor2\after\quad &0\\ 1&0&-X&0&\innerhdotsfor2\after\quad &0\\ 0&X&0&-X^2&\innerhdotsfor2\after\quad &0\\ 0&0&X^2&0&\ddots&&0\\ &&\ddots&\ddots&\ddots&\ddots&\vdots\\ 0&\innerhdotsfor3\after\quad &X^{2n-3}&0&-X^{2n-2}\\ 0&\innerhdotsfor3\after\quad &0&X^{2n-2}&0 \endpmatrix$$ is obtained, where we have written $X$ for $x+1$. The theorem follows now immediately.\quad \quad \qed \enddemo \subhead 3. Some binomial determinants\endsubhead In this section we prove some evaluations of binomial determinants which arose from conjectures in \cite{\BacRAZ} and subsequent private discussion with Roland Bacher. The first theorem in this section proves a common extension of two conjectures on page~15 in \cite{\BacRAZ}. It is also stated (without proof) in \cite{\BacRAA, p.~14}. \proclaim{Theorem 4}We have $$\det_{0\le i,j\le n-1}\(\binom {2i+2j+a}i-\binom {2i+2j+a}{i-1}\) =2^{\binom n2}.\tag3.1$$ \endproclaim \demo{Proof} We apply LU-factorization. Let us denote the matrix in (3.1) by $M$. I claim that $$M\cdot U=L,$$ where $U=\((-1)^{j-i}\binom ji\)_{0\le i,j\le n-1}$, and where $L=(L_{i,j})_{0\le i,j\le n-1}$ is a lower triangular matrix with diagonal entries $1,2,4,8,\dots,2^{n-1}$. It is obvious that the claim immediately implies the theorem. Again, we verify the claim by a direct calculation. Let $0\le i\le j\le n-1$. The $(i,j)$-entry of $M\cdot U$ is $$(M\cdot U)_{i,j}=\sum _{k=0} ^{j} {{\left( -1 \right) }^{j - k}}{j\choose k} {{\left( 1 + a + 2\,k \right) \,\left( a + 2\,i + 2\,k \right) !}\over {i!\,\left( 1 + a + i + 2\,k \right) !}}. $$ In standard hypergeometric notation $${}_r F_s\!\left[\matrix a_1,\dots,a_r\\ b_1,\dots,b_s\endmatrix; z\right]=\sum _{m=0} ^{\infty}\frac {\po{a_1}{m}\cdots\po{a_r}{m}} {m!\,\po{b_1}{m}\cdots\po{b_s}{m}} z^m\ ,$$ where the Pochhammer symbol $(\alpha)_m$ is defined by $(\alpha)_m:=\alpha(\alpha+1)\cdots(\alpha+m-1)$, $m\ge1$, $(\alpha)_0:=1$, the above binomial sum can be written as $${{\left( -1 \right) }^j} {{\left( 1 + a \right) \, ({ \textstyle 2 + a + i}) _{i-1} }\over {i!}} {} _{4} F _{3} \!\left [ \matrix { {3\over 2} + {a\over 2}, {1\over 2} + {a\over 2} + i, 1 + {a\over 2} + i, -j}\\ { {3\over 2} + {a\over 2} + {i\over 2}, 1 + {a\over 2} + {i\over 2}, {1\over 2} + {a\over 2}}\endmatrix ; {\displaystyle 1}\right ]. $$ To this $_4F_3$-series we apply the contiguous relation $$ {} _{4} F _{3} \!\left [ \matrix { a, b,c,d}\\ {e,f,g}\endmatrix ; {\displaystyle z}\right ] = {} _{4} F _{3} \!\left [ \matrix { a - 1, b,c,d}\\ { e,f,g}\endmatrix ; {\displaystyle z}\right ] + {z } {{bcd }\over {efg}} {} _{4} F _{3} \!\left [ \matrix { a, b+1,c+1,d+1}\\ {e+1,f+1,g+1}\endmatrix ; {\displaystyle z}\right ]. $$ We obtain $$\multline {{\left( -1 \right) }^j} {{\left( 1 + a \right) \, ({ \textstyle 2 + a + i}) _{i-1} }\over {i!}} {} _{3} F _{2} \!\left [ \matrix { {1\over 2} + {a\over 2} + i, 1 + {a\over 2} + i,-j}\\ { {3\over 2} + {a\over 2} + {i\over 2}, 1 + {a\over 2} + {i\over 2}}\endmatrix ; {\displaystyle 1}\right ]\\ - {{\left( -1 \right) }^j} {{2j\, ({ \textstyle 4 + a + i}) _{i-1} }\over {i!}} {} _{3} F _{2} \!\left [ \matrix { {3\over 2} + {a\over 2} + i, 2 + {a\over 2} + i, 1 - j}\\ { {5\over 2} + {a\over 2} + {i\over 2}, 2 + {a\over 2} + {i\over 2}}\endmatrix ; {\displaystyle 1}\right ]. \endmultline$$ Now we apply the transformation formula (cf\. \cite{\BailAA, Ex.~7, p.~98}) $$ {} _{3} F _{2} \!\left [ \matrix { a, b, -n}\\ { d, e}\endmatrix ; {\displaystyle 1}\right ] = {{ ({ \textstyle -a - b + d + e}) _{n} }\over {({ \textstyle e}) _{n} }} {} _{3} F _{2} \!\left [ \matrix { -n, -a + d, -b + d}\\ { d, -a - b + d + e}\endmatrix ; {\displaystyle 1}\right ] , \tag3.2$$ where $n$ is a nonnegative integer, to both $_3F_2$-series. The above expression then becomes $$\multline {{\left( -1 \right) }^j} {{\left( 1 + a \right) \, ({ \textstyle 1 - i}) _{j} \, ({ \textstyle 2 + a + i}) _{i-1} }\over {i!\,({ \textstyle 1 + {a\over 2} + {i\over 2}}) _{j} }} {} _{3} F _{2} \!\left [ \matrix { -j, {1\over 2} - {i\over 2}, 1 - {i\over 2}}\\ { {3\over 2} + {a\over 2} + {i\over 2}, 1 - i}\endmatrix ; {\displaystyle 1}\right ] \\ - {{\left( -1 \right) }^j} {{2j\, ({ \textstyle 1 - i}) _{j-1} \, ({ \textstyle 4 + a + i}) _{i-1} }\over {i!\,({ \textstyle 2 + {a\over 2} + {i\over 2}}) _{j-1} }} {} _{3} F _{2} \!\left [ \matrix { 1 - j, {1\over 2} - {i\over 2}, 1 - {i\over 2}}\\ { {5\over 2} + {a\over 2} + {i\over 2}, 1 - i}\endmatrix ; {\displaystyle 1}\right ]. \endmultline$$ If $j\ge i$, the first summand is always 0 because of the term $({ \textstyle 1 - i}) _{j}$. The second summand vanishes as long as $j>i$ because of the term $({ \textstyle 1 - i}) _{j-1} $. Thus, the matrix $L$ is indeed a lower triangular matrix. It remains to evaluate the above expression for $i=j$. As we already mentioned, in that case it reduces to the second term, which itself reduces to $$-{{\left( -1 \right) }^i}{{2i\, ({ \textstyle 1 - i}) _{i-1} \, ({ \textstyle 4 + a + i}) _{i-1} }\over {i!\,({ \textstyle 2 + {a\over 2} + {i\over 2}}) _{i-1} }} {} _{2} F _{1} \!\left [ \matrix { {1\over 2} - {i\over 2}, 1 - {i\over 2}}\\ { {5\over 2} + {a\over 2} + {i\over 2}}\endmatrix ; {\displaystyle 1}\right ]. $$ This $_2F_1$-series can be evaluated by means of Gau\3' $_2F_1$-summation (cf\. \cite{\SlatAC, (1.7.6); Appendix (III.3)}) $$ {} _{2} F _{1} \!\left [ \matrix { a, b}\\ { c}\endmatrix ; {\displaystyle 1}\right ] = \frac {\Gamma( c)\,\Gamma( c-a-b)} {\Gamma( c-a)\,\Gamma( c-b )}. \tag3.3$$ After some further simplification we obtain $2^i$, as desired.\quad \quad \qed \enddemo Our next theorem presents a variation of the previous theorem. It is again a common extension of two conjectures in \cite{\BacRAZ, p.~15}. It is also stated (without proof) in \cite{\BacRAA, p.~14}. \proclaim{Theorem 5}We have $$\det_{0\le i,j\le n-1}\(\binom {2i+2j+a}{i+1}-\binom {2i+2j+a}{i}\) =2^{\binom n2}\frac {\prod _{i=0} ^{n-1}(a+2i-1)} {n!}.\tag3.4$$ \endproclaim \demo{Proof} We apply LU-factorization. Let us denote the matrix in (3.4) by $M$. I claim that $$M\cdot U=L,$$ where $$U=\((-1)^{j-i}\binom ji\frac {(a+2j-1)} {(a+2i-1)}\)_{0\le i,j\le n-1},$$ and where $L=(L_{i,j})_{0\le i,j\le n-1}$ is a lower triangular matrix with diagonal entries $${a-1},\frac {2} {2}(a+1),\frac {4} {3}(a+3),\frac {8} {4}(a+5),\dots, \frac {2^{n-1}}{n}(a+2n-3).$$ It is obvious that the claim immediately implies the theorem. Let us do the required calculations. Let $0\le i\le j\le 2n-1$. The $(i,j)$-entry of $M\cdot U$ is $$(M\cdot U)_{i,j}=\sum _{k=0} ^{j} {{\left( -1 \right) }^{j - k}}{j\choose k} {{\left( a + 2\,j-1 \right) \,\left( a + 2\,i + 2\,k \right) !}\over {\left(i+1 \right) !\,\left( a + i + 2\,k \right) !}}. $$ We convert this binomial sum into hypergeometric notation. The result is $${{\left( -1 \right) }^j}{{\left( a + 2\,j -1\right) \, (a + 2\,i)! }\over {(i+1)!\,(a+i)!}} {} _{3} F _{2} \!\left [ \matrix { -j, {1\over 2} + {a\over 2} + i, 1 + {a\over 2} + i}\\ { 1 + {a\over 2} + {i\over 2}, {1\over 2} + {a\over 2} + {i\over 2}}\endmatrix ; {\displaystyle 1}\right ]. $$ This time it is not even necessary to apply a contiguous relation. We may directly apply the transformation formula (3.2) and obtain $${{\left( -1 \right) }^j} {{\left( a + 2\,j -1\right) \, (a + 2\,i)! \, ({ \textstyle -i}) _{j} }\over {(i+1)!\,(a+i)!\, ({ \textstyle {1\over 2} + {a\over 2} + {i\over 2}}) _{j} }} {} _{3} F _{2} \!\left [ \matrix { -j, -{{i}\over 2}, {1\over 2} - {i\over 2}}\\ { 1 + {a\over 2} + {i\over 2}, -i}\endmatrix ; {\displaystyle 1}\right ]. $$ Again, this expression is 0 if $j>i$ since it contains the term $(-i)_j$. If $j=i$ then it reduces to $${{\left( -1 \right) }^i} {{\left( a + 2\,i -1\right) \, (a + 2\,i)! \,({ \textstyle -i}) _{i} }\over {(i+1)!\,(a+i)!\, ({ \textstyle {1\over 2} + {a\over 2} + {i\over 2}}) _{i} }} {} _{2} F _{1} \!\left [ \matrix {- {{i}\over 2}, {1\over 2} - {i\over 2}}\\ { 1 + {a\over 2} + {i\over 2}}\endmatrix ; {\displaystyle 1}\right ]. $$ The $_2F_1$-series can again be evaluated by means of the Gau\3 summation formula (3.3). After further simplification we obtain $2^i(a+2i-1)/(i+1)$, as desired.\quad \quad \qed \enddemo The final determinant evaluation that we present here proves a common generalization of determinant evaluations that arose in discussion with Roland Bacher. It is interesting to note that the limiting case where $Y\to\infty$, i.e., the evaluation of the determinant $\det_{0\le i,j\le 2n-1}\((i-j)\,(X+i+j)! \)$ is covered by a theorem by Mehta and Wang \cite{\MeWaAA}. In fact, the main theorem of \cite{\MeWaAA} gives the evaluation of the more general determinant $\det_{0\le i,j\le n-1}\((Z+i-j)\,(X+i+j)! \)$. It seems difficult to find a common extension of the Mehta and Wang evaluation and the evaluation below, i.e., an evaluation of $$\det_{0\le i,j\le n-1}\((Z+i-j)\frac {(X+i+j)!} {(Y+i+j)!} \).$$ Neither the method used by Mehta and Wang nor the ``identification of factors method" used below seem to help. \proclaim{Theorem 6}Let $X$ and $Y$ be arbitrary nonnegative integers. Then $$\multline \det_{0\le i,j\le 2n-1}\((i-j)\frac {(X+i+j)!} {(Y+i+j)!} \) =\prod _{i=0} ^{2n-1}\frac {(X+i)!} {(Y+i+2n-1)!}\\ \times \prod _{i=0} ^{n-1}(2i+1)!^2 (X+2i+1) (Y+4n-2i-2) (X-Y-2i)^{2n-2i-2} (X-Y-2i-1)^{2n-2i-2}. \endmultline\tag3.5$$ \endproclaim \remark{Remark} The theorem is formulated only for integral $X$ and $Y$. But in fact, if we replace the factorials by the appropriate gamma functions then the theorem would also make sense and be true for complex $X$ and $Y$. \endremark \demo{Proof} To begin with, we take some factors out of the determinant: $$\multline \det_{0\le i,j\le 2n-1}\((i-j)\frac {(X+i+j)!} {(Y+i+j)!} \)= \prod _{i=0} ^{2n-1}\frac {(X+i)!} {(Y+i+2n-1)!}\\ \times \det_{0\le i,j\le 2n-1}\big((i-j)\,(X+i+1)_{j}\,(Y+i+j+1)_{2n-j-1} \big). \endmultline$$ If we compare with (3.5), then we see that it suffices to establish that $$\multline \det_{0\le i,j\le 2n-1}\big((i-j)\,(X+i+1)_{j}\,(Y+i+j+1)_{2n-j-1} \big) \\= \prod _{a=0} ^{n-1}(2a+1)!^2 (X+2a+1) (Y+4n-2a-2) \\ \cdot (X-Y-2a)^{2n-2a-2} (X-Y-2a-1)^{2n-2a-2} . \endmultline\tag3.6$$ This time we apply a different method. We make use of the {\it ``identification of factors" method\/} as explained in \cite{\KratBN, Sec.~2.4}. Let us denote the determinant in (3.6) by $T_n(X,Y)$. Here is a brief outline of how we proceed. The determinant $T_n(X,Y)$ is a polynomial in $X$ and $Y$. In the first step we shall show that the product on the right-hand side of (3.6) divides the determinant as a polynomial in $X$ and $Y$. Then we compare the degrees in $X$ and $Y$ of the determinant and the right-hand side of (3.6). As it turns out, the degree of the determinant is at most the degree of the product. Hence, the determinant must be equal to a constant multiple of the right-hand side of (3.6). Finally, in the last step, this constant is computed by setting $X=-2n$ on both sides of (3.6). \smallskip {\it Step 1. The product $\prod _{a=0} ^{n-1}(X+2a+1)$ divides $T_n(X,Y)$}. Let $a$ be fixed with $0\le a\le n-1$. In order to show that $X+2a+1$ divides the determinant, it suffices to show that the determinant vanishes for $X=-2a-1$. The latter would follow immediately if we could show that the rows of the matrix underlying the determinant are linearly dependent for $X=-2a-1$. I claim that in fact we have $$\sum _{i=0} ^{a}(-1)^i\binom ai \frac {({ \textstyle Y+2n+2a - i }) _{i}} {({ \textstyle \frac Y 2+ 2a - i }) _{i}} \big(\text {row $(2a-i)$ of $T_n(-2a-1,Y)$}\big)=0.$$ This claim is easily verified. Indeed, if we restrict it to the $j$-th column, we see that we must check $$\sum _{i=0} ^{a}(-1)^i\binom ai \frac {({ \textstyle Y+2n+2a - i }) _{i}} {({ \textstyle \frac Y 2+ 2a - i }) _{i}} {\left( 2\,a - i - j \right)\, ({ \textstyle -i}) _{j} \,({ \textstyle Y + 2a - i + j + 1}) _{2n-j-1}} =0. $$ Because of the term $(-i)_j$ this equation is certainly true if $a