11 A curious class of Hankel determinants
Johann Cigler Fakultät für Mathematik, Universität Wien [email protected]
Abstract
We consider Hankel determinants of the sequence of Catalan numbers modulo 2 (interpreted as integers 0 and 1) and more generally Hankel determinants where the sum over all permutations reduces to a single signed permutation.
0. Introduction
Let 21 1 n nC nn be a Catalan number. It is well known that
1, 0 det 1 ni j i j C for all . n Of course this remains true if we consider all terms modulo It is also well known that 1mod 2 n C if and only if k n for some . k But what happens if we consider the sequence mod 2 n C as a sequence of integers from The attempt to answer this question gave rise to the present paper. After completion of a first version I discovered the paper [1] by Roland Bacher, where similar questions are considered from a different point of view. There is some overlap between these approaches which I will be mention at the appropriate places. Let n n a satisfy 1 n a if n is a power of and 0 n a else. Computer experiments led me to guess that ( ) det ( 1) nni j i j d n a (0.1) and that the determinant det sgn( ) ni j n ni j a a a a (0.2) which in general is a sum over all permutations of n is reduced to a single term sgn 0 n n n n n n a a a (0.3) for a uniquely determined permutation . n For example
4, 0 i j i j d a (0.4) reduces to the term a a a corresponding to The Hankel determinants
11 , 0 ( ) det ni j i j
D n a also reduce to single permutations and thus satisfy ( ) 1. D n In this case the sequence ( ) 1,1,1, 1, 1, 1,1, 1, 1, 1, 1,1, n D n is rather interesting. For example [1], Theorem 10.1 implies that ( ) ( ) nj D n S j where ( ) 1,1, 1,1,1, 1, 1,1,1,1, n S n is the famous paperfolding sequence defined by (2 ) ( 1) , (2 1) ( ) n S n S n S n and (0) 1. S We shall also show that ( ) ( ) ( 1) , n D n where ( ) n is the number of pairs i i in the binary expansion of n with i i if i and Thus ( ) i ii n for k n with i i if i i for some i or if i and and i i else. Thus
01 ,
010 ,
111 ,
101 1,
110 0, .111 1, Consider for example i j i j
D a (0.5) The determinant reduces to
31 3 7 . a a a More generally we study Hankel determinants for sequences n n a such that n n a x if n is a power of and 0 n a else, where n x are arbitrary numbers. For some choices of n x we get curious results. For example for k k x x we get ( ) ( 1) , n a n d n x where ( ) a n is the total number of s in the binary expansions of the numbers n For example
22 33 2 (5) 10 a xx xx xx xx (0.6) The total number of 1’s in the binary expansions of is (5) 5. a If we choose x and ( 1) k k x for k we get the Golay-Rudin-Shapiro sequence ( ) ( ) D n r n which satisfies ( ) ( ) ( 1) n r n where ( ) n denotes the number of pairs in the binary expansion of . n For example (6) (0.7) Here gives (6) 1. This choice of n x also leads to the continued fraction k kk z r r z zr r z zr r z z (0.8)
1. Hankel determinants of Catalan numbers modulo 2
Let {0,1} n a satisfy mod 2 n n a C or with other words let 1 n a if n is a power of and 0 n a else. Then ( ) det ( 1) . nni j i j d n a (1.1) The following proof uses an idea due to Darij Grinberg [5] who called a permutation nimble if for each i in its domain the number ( ) 1 i i is a power of Thus a permutation is nimble if and only if n n a a a Theorem 1.1
For each n there exists a unique nimble permutation n of n such that ( ) det ( ) ( 1) . n n n nni j n n ni j d n a sgn a a a (1.2) Proof
For n we set (0) 1 d by convention. Let k and k k n Let us try to construct a nimble permutation . By definition we must have 1 ( 1) 2 1 n n for some . Since k n we get k and therefore k which implies ( 1) 2 1 ( 1) 2 . k k n n n (Since
2( 1) 2 1 k n we have k ). If we define ( 1 ) 2 1 1 k n j n j for all j for which 2 1, k j n n i.e. for 1 2 , 1 , k n j n n we get a nimble order reversing permutation of the interval 2 , 1 . k n n Let us show that each nimble permutation on [0, 1] n reduces to this permutation on 2 , 1 . k n n If k n j the same argument as above gives that ( 1 ) n j must be ( 1 ) 2 1 1 2 . k k n j n j j n If k k n n j then k k k k n j Choose i such that ( ) 2 1 1 . k i n j Then ( ) 2 1 1 2 1 k i i i n j for some and k This implies that k and thus 1 i n j and ( 1 ) ( 1 ). n j n j Since is order reversing on [2 , 1] k n n its sign is sgn ( 1) ( 1) ( 1) . k k n n n n Thus we have seen that for k k n there exists a uniquely determined nimble permutation on the interval 2 , 1 . k n n Since k k n we can suppose by induction that there is a unique nimble permutation of the interval 0, 2 1 . k n This gives us the desired nimble permutation . n It remains to show that sgn ( 1) . nn This follows by induction because k k k k nn n n n is even. If we write a permutation in the notation (0) (1) ( 1) n the first nimble permutations are For example choose n Since we have k
2, 0 i j i j a The above construction gives the permutation on
1, 2 with ( ) 3 2 1. i i There remains
0, 0 i j i j a with (0) 0 and ( ) 2 1 0. i i Thus with sgn( ) 1 1 .
2. Hankel determinants of the sequence n C modulo 2 Let as above 1 n a if k n for some k and 0 n a else. Note that n a and n a if and only if k n for some k or equivalently k n Thus . n n a a Therefore we get , , , 0, , 0, , 0, . a a a a a a This means that in this case the shifted sequence n n a coincides with the aerated sequence , 0, , 0, , 0, . n n A a a a Here we have 1 n A if k k n and 0 n A else. If ( ) k nnn k f x a x x is the generating function of the sequence n n a then the generating function of the aerated sequence n n A is . f x Since n n A a we even have ( ) 1 . f x xf x All Hankel determinants of the sequence n n C are Therefore we know in advance that no Hankel determinant
1, 0 ( ) det ni j i j
D n A vanishes. The first Hankel determinants of the sequence n n A are ( ) 1,1,1, 1, 1, 1,1, 1, 1, 1, 1,1, . n D n Theorem 2.1
Let k A for each k and n A else and let
1, 0 ( ) det . ni j i j
D n A If k k n then ( ) ( 1) 2 1 . n k D n D n (2.1)
Proof
Let us call a permutation m nimble if ( ) 2 1 i i m for some for each i in its domain. Then n n A A A if and only if is nimble. Since k k n we have k k k k n n n n and therefore k which implies ( 1) 2 1 k n n or ( 1) 2 1 . k n n We can now define a nimble permutation of the interval [2 1 , 1] k n n by ( 1 ) 2 1 k n j n j for 0 2 2 . k j n Then is an order reversing permutation of the interval [2 1 , 1]. k n n Let be any nimble permutation on [0, 1]. n Then on [2 1 , 1]. k n n If k n j we get k n j n j which implies . k If k n j then k k n j Let be a nimble permutation. Then ( ) 2 1 k i n j for some i and 2 1 2 2 k i n j for some . This implies k and i n j Thus we get a uniquely determined nimble permutation of the interval [2 1 , 1]. k n n The sign of this permutation is ( 1) ( 1) . k n n By induction we get (2.1).
For example for n we have k and
11 0 00 0 0 .0 0 00 1 101 0 i j i ji j i j
A a
The corresponding permutation is with and ( ) 2 2 2 i i on {0,1, 2}. Let us now try to find some regularities of the sequence of determinants ( ).
D n
Corollary 2.2
For k the sequence ( ) D n satisfies k n k
D n D n (2.2) for k n with initial values (0) (1) 1. D D
For example for k we get D nD n
Corollary 2.3
Let k For k n we get k D n D n (2.3) For k k n we get k D n D n (2.4) Proof
By (2.2) with k instead of k we get k k n k n k k n n k k D n D n D n D n which gives (2.3). Again by (2.2) we have k k k i k k i k i i k D i D i D i D i which gives (2.4).
For example for k we get for n ( ) 1 1 1 1 1 1 1 18 1 1 1 1 1 1 1 1 D nD n
Corollary 2.4
Let ( ) n be the number of pairs i i in the binary expansion of n such that i i for i and Then ( ) ( ) ( 1) . n D n (2.5) Proof
This is true for n If it is true for k n then by (2.3) it is true for k n with 2 k n because for n v we get k n v and k n n By (2.4) it is also true for k n v because k k D n D n and [11 ] [1 ] . v v Examples
For n we have and thus (9) 1. For n we get (15) 1 because
15 1111 . There is no pair 10 for i but 1 pair for i Theorem
The Hankel determinants
1, 0 ( ) det ni j i j
D n A satisfy (2 ) ( 1) ( ),(2 1) ( 1) ( ),(0) 1 n n D n D nD n D nD (2.6)
Proof
Let us give two different proofs. Let
1, , 0 ni j i j M be a matrix for which , i j M whenever i j is odd. Then (cf. e.g.[4]) det det det . n nni j i j i ji j i j i j M M M (2.7) Choose , . i j i j M A Then , n n ni j i j i ji j i j i jn n ni j i j i ji j i j i j M A aM A A because k i j implies 2 1 k i j and k i j implies 2 2. k i j Thus (2.7) gives 1( ) 2 2 n nD n d D which gives (2.6), because ( ) ( 1) . n d n Another proof uses Corollary 2.4. Let us first give another formulation of (2.6): (2 ) ( ) if 0,1mod 4, (2 ) ( ) if 2, 3 mod 4,(2 1) ( ) if 0, 3 mod 4, (2 1) ( ) if 1, 2 mod 4, (0) 1
D n D n n D n D n nD n D n n D n D n nD (2.8)
Let now [ ] . n v Then n v and n v The assertion for n follows from [ 0 ] [ 0 0] , v v [ 10] 1 [ 100] v v and [ 11] 1 [ 110] . v v The assertion for n follows from [ 00] [ 001] , v v [ 01] 1 [ 011] , v v [ 10] 1 [ 101] , v v and [ 11] [ 111] . v v Remark
As already mentioned a result by R. Bacher [1] implies that ( ) ( ) nj D n S j where ( ) 1,1, 1,1,1, 1, 1,1,1,1, n S n is the paperfolding sequence defined by 0 (2 ) ( 1) ,(2 1) ( ),(0) 1. n S nS n S nS (2.9)
This can easily be verified since ( ) 1
D n implies ( ) ( ) ( 1). S n D n D n
By (2.6) we get (2 ) (2 ) (2 1) ( 1) ( ) ( 1) ,(2 1) (2 1) (2 2) ( ) ( 1) ( ). n n S n D n D n D nS n D n D n D n D n S n
To obtain further information let us compare the above approach to Hankel determinants with the approach via orthogonal polynomials and continued fractions (cf. e.g. [3], [7]). Let me sketch the relevant results: Let n n u be a given sequence. Define a linear functional L on the polynomials by . n n uL x u If
1, 0 det 0 nn i j i j
H u for each , n then there exists a (uniquely determined) sequence of monic polynomials ( ) n n p x with deg n p n such that n m L p p for m n and n L p We call these polynomials n p orthogonal with respect to . L By Favard’s theorem there exist (uniquely determined) numbers n s and n t such that ( ) ( ) ( ) n n n n n p x x s p x t p x for all . n The numbers n t are given by
22 1 . n nn n H Ht H These give rise to the continued fraction .1 1 1 nnn uu z t zs z t zs z (2.10) Let us suppose that u Then the matrix
1, 0 nn i j i j
H u has a unique canonical decomposition ( ) tn n n n H A D t A (2.11) where
1, 0 ( , ) nn i j
A a i j is a lower triangular matrix with diagonal ( , ) 1 a i i and ( ) n D t is the diagonal matrix with entries
1, 0 ( ) . ii i kk d t t The entries ( , ) a i j satisfy ( , ) ( 1, 1) ( 1, ) ( 1, 1) j j a i j a i j s a i j t a i j (2.12) with (0, ) [ 0] a j j and ( , 1) 0. a n
See e.g. [7], (2.30). 1 Let us consider the decomposition (2.11) of the matrix
1, 0 . nn i j i j H a Let ( ) 1,1, 1,1, 1, 1, 1,1, n s n satisfy (2 ) ( 1) ( ) n s n s n and (2 1) ( ) s n s n with (0) 1 s and let ( ) n D s be the diagonal matrix
1, 0 ( ) ( )[ ] . nn i j
D s s i i j Let
1, 0 ( ) ( 1) [ ] . nin i j
D t i j Let x be the residue modulo of the number x and define
11, 0 , 0 nnn i j i j iB b i j s i i j
In [1], Theorem 1.2 it is shown that ( ) ( ) ( ). tn n n n n n
H D s B D t B D s Note that ( , ) ( ). b i i s i Therefore we get the canonical decomposition ( ) tn n n n
H A D t A (2.13) with
1, 0 ( , ) ( ) ( ), nn n n ni j
A a i j D s B D s (2.14) i.e. ia i j s i s j i j For example we have for H It is clear that (2.13) implies (0.1). For the aerated sequence , 0, , 0, , 0, u u u we get 0 n s for all . n In this case we write n T instead of . n t Since ( ) 0
D n for all n and moreover ( ) 1 D n we write in this case ( ) ( 2) ( 1) ( ) ( 2) ( 1) n n nn T D n D n (2.15) for n We also have ( ) ( 2) ( ) ( 1) ( 1) ( 2) ( ) ( 1). n T D n D n D n D n D n D n S n S n
The first terms are
1, 1, 1,1, 1,1, 1,1,1, 1,1, 1, 1,1, 1,1, . n n T
2 [1] Theorem 10.1 implies that if we define a sequence ( ) 1,1,1,1, 1,1,1,1, 1, 1, v n satisfying (2 1) ( ), (4 ) ( 1) (2 ), (4 2) (2 ) n v n v n v n v n v n v n and (0) 1 v then ( ) n n ti j i j n n ni j i j A a C D t C (2.16)
Here ( ) n D t is the diagonal matrix with entries ( )
S i and
1, 0 ( , ) nn i j
C c i j with 2 2( , ) ( ) ( ). ic i j v i v ji j By (2.10) and z in place of z we get the continued fraction (cf. [1]) k k z T zT z (2.17) Remark
The corresponding orthogonal polynomials ( ) n p x are , x x , x x x , x x x x , . x They satisfy ( ) ( ) ( ) n n n n p x xp x T p x and . n n L x A Theorem 2.6
The numbers n T satisfy ,,1, 1. n n nn n
T T TT TT T (2.18)
Proof
By (2.6) we get
12 22 12 22 1 (2 ) (2 2) ( 1) ( ) ( 1)(2 1) (2 1) ( 1) ( 1) ( ) n nn n nn
T D n D n D n D nT D n D n D n D n implies ( 1) ( 1) . n n n
T T D n D n T
The second assertion follows from ( 1) ( ) ( 1)( 1) ( ) ( 1) ( 1) 1. n n n n n nn n
T T D n D n D n D n
Remark
OEIS A104977 states that the numbers n T which occur in the continued fraction (2.17) satisfy ( 2) 1, ( 1) , b nn T if ( ) b n denotes the number of “non-squashing partitions of n into distinct 3 parts”. As has been shown in [9] the numbers ( ) b n of non-squashing partitions of n into distinct parts satisfy (2 ) (2 1) ( ) 1,(2 1) (2 ) 1. b m b m b mb m b m Since (2) 1 b and (3) 2 b we get ( 2) 1 ( 1) b nn T by comparing with (2.18). Let us now obtain some further properties of the sequence . n n T For k k n we have by (2.1)
11 11 ( ) ( 1) 2 1 ,( 1) ( 1) 2 2 ,( 2) ( 1) 2 3 . n kn kn k
D n D nD n D nD n D n
This implies k kk
D n D nD n D nD n D n
Therefore we have k n n T T (2.19) for k k n By n n T T we only need to consider n or
0, 2 mod 4. n For n we get ( 1) , nn T (2.20) because (4 ) (4 2) ( 1) (2 )( 1) (2 1) ( 1) (2 )( 1) ( 1) ( )( 1) ( 1) . n n nn nn n nn n n T D n D n D n D n D n D n
Then we get
14 2 2 ( 1) . nn n
T T (2.21) for
14 2 4 1 2 2 ( 1) . nn n n n
T T T T
To compute n T we look at n T and . n T
18 2 ( 1) . nn T (2.22)
4 because by (2.20)
18 2 8 1 4 8 4 ( 1) . nn n n n n
T T T T T
Now we claim that for k
12 2 2 ( 1) . k k nn T (2.23) By induction, (2.18) and (2.20) we get
12 2 2 2 2 3 2 2 2 2 2 4 2 2 2 4 2 2 1 ( 1) ( 1) . k k k k k k k k k k k k n nn n n n n n
T T T T T T
As special case we get k T for k This gives
Theorem 2.7
The numbers n T satisfy
12 3 for 2 2 3, 2, k k kn n T T n k (2.24) and ( 1) ,( 1) for 2. k k nn nn TT k (2.25) Together with T and T this gives another view on the structure of the sequence . n T The sequence begins with , , , , , ,1, , ,11 1 ,,1 1,11, , , , .
T T T T T T T T ( 1) nn T and n n T T gives a part , , , , , , , , , , , , , , , , . T T T T T T T T with period
18 2 ( 1) nn T gives , , , , , , , , ,1 ., , 1,1 1 with period 16,
116 8 2 ( 1) nn T gives a periodic part with period 32, etc. Let more generally , , k k M T T be a beginning block and
02 3 , , k k M T T this block in reverse order then we get k k k k M M M M R. Bacher [1] gives a simpler formulation of (2.23) which (in our notation) can be summarized as 5 ,( 1) ,( 1) ,. n nnn nnn n
T TTTT T (2.26)
To show that this is equivalent it suffices to show that this implies
12 2 2 ( 1) k k nn T for k This follows by induction from . k k k kk k k k n nn n T T T T
Let us recall (cf. [3]) that there is a simple relation between n t and . n T ,, . n n nn n n t T Ts Ts T T (2.27) This gives 1 n t and s The first terms of the sequence n s are
1, 2, 0, 0, 2, 0, 2, 0, 2, 2, 0, . n n s In terms of the paperfolding sequence ( )
S n we get for n (2 1) (2 1) (2 ) (2 2) (2 1) (2 ) (2 ) (2 1)(2 ) (2 1) (2 1) . n s D n D n D n D n S n S n S n S nS n S n S n By (2.10) this gives another continued fraction for k k z (cf. [1], Theorem1.4): k k z zz zz z (2.28)
3. Hankel determinants of shifted Catalan numbers modulo 2
Let m Consider the Hankel determinants
1, 0 ( , ) det ni j m i j d n m a (3.1) of the sequence . n m n a Let us give some examples: 6 ( , 2) 1, 0, 1, 0,1, 0, 1, 0, ,( , 3) 1,1, 0, 0, 1,1, 0, 0, 1, 1, 0, 0 . nn d nd n Theorem 3.1.
Let K K m Then
1, 0 ( , ) det 1 ni j m i j d n m a if K n or mod 2 K n m and ( , ) 0 d n m else. Remark
It is well known that
11 ,, 0 1 1 jmni j m n mi j j i n i jC H i j
Therefore we get
Corollary 3.2
Let K K m Then
11 1 jmj i n i ji j if and only if K n or mod 2 . K n m It would be nice to find a direct proof of this Corollary. For the proof of Theorem 3.1 we need some more information.
Lemma 3.3.
Let k k m n for some . k Then ( , ) 0. d n m (3.2) Proof.
The matrix
1, 0 ni j m i j a contains the vanishing row , , , k k k n a a a because k m and k k n Let for example m and n Then 7
63 , 0 i j i j a (3.3) Recall that a permutation is m nimble if for all i in its domain ( ) 2 1 k i i m for some . k An m nimble permutation can only exist if the last row of i j m a contains an element of the form 2 1 . k m For fixed , m n there can be at most one of the numbers m n i with i n such that k m n i because the extreme case would be k m n and k m n which is impossible. But it is possible that all elements of the last row are For example in
53 , 0 i j i j a the last row is , , , , 0, 0, 0, 0, 0 a a a a a because none of the numbers i for i is a power of Let us first consider the case m m Let m r
For k n r an m nimble permutation gives ( 1) 1 2 1 k n n r and for k n r j it implies ( 1) 2 1 1 2 . k n r j n j Thus we get an m nimble permutation on the interval n j n which satisfies n i n i for i j As above is uniquely determined and therefore we get Lemma 3.4.
Let m r and k be given. Then for k j r jk k d r j m d r j m (3.4) In example (3.3) we have m and n r Thus is the identity on the element In this case (7, 3) 0 d because (6, 3) 0 d since this matrix has a row of zeroes. Lemma 3.5. Let m r If k a m for some k we have a rk k d a m d a m m Proof a rk k ka r k d a m d r a r m d r a r md a m m
Corollary 3.6.
Let m r
Then ( , ) 0 d n m if and only if
0, mod 2 , K n m where K K m Proof
By Lemma 3.3 and Lemma 3.4 the determinants ( , ) d n m for k k r n r can be reduced those for k k r n r By induction we need only consider the case k K If K n m then the first row of
1, 0 ni j m i j a is , , , m m m n a a a and since K m n all terms vanish. If K m then this is trivially true because there is no such . n If K K m n then the row , , ,
K K K n a a a vanishes because K K K K n For K n m we have by Lemma 3.4 K mK K K d m m d r r m For K n we get m mK K K K d m d r r m d r r m d m m m Consider now the case m r If k n r j for some k with k j r we get ( 1) 2 2 k n r j n j because ( ) 2 (2 ) 1. k i i r
9 Now define on the interval n j n by n i n i for i j This implies that
21 12 jk k d r j m d r j m (3.5) for k j r Therefore the determinants ( , ) d n m for k k r n r can be reduced to those for k k r n r Therefore it suffices to consider the case . k K For m we have K and (0, 2) 1 d and (1, 2) 0. d If n then ( , 2) d n can be reduced by (3.5) to (0, 2) 1 d and if n to (1, 2) 0. d For m r with r we get K mK K K r d m m d r r m For K n we get m mK K K K d m d r r m d r r m d m m This gives
Lemma 3.7.
Let K K m r Then
1, 0 ( , ) det 1 ni j m i j d n m a if K n or mod 2 , K n m and ( , ) 0 d n m else. Theorem 3.7 and Corollary 3.6 imply Theorem 3.1.
Remark
With the condensation method (cf. [7], (2.16)) we get more precisely ( , 2) 1, 0, 1, 0,1, 0, 1, . d n (3.6) This method gives ( , 0) ( 2, 2) ( 1, 2) ( 1, 0) ( 1,1) . d n d n d n d n d n Since ( , 0) ( 1) n d n and ( 1,1) 1 d n we get ( , 2) ( 1) ( 1, 2) ( 1) nn d n d n with initial value (0, 2) 1. d
0 This gives (2 , 2) ( 1) n d n and (2 1, 2) 0. d n In the general case computer experiments lead to
Conjecture 3.8
Let K K m For m r we have KK r d n md n m m (3.7) For m r we have K KK n m K d n m d nd n m m d n m (3.8) where ( ) 0,1 . m
4. A slightly more general case
Let k k x be an arbitrary sequence of numbers or indeterminates and define a sequence n n a by n n a x if 2 1 k n and 0 n a else. Theorem 4.1.
Let n n a x if k n and n a else. Let
1, 0 ( ) det , ni j i j d n a log ( ) ( ) 2 1 n n and ( ) 2 1 ( ). n n n Then ( )2 ( )( ) ( ) ( 1) ( ( )). n nn d n x d n n (4.1) Proof
By convention (0) 1. d For n we have (1) 0, (1) 1 and
12 (1)0 (1) (1) ( 1) 1 (1) . d x x d Therefore (4.1) is true. For n choose k such that k k n Then k n k and ( ) 2 1. k n As in the proof of Theorem 1 we find a permutation of the interval k n n n n n such that ( ) ( ), i i n which implies that ( ) ( ) . i i n a x Since there are ( ) 2 1 ( ) n n n elements in the interval ( ) 1 , 1 n n n we get (4.1). Let us consider an example. The Hankel matrix
4, 0 i j i j a is 1 x x xx xxx xx We have (5) 7 because and (5) 10 1 7 2. We get the permutation with sgn 1 and (4) (3). d x d For n we get (3) 3 and (3) 6 1 3 2. This gives (3) (1). d x d Thus we finally get (5) . d x x x The sign is ( 1) 1. This can also be obtained from x x x as ( 1) ( 1) 1. The first terms of the sequence ( ) n d n are 1, , x , x
20 3 , x x , x , x x x , x x
60 7 , x x , . x By (4.1) see that (1) , d x (2) d x and
22 1 kk k d x for k Lemma 4.2
For k we get
22 1 k k n n k d n x d n (4.2) for k n Proof
By assumption we have k k k n Therefore k k n and k n n By (4.1) we get (4.2).
Lemma 4.3
For k we have (2 ) ( 1) kk k n n d n x d n (4.3) for k n Proof If k n then k k k n and we see that (2 ) 2 2 k k n n and (2 ) ( 1) . k kk nk n d n x d n
2 If k n then (4.3) is trivially true. If k n then k k i n and therefore (2 ) ( 1) k kk ik i d i x d i or equivalently ( ) ( 1) 2 k kk n n k d n x d n which equals (4.3). Let e.g. k (8 ) 1( ) 1(8 ) / ( ) 1 d n x x x x x x x x x x x x xd n x x x x x x x x x x x x xd n d n x x x x x x x x Let us write ( ) d n in the form ( )2 ( )2 10 ( ) ( 1) ii ii n ni d n x for some integers ( ). i n Then we get
Theorem 4.4
11 1 11 1 ( ) 0 for 0 2 ,(2 ) 2 for 0 2 ,(2 ) 2 2 for 0 2 ,( ) 0 for 2 2 2 . kk k kk k k kk k k kk n ni i ii i in i (4.4) and for k n ( ) ( mod 2 ). kk k n n (4.5) For example we get ( ) 0,1, ,( ) 0, 0, 2, 0 ,( ) 0, 0, 0, 2, 4, 2, 0, 0, . nnn nnn
Proof
Formula (4.4) is obvious from the above considerations. For example (2 ) 2 2 k kk i i follows from Lemma 4.2 and Lemma 4.3 because (2 ) (2 ) 2 2 k k kk k i i i and ( ) 0 k i for k i Again from Lemma 4.2 and Lemma 4.3 we get
R Rk k k n n n for 2 . R n By applying this several times we get (4.5).
3 Consider for example n
11 1 2 mod 2 implies (11) 1,
11 3 2 1mod 2 implies (11) max 2 2, 0 0,
11 3 2 1mod 2 implies (11) max 2 2, 0 2,
11 11 2 3 mod 2 implies (11) max 2 6, 0 2,
11 11 2 5 mod 2 implies (11) max 2 10, 0 6. For k we have
11 2 2 11 k k and thus (11) 0. k Therefore we get (11) . d x x x x
We know already that the sign is ( 1) 1, but now we can also derive this from the k because ( 1) ( 1) 1. An immediate Corollary of Lemma 4.2 is
Theorem 4.5
The sequence ( ) d n satisfies the recurrence ( ) ( 1) 2 kk n n k d n x d n (4.6) for k k n with initial values (0) 1, d (1) d x and (2) . d x Corollary 4.6
Let ( ) ( 2) .( 1) n d n d nt d n Then we have
212 20 2 20 2 142 2 1 2 1 , for 1. kk k k nn xt x x xt kx (4.7) This can be proved in an analogous way as Theorem 5.7 4 As special case of Theorem 4.5 let us choose . k k x x Then we get
Corollary 4.7
Let k k b x and n b else. Then det ( 1) , nn a nn i j i j d b x (4.8) where ( ) a n is the total number of 1’s in the binary expansions of the numbers n Proof
A search in OEIS led to the conjecture that ( 1) , n a nn d x where ( ) a n is the total number of 1’s in the binary expansions of the numbers , n ( OEIS A000788). The following proof follows Darij Grinberg [6]. By (4.6) we have ( 1) k k k nnn n d x d for k k n Therefore it suffices to show that k k k n a n a n the total number of 1’s in the binary expansions of the numbers 2 , k n k n , n Let the binary expansion of n be . k n Let us write all binary expansions with k digits 0,1. i The total number of 1’s in the binary expansions of , , 2 1 k n is the total number of 0’s in the binary expansions of k n of length k which is k k n k a n Thus k k k a a n n k a n
Now k k a k since each i occurs k times. Therefore we have ( ) (2 ) 2 2 2 . k k k k a n a n k k n k n This proves (4.8) by induction since the initial values (0) (1) 0 a a and (2) 1 a give d d and . d x For example (3) a is the number of 1’s in i.e. (3) 2. a Thus xxxd x x x Example 4.8
Let (2 1) k k c x and ( ) 0 c n else. Then
21 2 2, 0 det ( ) ( 1) . n nnn i j c i j x d (4.9) For example x xx x x xxx d For the proof observe that ( 1) k k k nnn n x d d for k k n It suffices to verify that
22 2 1 2 2 2 .22 k k k nn n
5. The matrices
11 , 0 . ni j i j a Theorem 5.1
Let
11 , 0 ( ) det , ni j i j
D n a log ( ) ( ) 2 1 n n and ( ) 2 ( ). n n n Then ( )( ) ( ) ( 1) ( ) . n nn
D n x D n n (5.1) Proof
For given n choose k such that k k n Then k n k and ( ) 2 1. k n Let be a 1-nimble permutation. Then as above we see that induces an order reversing permutation on the interval [2 1 , 1]. k n n Here we have ( ) 2 2. k i i Since there are ( ) 2 ( ) n n n elements in the interval [2 1 , 1] k n n we get (5.1) by induction. Corollary 5.2
The sequence ( )
D n satisfies k k n n k D n x D n (5.2) for k n
6 Let us compute the first values with this recursion:
133 1 3 (0) 1 (1)(3) (2)
D D xD x D x x
31 1 3 37 5 3 37 1 7 1 3 7 3 7 (0) 1 (1) (2) (3)(7) (6) (5) (4)
D D x D x x D xD x D x x D x x x D x x
For example for k n we have ( ) n n and ( ) . n n This implies for k kk k D x (5.3) Lemma 5.3
For k n we get kk k k n n D n x x D n (5.4) Proof
By (5.2) we get k k k n n k n n k k
D n x D n x D n
Again by (5.2) we have kk k k n n D n x D n
Thus kk k k n n D n x x D n Lemma 5.4
For k k n we get kk k k n n D n x x D n (5.5) Proof k kk kkk k k k i i k k i i ki i i i k
D i x D i x D ix x D i which is equivalent with (5.5). Let us write ( )2 ( )2 11 ( ) ( 1) . ii ii n ni
D n x (5.6) Then ( ) k n only depends on the residue class modulo k Theorem 5.5
Let k i Then ( ) 0,(2 ) 2 1,(2 ) 2 2 1,(2 2 ) 0. k kk k kk k kk i i ii ii (5.7) For k n we have ( ) ( mod 2 ). kk k n n Proof
Formula (5.7) follows from (5.1). By (5.4) we have Rk k n n for R k which implies ( ) ( mod 2 ). kk k n n Thus for k n we have ( ) k i Let us for example compute (11). D
11 3 2 1mod 2 implies (11) max 2 1 2, 0 0,
11 3 2 1mod 2 implies (11) max 2 1, 0 3,
11 11 2 3 mod 2 implies (11) max 2 1 6, 0 1,
11 11 8 3mod 2 implies (11) max(6 1, 0) 7. Therefore we get (11) ( 1) .
D x x x x x x
Let us now determine the numbers ( ) ( 2) .( 1) n D n D nT D n (5.8)
From
111 1 kk kkk kkk kkk k k k kk k kk k kk k k
D x x DD x x DD x x DD x x D we get k k k
T T and . k k k T T
8 For m and j we get by (5.4) k k m kk m k m k m k j j k D j x x D j which implies k m k k
T T and . k m k k T T The same argument using (5.4) gives k m k k k k k j j j
T T T for m and 0,1. j There remains . k k k j T By (5.5) we get . k k k k k k j j j T T T
This gives
Theorem 5.6
The numbers k k n j T satisfy ( 1) k k k nn j j T T (5.9) for j The first terms of the sequence n n T are , xx , xx , x xx , x xx , xx , xx , x xx , . x xx Theorem 5.7
The numbers , n T n satisfy ,( 1) ( 1) for 2. kk k k n nnn nn T T xT x x xT kx (5.10) Proof
For k k n we have ( ) ( 1) 2 1 ,( 1) ( 1) 2 2 ,( 2) ( 1) 2 3 . kk kk kk n n kn n kn n k D n x D nD n x D nD n x D n
9 This implies k kk kk kk k n k n k k kn k n k k x D n x D n D n D nD n D nD n x D n x D n D n
Therefore we have k n n T T (5.11) for k k n It remains to prove , kk kkk k x xT xx xT x (5.12) for k By (5.4) we have kk kkk k kk D x x xD x x By (5.3) kk k D x and by (5.2) kk k D x x This gives (5.12). Let us prove that
34 1 ( 1) nn xT x and ( 1) . nn xT x By induction we get using (5.11) ( 1) . kk k k k jj j j xT T T x ( 1) . kk k k k jj j j xT T T x Remarks
Let us derive some connections between ( )
D n and ( ). d n
Lemma 5.8 ( ) ( 1) ( 1) .( ) n d n d n xD n (5.13) Proof
For k n we have by (4.2)
22 11 2 22 1 k k k n n kk n n k d n x d nd n x d n and therefore k k k n k k d n d n x d n d n By (5.2) we have k k n k D n x D n This implies k k k kk kk k d n d n d n d nh n h nD n D n for 0 2 . k n Further we have
22 12 222 1 kk k k k kk kk k d d d x dh hD x D Therefore we get k k h n h n for 0 2 k n and k This gives by induction (2 ) 2 1 ( 1) ( 1) . k k k k n n h n h n x x
Example 5.9
Let k k b x and n b else. Then det 1 , n n a n a nn i j i j D b x (5.14) where ( ) a n denotes the total number of 1’s in the binary expansions of the first n positive integers. This follow immediately from (4.8) and (5.13). For the numbers n T we get ( 2) ( ) ( 1) , n s n s nn T x (5.15) where ( ) s n denotes the sum of digits of the binary expansion of . n
1 For ( ) ( 1) ( 2) ( 3) ( 3) ( 2) ( 1) ( )2 2 ( 1) 2 ( 2)( 2) ( ) ( ) ( 2) ( 1) ( 1)( 1)( 1) . n nn a n a n a n a n a n a n a n a nn a n a ns n s n
D n D n x xT xD n xx
The first terms of n n T are x x x x x x x x x xx x x x By (5.10) we have ,( 1) ( 1) ( 1) for 2. k k n nnn kn n kkn T TT x xT x kx This is in accord with k k k k k s n s n k Example 5.10
Let k k c x and ( ) 0 c n else. Then
1, 0 det ( 1) 1 . n n nn i j c i j x D (5.16) Proof
We know that
22 2 ( 1) . n nn x d Since
12 2 n n n we get (5.16). In this case we get ( 1) . n n T x Theorem 5.11
If we choose (1) 1, g and k k g for k and ( ) 0 g n else then
1, 0 det ( 1) ( ) ni j g i j r n (5.17) where ( ) r n denotes the Golay-Rudin-Shapiro sequence , which is defined by (2 ) ( ), (2 1) ( 1) ( ),(0) 1. n r n r n r n r nr (5.18) Proof
Some information about the Golay-Rudin-Shapiro sequence can be found in OEIS [8] A020985. As has been observed in [2] the Golay-Rudin-Shapiro sequence counts the number of pairs 11 in the binary expansion of n modulo 2: ( ) ( 1) if . k k k k r n n (5.19) Thus (0) ( 1) 1. r If k k n then k k n and k k n which implies (2 ) ( ) r n r n and (2 1) ( )( 1) n r n r n because if n is odd. The Golay-Rudin-Shapiro sequence can also be characterized by the recursion k kk k k r n r n nr n r n nk r r (5.20) If k n then k n and thus k r n r n If k k n then k n and thus k r n r n The proof of (5.17) now follows from (5.4) and (5.5). From (2.10) we get
Corollary 5.12
1( 1) .(0) (2)1 (1) (3)1 (2) (4)1 1 k kk z r r zr r zr r z (5.21)
6. Hankel determinants of shifted sequences . n m n a All results are very similar to the case 1. n x Therefore we only need to make slight alterations. Let us state the first terms of
1, 0 ( , ) det ni j m i j d n m a for m and m ( , 3) 1, , 0, 0, , , 0, 0, , , 0, 0, , , d n x x x x x x x x x x x ( , 5) 1, 0, 0, , 0, 0, 0, 0, , 0, 0, , . d n x x x x
3 As in Lemma 3.3 we see that
1, 0 ( , ) det 0 ni j m i j d n m a (6.1) if k k m n for some . k Lemma 6.1
Let m r and k be given. Then for k j r k jk j k d r j m x d r j m (6.2) For example
10 5 515 15 (9, 3) (8 1 2, 3) ( 1) 8 1 2 1, 3 (4, 3). d d x d x d
Lemma 6.2
Let m r
Then for R a m for some R we have R a rR a m R d a m x d a m m (6.3) if R a m Lemma 6.3
Let m r and k be given. Then for k j r
221 2 12 1 k jk j k d r j m x d r j m (6.4) Lemma 6.4
Let m r Then for R a m for some R we have R a mR a m R d a m x d a m m (6.5) Theorem 6.5 ( , ) 0 d n m if and only if