Advanced Computer Algebra for Determinants

Adv anced Compute r Algebra for Determinants Christoph K outsc han and Thotsap orn “Aek” Thanatipanonda Abstract. W e pro ve three conjectures concerning the eva luation of de- terminants, which are related to the counting of p lane partitions and rhombus tilings. One of them was p osed by George Andrews in 1980, the other tw o were by Guo ce Xin and Christian K rattenthaler. Our proofs emplo y computer a lgebra method s, namely , the holonomic ansatz prop osed by Doron Zeilb erger and vari ations thereof. These v ariations make Zeilb erger’s original approach even more p ow erful and allow for addressing a wider v ariet y of determinants. Finally , we present, as a chal lenge problem, a conjecture about a closed-form ev aluation of An- drews’s determinan t. Mathematics Sub ject Classification (2010). Primary 33F10; Secondary 15A15, 05B4 5. Keywords. determinan t, comput er algebra, h olonomic ansatz, rhombus tiling. 1. In tro du ction The concept o f determinants evolved as e a rly as 154 5 when Girola mo Car- dano tr ie d to solve systems of linear equations. The ma thematics communit y slowly r ealized the imp or tance of determinants; we had to wait for mo re tha n 200 years b efore s omeone formally defined the term “determinant”. It was first introduced b y Ca rl F r iedrich Gauß in his D isquisitiones Arithmetic ae in 1801. Determinants pos sess many nice prop erties and formulas such as multi- plicativity , in v ariance under row operations, Cramer ’s rule, etc. Every studen t now adays lear ns how to compute the determinant of a sp ecific g iven matrix, say , with fixed dimensio n a nd co nt aining numeric quantities as entries. O n the other hand, there are lots of matric e s with symbolic entries that hav e a nice clos ed-form formula for arbitrar y dimension. The fir st exa mple in the CK was supp or ted by the Austrian Science F und (FWF): P20162-N18, and in part by the gr an t DMU 03/17 of the Bulgarian National Science F und. TT was supp orted by the strategic program “Inno v ativ es O ¨ O 2010plus” by the Upper Austrian Go vernmen t. The final publi cation is av ailable at www.link. s pringer.com (Annals of Combinato ri cs, DOI 10.1007/s00026-013-0183-8). 2 Christoph K outschan and Thotsa p orn “Aek” Thana tipanonda history of mathematics and still the most pr ominent o ne is the V andermonde matrix. Starting in the mid 1 970’s , the impo rtance of ev aluating determinants bec ame even more significa nt when p eo ple related co untin g problems fro m combinatorics to the ev aluation of certain determinants. E v aluating determi- nants of matrices whose dimension v aries a ccording to a parameter is a deli- cate problem but at the same time a very imp ortant o ne. Around the s a me time the field of computer alg e br a emer ged. Dor on Zeilb erger was amo ng st the first mathematicians to rea liz e the imp ortance of c o mputer alge bra algo- rithms for combinatorial proble ms , sp ecial function identities, symbolic sum- mation, a nd many mor e. In his pa p er [17] he built the bridg e b etw een these t wo topics, namely , symbolic determinant ev aluation a nd computer a lgebra, and s ince then his holonomic ansatz has b een success fully applied to many problems related to the ev aluation of determinants. The most prominent one is pr obably a notor ious co njecture from enumerative combinatorics, the so- called q -TSPP conjecture, which was the only remaining op en problem from the famous list [14] by Richard Sta nley (it also app eared in [1 2]) un til it was recently proved [10] using Z eilb erger’s holo nomic a nsatz. In this pap er , we so lve some of the problems that a re lis ted in Chris tian Krattenthaler’s co mplement [13] to his celebrated essay [12] (the a ttentiv e reader may alrea dy hav e obse r ved that our title is an allusio n to this refer- ence). At the same time we show that the ho lonomic ansatz ca n b e mo dified in v arious wa ys in order to apply it to pa rticular pr oblems that co uld not b e addressed with the origina l metho d. 2. Zeilb erger’s Holonomic Ansatz F or sa ke of self-containedness we reca ll br iefly the or iginal holonomic a nsatz for determinant ev aluations as it w as pro p osed by Zeilber ger [17]. Its steps a re completely automa tic and pro duce a rigo rous pro of—provided that they can be successfully carr ied out in the example at hand. In particular , the approach relies on the existence of a “nice ” description for a n auxiliary function (it app ears as c n,j below); if such a description do es not e xist then the holonomic ansatz fails. That’s why we call it an “a ppr oach” or a n “ansatz”, rather than an algor ithm. Generally s pe aking, Z e ilb er ger’s holo nomic ansa tz addres s es determi- nant ev aluations of the t yp e det A n = det 1 6 i,j 6 n ( a i,j ) = b n ( n > 1) where the entries a i,j of the n × n matrix A n and the (conjectured) ev alu- ation b n (where b n 6 = 0 is required for all n > 1) ar e e x plicitly given. The underlying pr inciple is an induction arg ument on n . The base case a 1 , 1 = b 1 is easily chec ked. Now ass ume that the determinant ev aluation has b een prov en for n − 1. In particula r, it follows that det A n − 1 is nonze r o by the gener a l Adv anced Computer Alg e bra for Determinants 3 assumption o n b n . Hence the r ows of A n − 1 are linearly indep endent and thus the linear system      a 1 , 1 · · · a 1 ,n − 1 a 1 ,n . . . . . . . . . . . . a n − 1 , 1 · · · a n − 1 ,n − 1 a n − 1 ,n 0 · · · 0 1           c n, 1 . . . c n,n − 1 c n,n      =      0 . . . 0 1      has a unique solution. In the well-known Laplace expa nsion formula (her e with resp ect to the la st row) det 1 6 i,j 6 n ( a i,j ) = n X j =1 ( − 1) n + j M n,j a n,j the expressio n ( − 1) n + j M n,j is called the ( n, j )-cofactor of A n . The minor M n,j is the deter mina nt of the ma trix obtained by removing the n -th row and the j -th c olumn. The ab ove linea r system has b een constructed in such a wa y that the en try c n,j in its solutio n is precisely the ( n, j )-cofactor of A n divided by its ( n, n )-co factor (which is just the ( n − 1 )-determinant). This fact can e a sily be seen by Cramer’s rule, i.e., by considering the matrix A ( i ) n that we obtain from A n by replacing the last row by the i -th row. F or 1 6 i < n , the Laplace expansion of A ( i ) n corres p o nds exactly to the i -th equation in our linear system. Now the determina nt of A n is given by b n − 1 n X j =1 c n,j a n,j . T o complete the induction step it r emains to show that this expr e ssion is equal to b n . The pr oblem is that we cannot exp ect to o btain a clos e d- form expres- sion for the quantit y c n,j (otherwise, we certa inly would b e able to derive a closed form for the determinant and we were done). Instead, we will guess an implicit, r ecursive definition for a biv ariate se q uence c n,j and then pr ov e that it s atisfies c n,n = 1 ( n > 1) , (1) n X j =1 c n,j a i,j = 0 (1 6 i < n ) , (2) n X j =1 c n,j a n,j = b n b n − 1 ( n > 1) . (3) F rom the fir st tw o identit ies which co rresp ond to the linear sy stem given ab ov e, it follows that our gues sed c n,j is indeed the normalized ( n, j )-cofactor . Ident ity (3) then certifies that the determinant ev aluates to b n . Hence the sequence c n,j plays the rˆ ole of a cer tificate for the determinant ev aluation. Now what kind of implicit definition for c n,j could we think of ? Of course, there was a go o d reaso n for Z eilb erger to na me his approach the 4 Christoph K outschan and Thotsa p orn “Aek” Thana tipanonda “holonomic ansatz”. That is to say , b ecaus e he had the clas s of holo nomic functions (or b etter, sequences ) in mind when he formulated his approach. In short, this clas s consists of multi-dimensional sequences that satisfy linear recurrence relations with p o lynomial co efficients, such that the sequence is uniquely determined by sp ecifying fi nitely many initial v alues (we omit some additional technical co nditio ns here). What makes the use o f this class of se- quences co nv enient is the fact that it is not only clo s ed under the ba sic a rith- metic oper ations (addition, m ultiplication), but also under more adv anced op erations such as sp ecia lization, diagonaliza tion, and definite s umma tion; all these are needed to prov e (1), (2), a nd (3). Mor eov er, there exist known algorithms for p erforming zero tes ts and the pr eviously men tioned op era - tions. F or more details o n holonomic functions and related alg orithms see, e.g., [15, 3 , 8]). T hus if the matrix entries a i,j are ho lo nomic and if luckily the auxiliary function c n,j turns out to b e holo nomic, then the approa ch will succeed to pro duce a r e c urrence for the quotient b n /b n − 1 . 3. An O ld Pr oblem by George Andrews In the context of en umerating cer tain cla sses of plane partitions, namely , cyclically symmetr ic ones and descending ones, Geor g e Andr e ws [2] encoun- tered an intriguing determinant which he p osed as a challenging problem. In Krattenthaler’s survey [13], it app ear s as Pr oblem 34 . This conjecture do es not even deal with a closed-for m ev aluation, but it is “ only” a b o ut the quo- tien t o f tw o consecutive determinants; a s ituation that strong ly sugge sts to employ Zeilb erg er’s holonomic ansa tz! Theorem 1. L et the determinant D 1 ( n ) b e define d by D 1 ( n ) := det 1 6 i,j 6 n  δ i,j +  µ + i + j − 2 j  wher e µ is an indeterminate and δ i,j is the Kr one cker delta fun ction. Then the fol lowing r elation holds: D 1 (2 n ) D 1 (2 n − 1) = ( − 1) ( n − 1)( n − 2) / 2 2 n  1 2 ( µ + 2 n )  ⌊ ( n +1) / 2 ⌋  1 2 ( µ + 4 n + 1)  n − 1 ( n ) n  1 2 ( − µ − 4 n + 3)  ⌊ ( n − 1) / 2 ⌋ . Pr o o f. By lo oking at the firs t few ev aluations of D 1 ( n ) (for 1 6 n 6 8 they are explicitly displayed in [1 3]), we see that only the quotient D 1 (2 n ) /D 1 (2 n − 1) is nice, but not D 1 (2 n + 1) /D 1 (2 n ). The r e ason is the o ccurrence of irr educible nonlinear factors that change every tw o steps, i.e ., D 1 (2 n ) and D 1 (2 n − 1) share the same “ugly” factor (thus their q uotient is nice), but in D 1 (2 n +1) the “ugly” part will b e different (and therefor e the quotient D 1 (2 n + 1) /D 1 (2 n ) do es not factor nicely). W e fir st tr ie d Zeilb erger ’s original approach o n the deter minant D 1 ( n ). But we didn’t even succeed to guess the r ecurrences for c n,j as they either ar e extraor dinarily larg e or do not exist at all. Moreov er, we have go o d reasons to Adv anced Computer Alg e bra for Determinants 5 belie ve that the quotient D 1 (2 n + 1) /D 1 (2 n ) is not holono mic (see Section 6), in which ca se we know a prior i that the appr oach cannot succe ed. Therefore we hav e to come up with a v ariation of Zeilb erg er’s appr oach, that pays attention to even n o nly . This means that w e c o nsider the normal- ized co factors c n,j only for matrices of even size. The new identities, to b e verified, will be c n, 2 n = 1 ( n > 1) , (1a) 2 n X j =1 c n,j a i,j = 0 (1 6 i < 2 n ) , (2a) 2 n X j =1 c n,j a 2 n,j = b 2 n b 2 n − 1 ( n > 1) . (3a) In order to come up with an appro priate guess for the yet unknown function c n,j , we compute the nor malized cofa c tors for all even-size matrices up to dimension 3 0. This gives a 15 × 30 array with v alues in Q ( µ ) that is used for g uessing linear recur rences for c n,j . F or this step, Kauer s’s Mathematica pack ag e G uess ha s b een employ ed, see [7] for mor e details . T o give the reader an impr ession o f what the output lo oks like, we display the results here in truncated form: Whenever the abbreviatio n h k terms i a ppe a rs, it indica tes that this p oly nomial cannot b e factored into smalle r pieces, and tha t for better rea dability it is not display ed in full size here. 2 n ( j + 1)(2 n − 1)(2 j + µ )( j − 2 n )( j − 2 n + 1) × ( µ + 4 n − 5)( µ + 4 n − 3)( j + µ + 2 n − 1) c n,j = j ( j + µ − 1 )(2 j + µ − 1)( j − 2 n + 3)( µ + 4 n − 3 ) × ( j 4 + 2 j 3 µ + . . . h 24 terms i + 1 2 ) c n − 1 ,j +1 − ( j + 1 )( j + µ + 2 n − 3)(2 j 6 µ + 8 j 6 n + . . . h 92 terms i − 2 10 µn ) c n − 1 ,j (4) ( j − 1)( j + µ − 3)( j + µ − 2 )(2 j + µ − 4)( j − 2 n )( j + µ + 2 n − 1) c n,j = j ( j + µ − 3)(4 j 4 + 8 j 3 µ + . . . h 26 terms i + 16) c n,j − 1 − j ( j − 1 )( j + µ − 2)(2 j + µ − 2 )( j − 2 n − 2)( j + µ + 2 n − 3) c n,j − 2 (5) When transla ted into op erator notation, the tw o recurr ences (4) and (5) constitute a Gr ¨ obner bas is of the left ideal that they ge nerate in the co rre- sp onding op era tor algebra . As a consequence it fo llows that they ar e compa t- ible; this means that s tarting fro m s ome given initial v alues, thes e recur rences will a lways pr o duce the sa me v alue fo r a particula r c n 0 ,j 0 , indep endently from the or der in which they are a pplied. The supp o rt o f the re c ur rences sugg ests that fixing the initial v alues c 1 , 1 = − µ/ 2 and c 1 , 2 = 1 is sufficient: (5 ) would pro duce c 1 ,j for all j > 2, and then (4 ) could be used to obtain the full ar ray of v alues ( c n,j ) n,j > 1 . 6 Christoph K outschan and Thotsa p orn “Aek” Thana tipanonda Unfortunately it is not that easy , since there are tw o disturbing pheno- mena. The first is the factor j − 2 n that a pp e ars in b oth leading co e fficients. Hence for computing c n, 2 n none of the recurre nces is applica ble and we are stuck. Note that the factor j − 2 n + 1 in the lea ding co efficien t of (4) is not a problem since we can use (5) instead. W e ov erc o me the problem with c n, 2 n by finding a recurre nc e of the form p 2 c n +2 ,j +4 + p 1 c n +1 ,j +2 + p 0 c n,j = 0 , p 0 , p 1 , p 2 ∈ Q [ n, j, µ ] (6) in the ideal genera ted by (4) and (5). The co efficient p 2 do es not v anish for j = 2 n , and thu s, together with the additional initial v alue c 2 , 4 = 1 , the recurrence (6) allows to compute the v alues c n, 2 n . The second unplea sant phenomenon is the free parameter µ which can cause the same effect for certain choices. F or example, if µ = − 6 then we can- not compute c 2 , 3 from the given initial v alues since b oth leading co efficients v anish simultaneously by virtue of the factors j − 2 n + 1 and j + µ + 2 n − 1. W e handle it by restricting the par ameter µ to the r eal num ber s > µ 0 for a certain µ 0 ∈ R , and by showing that all our calculations a re sound under this assumption (in most cases we use µ > 0 or µ > 2 ). But since the determina nt for every n ∈ N is a p olyno mial in µ we can extend our r esult afterwards to all µ ∈ C . Alternatively , one ca n argue that µ is a formal parameter and therefore e x pressions like µ + 6 ar e not zer o as they are considered as elemen ts in the p o lynomial ring C [ µ ]. W e hav e now prepa r ed the stage for ex e cuting the main part of Zeil- ber ger’s a pproach. Identit y (1a) is e a sily shown with the help of r ecurrence (6): Substituting j → 2 n pro duces a recurrence for the entries c n, 2 n , which to - gether with the initial v alues implies that c n, 2 n = 1 for all n . Iden tities (2a) and (3 a) a re proven a utomatically as well, since they are standard appli- cations of holonomic closure prop erties and summation techniques (cre a tive telescoping). F or these tasks we hav e used the first author’s Ma thematica pack ag e Hol onomi cFunc tions [9]. The int eres ted r eader is in vited to study our computations in detail, by downloading the electronic supplementary material fro m o ur webpage [1 1].  4. In terlude: The Double-Step V arian t In this section, we prop ose a v ariant of Z eilb erger’s holonomic ansatz, de- scrib ed in Section 2, that enlarg es the class o f determinants which can b e treated b y this kind of ansatz. The condition b n 6 = 0 for all n > 1 im- po ses alrea dy so me restr ic tion. F or exa mple, when studying a Pfaffian Pf ( A ) for some skew-symmetric ma trix A , one could b e tempted to apply Zeil- ber ger’s approa ch to the determinant of A ; recall that P f ( A ) 2 = det( A ). The problem then is that det( A ) is zero whenever the dimension of A is o dd. Hence one would like to study the quotient b n /b n − 2 instead of the forbidden b n /b n − 1 ; a s in the pr evious section, n has to b e r estricted to the even int eger s. This dilemma can b e so lved by mo difying Zeilb erg er’s ansatz sub ject to the Laplace expansion o f Pfaffia ns, s e e [6]. Adv anced Computer Alg e bra for Determinants 7 On the other hand there are deter minants which do factor nicely for even dimensions but not for o dd ones . Also her e , we exp ect the quotient b n /b n − 2 to b e nice, wherea s the expr ession b n /b n − 1 might not even sa tisfy a linear recurrence and hence could not b e ha ndled by Zeilb er ger’s holonomic ansatz a t all. Similar ly when the closed fo rm b n is differe nt for even and o dd n : While here the o riginal approach could proba bly work in principle, one may not succeed b ecause of the computational co mplex it y that is caused by studying the quotient b n /b n − 1 , which is exp ected to b e more co mplica ted than b n /b n − 2 . See Theorems 2 a nd 5 b elow for s uch examples, which actually hav e motiv ated us to pro p o se the following v ariant. As we announced we now generalize Zeilb erg er’s metho d in order to pro duce a r ecurrence for the quotient o f determinants who se dimensions differ by tw o. As b efore, let M = ( a i,j ) 1 6 i,j 6 n and let b n denote the conjectured ev aluation of det( M ), which for a ll n in question has to b e nonzero. W e pull out o f the hat tw o discre te functions c ′ n,j and c ′′ n,j and verify the following ident ities: c ′ n,n − 1 = c ′′ n,n = 1 , c ′ n,n = c ′′ n,n − 1 = 0 , (1b) n X j =1 a i,j c ′ n,j = n X j =1 a i,j c ′′ n,j = 0 (1 6 i 6 n − 2) , (2b)  n X j =1 a n − 1 ,j c ′ n,j   n X j =1 a n,j c ′′ n,j  −  n X j =1 a n − 1 ,j c ′′ n,j   n X j =1 a n,j c ′ n,j  = b n b n − 2 . (3b) Then the determinant ev aluation follows as a consequence, us ing a simila r induction ar gument as in Sectio n 2. Let’s tr y to give the motiv ation for this approach which als o e xplains why it works. The idea is ba sed on the formula for the deter minant o f a blo ck matrix: det( M ) = det  M 1 M 2 M 3 M 4  = det( M 1 ) det( M 4 − M 3 M − 1 1 M 2 ) . W e wan t to divide the matrix M in to blo cks such tha t M 4 is a 2 × 2 matrix. Let C = ( C ′ , C ′′ ) denote the ( n − 2 ) × 2 matrix whose first column is C ′ = ( c ′ n,j ) 1 6 j 6 n − 2 and whose second column is C ′′ = ( c ′′ n,j ) 1 6 j 6 n − 2 . With this notation, the tw o eq uations (2b) can b e wr itten as ( M 1 , M 2 )   C ′ C ′′ 1 0 0 1   = ( M 1 , M 2 )  C I 2  = M 1 C + M 2 = 0 , where the conditions of Eq ua tion (1b) hav e b een employ ed to constitute the ident ity matr ix I 2 . No w by the induction hypothesis we ma y ass ume that det 1 6 i,j 6 n − 2 ( a i,j ) = det( M 1 ) eq uals b n − 2 , whic h by our g eneral assumption is nonzer o. Th us the a b ov e s y stem determines C uniquely and we c a n wr ite C = − M − 1 1 M 2 . 8 Christoph K outschan and Thotsa p orn “Aek” Thana tipanonda Finally , we obtain the missing par t fro m the blo ck matrix for mula that gives us the quotient det( M ) / det( M 1 ), as the determinant of the 2 × 2 matrix ( M 3 , M 4 )  C I 2  = M 3 C + M 4 = M 4 − M 3 M − 1 1 M 2 . This is exactly wha t is ex pr essed in Equation (3 b), with all matr ix m ultipli- cations explicitly wr itten out. Even though it tur ne d out during our r esearch that there is no need to apply the double-s tep metho d in the pr e s ent co nt ext, we decide d to keep it, as we ar e sure that it will b e useful for future use. Let us also r emark that it is straight-forward to generalize this idea in order to pr o duce a triple-step metho d, etc. But since the ident ities to b e verified b ecome mo re and more complicated, we don’t b elieve that these lar ge-step metho ds are relev ant in practice. 5. Small Change, B ig Impact The next determinants we wan t to study a pp ear as Conje ctur e 35 and Con- je ctur e 36 in [13 ]. Kr attenthaler (pr iv ate communication) describ es the story of how they were ra ised: I wr ote this article dur ing a stay at the Mittag-Leffler Institut. Guo ce Xin was als o ther e , as well as Alain Lascoux. They follow ed the progres s on the article with interest. So, it was Guo ce Xin, who had b een lo oking at similar determinants a t the time (in the co urse of his work with Ira Gessel on his big article on determinants a nd path counting [5]), who told me w ha t b ecame Conjecture 3 5 . I made some exp er iments and discovered Co njecture 36. Alain Las- coux saw all this, and he came up with Conjecture 3 7. Xin’s o bserv ation was that a certain matrix, very similar to the one of Section 3, has a determinant that facto r s completely; the only change is the sign of the Kro neck er delta δ i,j . But still, the ev aluation is given as a cas e distinction b etw een even a nd o dd dimensions of the matrix. Theorem 2. L et µ b e an indeterminate and n b e a nonne gative inte ger. Then the determinant det 1 6 i,j 6 n  − δ i,j +  µ + i + j − 2 j  (7) e quals ( − 1) n/ 2 2 n ( n +2) / 4  µ 2  n/ 2  n 2  ! ( n − 2) / 2 Y i =0 i ! 2 (2 i )! 2 ! × ⌊ n/ 4 ⌋ Y i =1  1 2 ( µ + 6 i − 1)  2 ( n − 4 i +2) / 2  1 2 ( − µ − 3 n + 6 i )  2 ( n − 4 i ) / 2 ! Adv anced Computer Alg e bra for Determinants 9 if n is even, and it e quals ( − 1) ( n − 1) / 2 2 ( n +3)( n +1) / 4  1 2 ( µ − 1 )  ( n +1) / 2 ( n − 1) / 2 Y i =0 i !( i + 1)! (2 i )!(2 i + 2)! ! × ⌊ ( n +1) / 4 ⌋ Y i =1  1 2 ( µ + 6 i − 1)  2 ( n − 4 i +1) / 2  1 2 ( − µ − 3 n + 6 i − 3 )  2 ( n − 4 i +3) / 2 ! if n is o dd. Pr o o f. W e co uld try to solve this pro ble m directly , either by Zeilb erg er’s original ansatz or in the wa y we did in Section 3. How ever, this do es not work in practice a s the computations b eco me to o larg e (in the seco nd c a se w e were even able to gues s the r ecurrences for c n,j , but their size destroyed any hop e to prov e (2a) and (3a )). Next, we could g ive the double- step metho d a try , which we describ ed in Sectio n 4. W e succee ded to make the pro of for some concrete integer µ , but a gain, it seems intractable for symbolic µ . Instead we make a small detour and br eak this determina nt into pieces in order to ma ke the calculations smaller. T o obtain the desired result, we put these pieces tog e ther by the Desnanot-J acobi adjoint matrix theorem, also known as Do dgso n’s rule; this formula gav e r ise to a celebrated alg orithmic determinant ev aluation metho d as well [16, 1], but that approach is different from our usa ge of Do dgs on’s rule. Let us introduce the following no tation b n ( I , J ) := b n ( I , J, µ ) := det I 6 i 6 n − 1+ I J 6 j 6 n − 1+ J  − δ i,j +  µ + i + j − 2 j  so that o ur determinant (7) is denoted by b n (1 , 1). In this notation the Desnanot-Jac o bi identit y is stated as follows: b n (0 , 0) b n − 2 (1 , 1) = b n − 1 (0 , 0) b n − 1 (1 , 1) − b n − 1 (0 , 1) b n − 1 (1 , 0) . (8) After substituting n → 2 n + 2 and n → 2 n + 1 in (8) and using the fa c t that b 2 n (0 , 0) = − b 2 n − 1 (1 , 1) (to b e shown la ter) we hav e b 2 n +1 (1 , 1) = b 2 n +1 (0 , 1) b 2 n +1 (1 , 0) b 2 n (1 , 1) + b 2 n +1 (0 , 0) b 2 n − 1 (1 , 1) = − b 2 n (0 , 1) b 2 n (1 , 0) b 2 n (1 , 1) + b 2 n +1 (0 , 0) from which the desired q uo tient can b e obtained: b 2 n +1 (1 , 1) b 2 n − 1 (1 , 1) = − b 2 n +1 (0 , 1) b 2 n (0 , 1) · b 2 n +1 (1 , 0) b 2 n (1 , 0) . (9) Similarly , w e substitute n → 2 n + 1 and n → 2 n into (8) and us e the fact that b 2 n − 1 (0 , 0) = 0 (again, to b e shown later), to derive the quotient o f even determinants b 2 n (1 , 1) b 2 n − 2 (1 , 1) = − b 2 n (0 , 1) b 2 n − 1 (0 , 1) · b 2 n (1 , 0) b 2 n − 1 (1 , 0) . (10) W e now use the first v ariation of Zeilb er ger’s ansatz (see Sectio n 3) to de r ive recurrence s for the q uo tients b 2 n +1 (0 , 1) /b 2 n (0 , 1), etc. which app ear o n the 10 Christoph K outschan and Thotsa p orn “Aek” Thana tipanonda right-hand sides o f (9) a nd (10 ). Since the ar guments clo sely follow the lines of the pro o f of Theorem 1, we do n’t detail further this part and refer to the electronic materia l [11]. Although for o ur purp oses it would suffice to work with these recurrences , we succeed in s olving them in closed form: Q 1 ( n ) := b 2 n +1 (0 , 1) b 2 n (0 , 1) = 2  µ 2 + 2 n  n +1  µ + 2 n − 1  n +1  n + 2  n +1  µ 2 + n  n +1 , (11) Q 2 ( n ) := b 2 n (0 , 1) b 2 n − 1 (0 , 1) = ( µ + 2 n − 2)  µ 2 + 2 n − 1  n − 1  µ + 2 n + 1  n − 1 n  n  n − 1  µ 2 + n + 1  n − 1 , (12 ) Q 3 ( n ) := b 2 n +1 (1 , 0) b 2 n (1 , 0) = 2  µ 2 + 2 n  n +1  µ + 2 n + 1  n − 1  n + 1  n  µ 2 + n + 1  n , (13) Q 4 ( n ) := b 2 n (1 , 0) b 2 n − 1 (1 , 0) = 2  µ 2 + 2 n − 1  n − 1  µ + 2 n + 1  n − 1  n  n − 1  µ 2 + n + 1  n − 1 . (14) These quotients immediately give closed-form ev aluations o f the co r resp ond- ing determinants (see also Theor ems 3 and 4). It remains to justify the as- sumptions b 2 n (0 , 0) = − b 2 n − 1 (1 , 1) and b 2 n − 1 (0 , 0) = 0 that were use d to derive (9) a nd (10). In or der to ev aluate the quotient b 2 n (0 , 0) /b 2 n − 1 (1 , 1) w e need to mo dify the metho d pres ented in Sec tion 3: W e apply Lapla ce expa nsion with resp ec t to the first row of the matrix , instead of the n -th row, and we normalize the auxiliary function c n,j in such a way that c n, 0 = 1. If we come up with a recursive description of some function c n,j and are able to verify the following ident ities, then we a re done: c n, 0 = 1 ( n > 1) , (1c) 2 n − 1 X j =0 c n,j a i,j = 0 (0 < i 6 2 n − 1 ) , (2c) 2 n − 1 X j =0 c n,j a 0 ,j = b 2 n (0 , 0) b 2 n − 1 (1 , 1) = − 1 ( n > 1) . (3c) As b efore, the details can be found in [11]. Last but not least we have to a r gue that b 2 n − 1 (0 , 0) v anishes. Kra tten- thaler k indly p ointed us to [4, Theorem 1 1] (see also [12, Theor em 35 ]) which contains this s tatement. Anyw ay we have pr o duced an a lternative, comput- erized pro of: Actually it is very simple, sinc e we just have to c o me up with a guess for the co efficients o f a linear co mbination of the columns (or rows) that gives 0, and then pr ov e tha t our gues s do es the job. Hence we find a recursive de s cription of a function c n,j for n > 1 and 0 6 j 6 2 n − 2, such that c n, 0    a 0 , 0 . . . a 2 n − 2 , 0    + · · · + c n, 2 n − 2    a 0 , 2 n − 2 . . . a 2 n − 2 , 2 n − 2    =    0 . . . 0    Adv anced Computer Alg e bra for Determinants 11 and such that there is an index j for whic h c n,j 6 = 0. F or g uessing, we compute c n,j for all n up to some b ound a nd normalize them. Luckily the nullspace of the ab ove sys tem has alwa ys dimension 1, other wis e it would b e tr ickier to find a suita ble linear combination (howev er, in our pro blem this is no surprise since we ar e finally aiming at proving that the minor M 0 , 0 of this matrix ev aluates to some nonzero express ion). So we are s uccessful in guessing the recur rences of c n,j and use them to pr ov e c n, 2 n − 2 = 1 ( n > 1 ) , 2 n − 2 X j =0 c n,j a i,j = 0 (0 6 i 6 2 n − 2) . This concludes the pro of of Theorem 2.  Since the ev aluations of the determinants b n (0 , 1) and b n (1 , 0) are in- teresting results in their own right, but a re somehow hidden in the pro of of Theorem 2, we are go ing to state them explicitly her e. Theorem 3. L et µ b e an indeterminate and n b e a nonne gative inte ger. L et Q 1 ( n ) and Q 2 ( n ) b e define d as in (11) and (12) , r esp e ctively. Then the de- terminant b n (0 , 1) = det 1 6 i,j 6 n  − δ i − 1 ,j +  µ + i + j − 3 j  e quals n/ 2 − 1 Y k =0 Q 1 ( k ) ! n/ 2 Y k =1 Q 2 ( k ) ! if n is even, and it e quals ( µ − 1 ) ( n − 1) / 2 Y k =1  Q 1 ( k ) Q 2 ( k )  if n is o dd. Pr o o f. It is a nalogous to the pr o of of Theo rem 1 a nd ca n b e fo und in [1 1].  Theorem 4. L et µ b e an indeterminate and n b e a nonne gative inte ger. L et Q 3 ( n ) and Q 4 ( n ) b e define d as in (13) and (14) , r esp e ctively. Then the de- terminant b n (1 , 0) = det 1 6 i,j 6 n  − δ i,j − 1 +  µ + i + j − 3 j − 1  e quals n/ 2 − 1 Y k =0 Q 3 ( k ) ! n/ 2 Y k =1 Q 4 ( k ) ! 12 Christoph K outschan and Thotsa p orn “Aek” Thana tipanonda if n is even, and it e quals ( n − 1) / 2 Y k =1  Q 3 ( k ) Q 4 ( k )  if n is o dd. Pr o o f. It is a nalogous to the pr o of of Theo rem 1 a nd ca n b e fo und in [1 1].  By r eplacing j by j + 1 at the b ottom of the binomial co efficient in the ent ries of (7), we arrive a t our last determinant; it has b een discovered by Krattenthaler and a pp ears a s Conje ctur e 36 in his pap er [13] (note that the formula ther e is erroneo us, as one of the pr o duct quantors is missing so that the co rresp o nding factor slided into the previous pro duct). Also this problem has its own combinatorial interpretation in terms o f counting certa in rhombus tilings. Theorem 5. L et µ b e an indeterminate. F or any o dd nonne gative inte ger n ther e holds det 1 6 i,j 6 n  − δ i,j +  µ + i + j − 2 j + 1  = ( − 1) ( n − 1) / 2 2 ( n − 1)( n +5) / 4 ( µ + 1 )  1 2 ( µ − 2)  ( n +1) / 2  1 2 ( n + 1)  ! ( n − 1) / 2 Y i =0 i ! 2 (2 i )! 2 ! × ⌊ ( n +3) / 4 ⌋ Y i =1  1 2 ( µ + 6 i − 3)  2 ( n − 4 i +3) / 2 ! × ⌊ ( n +1) / 4 ⌋ Y i =1  1 2 ( − µ − 3 n + 6 i − 1 )  2 ( n − 4 i +1) / 2 ! . Pr o o f. Because of the similarity of this determinant with (7 ) , we ar e able to relate these tw o pro blems via shifting the star ting p oints: det 1 6 i,j 6 2 n − 1  − δ i,j +  µ + i + j − 2 j + 1  = det 2 6 i,j 6 2 n  − δ i,j +  ( µ − 2 ) + i + j − 2 j  . By us ing the notation from ab ove, the deter minant of Theorem 5 is denoted by b 2 n − 1 (2 , 2 , µ − 2 ). Analog ously to (1c)–(3c), we a pply a v ariation of Zeil- ber ger’s appro ach to derive a recurr ence fo r q n ( µ ) = b 2 n (1 , 1 , µ ) b 2 n − 1 (2 , 2 , µ ) . The r esult is q n +1 ( µ ) − q n ( µ ) = 0 which r e veals tha t the quotient q n ( µ ) is constant. T o gether with the initial v alue q 1 ( µ ) = − 4 / ( µ + 3) and the fact that b 2 n (1 , 1 , µ ) is alrea dy known from Theor em 2, we get the desired result. Once aga in, we refer to [11] for the details of the computatio ns .  Adv anced Computer Alg e bra for Determinants 13 As mentioned a b ove, Lascoux found that the mo re gener al determinant det 1 6 i,j 6 n  − δ i,j + r − 1 +  µ + i + j − 2 j + r − 1  factors c o mpletely fo r o dd natural num bers n and r , and its complica ted ev aluation, which was figure d out b y Kr attenthaler, a pp ears as Conje ctur e 37 in [13]. W e remark that the formula given there holds only for r 6 n ; other- wise the Kro ne cker delta do es not show up in the matr ix a nd the ev aluation is muc h simpler. W e c a nnot attack this determina nt direc tly with Zeilb erger’s ansatz since the ma trix e ntries do no t ev aluate to p oly nomials in µ for con- crete in teger s i and j , as long as r is kept s ymbolically . Ther efore the g uessing for c n,j will not work. A different strateg y would co nsist in finding some con- nection be t ween the cas es r and r + 2; then induction on r would provide a pro of, using Theo rem 2 as the ba se case r = 1 . Unfor tunately we were not able to achiev e this goal. 6. A Challenge Pr o blem W e want to conclude our a rticle with a challenge problem for the next genera- tion of computer algebra to ols. In Section 3, we hav e only prov en a statement ab out the quotient of tw o c o nsecutive determinants (Theorem 1). But so far nob o dy has come up with a closed form for the determinant D 1 ( n ). W e now present a conjectured closed form, which, howev er, we are unable to prove with the metho ds des crib ed in the present pa p er. W e have alrea dy remarked in Sectio n 3 that the quotient D 1 ( n ) /D 1 ( n − 1) most proba bly is not holo- nomic; in that case Zeilb er g er’s holonomic ansatz and our v ariations of it are not applicable. Conjecture 6. L et µ b e an indeterminate and let the se quenc es C ( n ) , F ( n ) , and G ( n ) b e define d as fol lows: C ( n ) = ( − 1) n + 3 2 n Y i =1  i 2  ! i ! , E ( n ) = ( µ + 1 ) n ⌊ 3 2 ⌊ 1 2 ( n − 1) ⌋ − 2 ⌋ Y i =1  µ + 2 i + 6  2 ⌊ 1 3 ( i +2) ⌋ ! × ⌊ 3 2 ⌊ n 2 ⌋ − 2 ⌋ Y i =1  µ + 2 i + 2  3 2  n 2 + 1  − 1  2 ⌊ 1 2 ⌊ n 2 ⌋ − 1 3 ( i − 1) ⌋ − 1 ! , F m ( n ) = ⌊ 1 4 ( n − 1) ⌋ Y i =1 ( µ + 2 i + n + m ) 1 − 2 i − m ! × ⌊ n 4 − 1 ⌋ Y i =1 ( µ − 2 i + 2 n − 2 m + 1 ) 1 − 2 i − m ! , 14 Christoph K outschan and Thotsa p orn “Aek” Thana tipanonda F ( n ) =        E ( n ) F 0 ( n ) , if n is even , E ( n ) F 1 ( n ) 1 2 ( n − 5) Y i =1 ( µ + 2 i + 2 n − 1) , if n is o dd , T ( k ) = 5529 6 k 6 + 41 472( µ − 1) k 5 + 38 4(30 µ 2 − 66 µ + 53 ) k 4 + 96 ( µ − 1)(15 µ 2 − 42 µ + 61 ) k 3 + 4(19 µ 4 − 12 2 µ 3 + 41 9 µ 2 − 54 4 µ + 7 2) k 2 + ( µ − 1 )( µ 4 − 14 µ 3 + 10 1 µ 2 − 16 0 µ − 8 4) k + 2( µ − 3)( µ − 2)( µ − 1)( µ + 1) , S 1 ( n ) = n − 1 X k =1  2 6 k ( µ + 8 k − 1)  1 2  2 2 k − 1  1 2 ( µ + 5 )  2 k − 3  1 2 ( µ + 4 k + 2)  k − 2 ×  1 2 ( µ + 4 k + 2 )  2 n − 2 k − 2 T ( k ) . (2 k )!  1 2 ( µ + 6 k − 3)  3 k +4  , S 2 ( n ) = n − 1 X k =1  2 6 k ( µ + 8 k + 3)  1 2  2 2 k  1 2 ( µ + 5 )  2 k − 2  1 2 ( µ + 4 k + 4)  k − 2 ×  1 2 ( µ + 4 k + 4 )  2 n − 2 k − 2 T  k + 1 2  . (2 k + 1)!  1 2 ( µ + 6 k + 1)  3 k +5  , P 1 ( n ) = 2 3 n − 1  1 2 ( µ + 6 n − 3)  3 n − 2  1 2 ( µ + 5 )  2 n − 3  1 2 ( µ + 2 )  2 n − 2 ( µ + 3) 2 + µ ( µ − 1) S 1 ( n ) 2 13 ! , P 2 ( n ) = 2 3 n − 1  1 2 ( µ + 6 n + 1)  3 n − 1  1 2 ( µ + 5 )  2 n − 2 ( µ + 14)  1 2 ( µ + 4)  2 n − 2 ( µ + 7)( µ + 9 ) + µ ( µ − 1) S 2 ( n ) 2 9 ! , G ( n ) = ( P 1  1 2 ( n + 1)  , if n is o dd , P 2  n 2  , if n is even . Then for every p ositive int e ger n we have det 1 6 i,j 6 n  δ i,j +  µ + i + j − 2 j  = C ( n ) F ( n ) G  1 2 ( n + 1)  . Let us add a few remarks on our co njectured closed form. The e le ments of the se quence C ( n ) a re rational n umbers of the fo rm 1 /k where k is a n int eger . The seq uenc e s F ( n ) a nd G ( n ) consist o f monic poly nomials in µ with integer co e fficie nts. The F ( n ) fa c tor completely into linear factor s of the form ( µ + k ) where k ∈ N , and thus hav e p ositive co efficients. The G ( n ) hav e po sitive co efficients as well, but turn out to b e mostly irr educible; the o nly counterexample we fo und is G (4) = ( µ + 34 )( µ 3 + 4 7 µ 2 + 9 54 µ + 5928). They corres p o nd to the “ ugly fac to rs” mentioned in Section 3. F or the conv enience of the r eader, we provide the Ma thematica co de for all quantities in tro duced in Conjecture 6 in the supplement ar y material [11]. In o rder to come up with this complicated conjecture, we computed the determinants D 1 ( n ) for 1 6 n 6 2 9 5 which g av e us the firs t 14 8 po lynomials of the sequence G ( n ). These data ena ble d us to g uess recurr ences for the Adv anced Computer Alg e bra for Determinants 15 subsequences P 1 ( n ) a nd P 2 ( n ); the r ecurrence s b y the wa y ar e used in [11] to provide a fast pro cedure for co mputing D 1 ( n ). The Maple pack age LREt ools was able to find “clo sed form” solutions which, after lots of a uto matic and manual simplifications , b ecame the fo r mulae for T ( k ), S i ( n ), and P i ( n ). Ackno wledgments W e would like to thank Christia n Kr attenthaler a nd Peter Paule fo r their many v aluable comments, Ma nuel Kauer s for his supp ort concerning the guessing softw are, Dor on Zeilb erg er for his enco uragement to tackle these conjectures, and the tw o anonymous refer ees for their diligent work. 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