Some Hankel determinants with nice evaluations. Johann Cigler Talk at the occasion of Peter Paule s 60 th birthday

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Transcription:

Some Hakel determiats with ice evaluatios Joha Cigler Talk at the occasio of Peter Paule s 6 th birthday

Itroductio For each we cosider the Hakel determiat H ( a ) + = det. i j i, j= We are iterested i the sequece ( ) H with H =. It is well kow that the sequece of Catala umbers C = + ca be characterized by the fact that all Hakel determiats of the sequeces ( C ) ad ( ) C + are. The geeratig fuctio of the Catala umbers 4 C ( ) = C = satisfies C ( ) = + C ( ). Let C = C The we get r ( r) ( ). C = C + ad () C + r =. + r ( r) r

I the first part of this talk I wat to give some overview about the Hakel determiats d r i+ j+ r ( ) = det i+ j i, j= = for r. ( r ) ad D ( ) det ( C ) r i+ j i, j= May of these determiats are easy to guess ad show a iterestig modular patter, but stragely eough I foud almost othig about them i the literature ecept for r = ad r =. Oly after I posted a questio i MathOverflow I leared that at least Egecioglu, Redmod ad Ryavec (arxiv:84.44) had cosidered d ( ). Proofs seem oly to be kow for r 3. 3 ( d ) = ( ) ( ),,,,,, ( d( ) ) = (,,,,, ), ( d( ) ) = (,,,,,,,, ), ( d3( ) ) = (,, 4, 3,3, 8, 5, 5,, 7, 7, 6, ), ( d4( ) ) = (,, 8, 8,,, 6, 6,,, 4, 4, ), ( d ( ) ) = (,, 3, 6, 6, 9,9, 78, 64,37, 5, 5, 695, 44,, 7 ) 5 r

It seems that ( ) ( ) k d (k+ ) = d (k+ ) + = (+ ), k+ k+ k + k + k ( + + + ) = ( + ) k ( ) = ( + ) = ( ), d (k ) k ( ) 4, d k d k k k k k+ k k k ( + ) = k ( + + ) = ( ) 4 ( + ). d k k d k k The other values are ot so ice. For eample k ( ) ( ) ( + )(+ ) 5+ 39 ( + )(+ 3) 5+ 6 d5(5+ ) =, d5(5+ 4) =. 3 3 But k k + dk(k ) + dk(k + ) = ( ), k + k dk+ ( (k+ ) ) + dk+ ( (k+ ) + ) = (+ ).

If ( ) d Some backgroud material = det a for each we ca defie the polyomials i+ j i, j= a a a a a a p a a a ( ) = 3 +. d a a+ a If we defie a liear fuctioal L o the polyomials by L( ) = a the ( ) L pp = for m m ad ( ) There eist s ad t such that p ( ) = ( s ) p ( ) t p ( ). dd + The umbers t are give by t =. d L p (Orthogoality). +

For arbitrary s ad t defie umbers a( j ) by The we get a ( j) = [ j = ], a() = sa () + ta (), a ( j) = a ( j ) + s a ( j) + t a ( j+ ). j j ( a + ) i det () t. If we start with the sequece ( ) = i j i, j= j i= j= a ad guess s ad t ad if we also ca guess a( j ) ad show that a () = a the all our guesses are correct ad the Hakel determiat is give by the above formula. For the aerated sequece (,,,,,,5,,4,, ) of Catala umbers it is easy to guess that s = ad t = ad that a ( k) C + a + () =. Therefore all Hakel determiats are. ( k ) + k = ad all other ( ). a j = Thus a () = C ad

There is a well-kow equivalece with cotiued fractios, so-called J-fractios: a = t s t s. For some sequeces this gives a simpler approach to Hakel determiats. The geeratig fuctio of the Catala umbers satisfies C ( ) = + C ( ). Therefore C ( ) = C( ) ad C ( ) ( ) = =. C This agai implies that the Hakel determiats of the aerated sequece of Catala umbers are ad also that D ( ) =.

Some other eamples of J-fractios C ( ) = C ( ) implies D ( ) =. 4 = C ( ) = C ( ) implies d = ( ). + + C ( ) = = = + 4 4 ad ( ) C ( ) 3 C ( ) = 4give ad thus d ( ) =. C ( ) 4 = 3 C ( )

d i+ j+ ( ) = det i+ j i, j= It is easy to guess that s k = 4, s k + = ad t k =. We also guess that a + ( k) =. k This implies a (k+ ) =. k k 3 It remais to verify the trivial idetity + = + 4 +. k k k k k k 4 Therefore we get ( d ) = ( ) ( ),,,,,,,,.

A proof with J-fractios By iductio we get r + r C ( ) Br ( ) = =. 4 This implies B( ) + B( ) =. 4 For C ( ) 4= C ( ) ad C ( ) C ( ) = ad therefore 4 C ( ) C ( ) 4 4 ( 4 ) + = C ( )( C ( ) 4 ) + ( C ( ) ) ( ) ( ) = C ( ) C ( ) + C ( ) =. This implies B ( ) = = =. ( ) ( ) 4 + B( ) 4+ 4 B( ) 4 + + B

For r 3 the situatio becomes more complicated. Sice o Hakel determiat vaishes the above method should i priciple be applicable. It seems that it is possible for each fied r to guess s ad t. But for r 5 I could ot guess a( j ). Let me sketch the case r = 3: Here we get d3(3 ) = d3(3+ ) = + ad d (3 + ) = 4( + ). 3 + + 3 s3 = 5, s3+ =, s3+ =, 4( + ) 4( + ) 4( + ) (+ )(+ 3) 4( + ) t3 =, t3+ =, t 3+ =. + 4 ( + ) + 3 + 3 a (3 k) =, 3k + k+ + k+ + a (3k+ ) = +, 3k 4( k+ ) 3k 4( k+ ) 3k 3 + + 4( k + ) + a (3k+ ) = +. 3k 3k 3 k + 3 3k 4

I have oly foud the followig curious regularities: Let r. The s r = r+, s + s + + s =, r r r+ r+ r t t t r r+ r+ r =. Furthermore it seems that ( ) + r a. rk = rk

( r ) ( C ) + D ( ) = det. r i j These determiats show a similar patter. But some of them vaish. For eample for r = 3 it is kow (C. Krattethaler ad J.C. ) that ( ( )) (,,,,, ) 3 i, j= k k D3 ( ) = ( ) k = k or D =, which is periodic with period 6. For r > 3 apparetly o results appear i the literature. But we will show that for odd r there are always vaishig determiats. Therefore the method of orthogoal polyomials is ot directly applicable. I have studied the case r = 3 i more detail ad looked for other tricks to compute these determiats. Guo-Niu Ha, arxiv:46.593, has show that each formal power series has a uique epasio as a so-called H-fractio k a = + k+ k t s( ) + k + k t s ( ) ad proved a formula for the o-vaishig Hakel determiats.

The case r=3 as H-fractio The powers of the geeratig fuctio of the Catala umbers satisfy where r r r ( ) r ( ) = + ( ), C L C ( L ( ) ) (,,, 3, 4, 5 5, r ) = + + + + + + r are Lucas polyomials. This gives rise to cotiued fractios. For r = 3 we get the H-fractio C ( ) 3 = 3 3 3 3 3 from which we get agai ( D ) = ( ) ( ),,,,,,. 3 Aalogously C ( ) k k ad C ( ) k k + give H-fractios.

A valuable Lemma Aother helpful trick is the followig Lemma (Szegö 939): Let p ( ) be moic polyomials which are orthogoal with respect to the liear fuctioal L with momet L( ) = a ad let r( ) = a a +. The ( + ) = ( + ) det r ( ) det a p ( ). i j i, j= i j i, j= For the proof let The we get B = + + + + ad p( ) b, b, b, =. b, b, b, + b, ( ri+ j ) = B ( ai+ j) ) i, j= i, j= (.

For a = C + we get s =, t = ad k k k ( ) = ( ) ( ). k = k p Sice C = C C the Lemma implies (3) + + k k D3 ( ) = ( ). k = k The Lemma also gives aother proof of the Theorem (Cvetkovic, Rajkovic ad Ivkovic) ( ) det C + C = F. i+ j i+ j+ i, j= +

Narayaa polyomials Aother trick is to itroduce aother parameter such that o determiat vaishes. The Narayaa polyomials C() t = t k = k k k + for > ad C () t = satisfy C () = C. The first terms are 3,,, 3, 6 6,. + t + t+ t + t+ t + t k For the sequece ( C () t ) + we get s = + t ad t = t. The orthogoal polyomials are By the Lemma we get ( i+ j+ i+ j+ ) k k k k (, ) = ( ) ( ). k = k p t t t k k k det C ( t) + C ( t) = t ( ) ( t ). i, j= + k = k Aother proof by the Lidström-Gessel-Vieot theorem has bee give by C. Krattethaler.

For t = we ca agai get D ( ). 3 More iterestig is the case t =. Here we get ( C ( ) + C ( ) ) = (,,,, 5, 5, 4, 4, ) + +, 4,4,. The correspodig Hakel determiats are Fiboacci umbers ( d ) = (, ),,, 3, 5, 8, 3,. For (,,,,, 5, 5, 4,4, ) we get the Hakel determiats ( d ) (, ) =,,,,,,,,,,,. These results ca also be obtaied directly with the method of orthogoal polyomials.

For odd r some Hakel determiats vaish We ca prove that D + ( k+ ) =. k A search for a liear relatio led to k+ j k+ j+ k k j (k+ ) (, ) = ( ) + C+ j = j= j+ j+ Rk for k if k >. More geerally we get Rk C k+ 4k+ (, ) = ( ). Christia Krattethaler has provided a proof usig hypergeometric idetities. It ca also be proved with Peter Paule s implemetatio of Zeilberger s algorithm. I wat to cogratulate Peter Paule ud his team for the very valuable Mathematica packages which were idispesible for my work sice my iterest tured to eperimetal mathematics.

DD kk+ () For r > 3 I have oly cojectures: ( D ( ) ) = (,, 5,,5,,,,, 5,,5, ) 5,,, ( D7 ) = ( ) ( ),, 4, 7,,7,39,,, 35,( 7),, ( 7), 687,. More geerally k D ((k+ ) ) = D ((k+ ) + ) = ( ), k+ k+ k + ( ) D (k+ ) + k+ =, k k + + k ( ) ( ) ( ) ( ) D (k+ ) + k+ = D (k+ ) + k = ( ) k+ ( + ), k+ k+ k+ k+ (( + ) ) + k+ (( + ) + ) = ( ) ( )( + ). D k D k k k

DD 4 () (4) For ( C ) = (, 4,4, 48,65,57,, ) we get ( D ) = ( ) ( ),,,,3,3, 4, 4,. 4 k + k Here we have s k = 4, s k + =, t k = ad t k. k + = + + k + The correspodig a ( j ) satisfy k 4k+ 4 ( ) = ( ), a k C k + a k+ = C C k + k+ 4k+ 4 k+ 3 4k+ 8 ( ) ( ) ( ).

DD kk (, tt) Defie C ( k ) () t by C () t = C () t : ( k) + k This implies that C () = C. ( k) ( k) Let k i j ( k ) ( ) + D ( t, ) = det C ( t). i, j= If we use the q otatio [ ] = + q+ + q q the we get D t t [ ] t [ ] ( ) 4(, ) = ( ) +, D t t 4( +, ) = ( ) +. t

The first terms of D ( ) are 6 DD 6 (, tt) ( + ), ( 3 ),, 3,,,3 3,3, 3 + +,. Cojecture: 6(3 ) = 6(3 + ) = ( ) ( + ), D D + + D6 ( 3+ ) = 3 ( ) j. 6 6 6 9 j= [ ] D(3 t, ) = ( ) t +, (3 ) 3 [ ] 3 t D (3+, t) = ( ) t +, (3+ ) 3 + [ ] t D (3+, t) = ( ) 3 3 t r ( t) with 3 6 3 6 9 ( r t ) ( t t t t t t ) t 3 ( ), 3, 3 6 3,. = + + + + + +

Some more cojectures ( ) k ( ) ( ) ( ) k k k ( ) ( ), D k = D k + = + k k + k + = ( 3)( + ). D k D k k k k k k k k ( ) = [ + ] D k, t ( ) t, k k + k k ( + ) = [ + ] D k, t ( ) t. t k t k

Catala umbers modulo k It is well kow that C mod iff = for some k : Let f( ) = C ( ) mod. The f ( ) f ( ) = + which implies Let ow a k = ad a = else. The d f = ( ). k ( ) a + = det = ( ). i j i, j= I this case the determiat is reduced to a sigle term π + π () + π ( ) k d = sg a a for a uiquely determied permutatio π. For eample π 5 = 43 ad d 5 = det =.

Similar determiats have previously bee cosidered by R. Bacher (4) from aother poit of view. I have posted some questios about such determiats o MO ad received some proofs from Darij Griberg. k More geerally let b = k ad b = else. For eample B 5 = 3 3. The correspodig determiats are a( ) det B = ( ), where a ( ) is the total umber of s i the biary epasios of the umbers,,,. I the above eample we get a (5) = 5 because the umber of s i,,, is 5.

The aerated sequece (aa,, aa,, aa,, ). Let a k = ad a = else ad let A = a ad A + = be the aerated sequece. It is easy to see that A. = a + For eample ( Ai+ j) ( ai+ j+ ) = =. 3 3 i, j= i, j= I this case too the determiat is reduced to a sigle permutatio. We get D ( A ) + δ = det = ( ), i j i, j= where δ is the umber of pairs εi+ εi = for i or εε = i the biary epasio of. For eample δ 4 = because 4 = or δ 75 = 3 because 75 =. The determiats satisfy D = ( ) D ad + D = ( + ) D.

A approach via orthogoal polyomials These determiats have also bee studied by R.Bacher who foud the iterestig formula D = S( j), j= S = is the so-called paperfoldig sequece where ( ( )) (,,,,,,,,,, ) which satisfies S( ) = ( ), S(+ ) = S( ) ad S () =. The method of orthogoal polyomials gives s = ad T = SS ( ) ( + ). The umbers T are uiquely determied by the recursio T = T T T T = T +,, =, T =.

Let () The g = ad ( k ) Golay-Rudi-Shapiro sequece k g = ( ) for k > ad g ( ) = else. ( ) det gi ( + j + ) = r ( ), i, j= where r ( ) is the Golay-Rudi-Shapiro sequece defied by Equivaletly epasio of. r( ) = r ( ), r(+ ) = ( ) r ( ), r() =. R( ) r ( ) = ( ), where ( ) R deotes the umber of pairs i the biary

Associated cotiued fractios Let me fially state two associated cotiued fractios: k k = S() S() S() S() ad k k = r() r() + r() r(3) + + k ( ).