Bounds for the Length of Recurrence Relations for Convolutions of P-recursive Sequences

Bounds for the Length of Recurrence Relations for Convolutions of P-recursive Sequences

Europ. J. Combinatorics (1997) 18 , 707 – 712 Bounds for the Length of Recurrence Relations for Convolutions of P-recursive Sequences MICHAEL STOLL W...

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Europ. J. Combinatorics (1997) 18 , 707 – 712

Bounds for the Length of Recurrence Relations for Convolutions of P-recursive Sequences MICHAEL STOLL We derive an upper bound for the minimal length of a linear recurrence satisfied by a sequence with generating function of the form f e11 ? ? ? f errg , where f1 , . . . , fr , g are D-finite power series, e1 , . . . , er are positive integers, and the fj satisfy differential equations of a certain type. By way of introducing our ideas, we give a simple proof of Franel’s Conjecture: The sequence (Sr (n ))n PN given by Sr (n ) 5 onk 50 ( nk )r satisfies a linear recurrence relation of length  1–2 (r 1 1). ÷ 1997 Academic Press Limited

1. INTRODUCTION We consider the problem of finding a bound for the minimal length of a linear recurrence relation (with polynomial coefficients) satisfied by the convolution of two P-recursive sequences: Given sequences (an ) and (bn ) with (non-trivial) recurrence relations P0(n )an 1 P1(n )an 21 1 ? ? ? 1 Pr (n )an 2r 5 0 , Q0(n )bn 1 Q 1(n )bn 21 1 ? ? ? 1 Qs (n )bn 2s 5 0 , to find a bound for the minimal t such that the sequence (cn ) , which is given by cn 5 onk50 ak bn 2k , satisfies a non-trivial relation of length t : R0(n )cn 1 R 1(n )cn 21 1 ? ? ? 1 Rt (n )cn 2t 5 0 . (Pj , Qj and Rj are polynomials, and the leading coefficients P0 , Q 0 and R0 are not the zero polynomial.) If one measures the complexity of a recurrence relation not by its length, but by the maximum degree of its coefficients, the answer is easily obtained: the product of the degrees of the given recurrences is a bound for the minimal degree of a recurrence for the convolution. To obtain bounds for the length, however, appears to be more difficult. See Section 2 for the history of a very special case of this problem. In this paper, we obtain a (hopefully) useful bound for the length in the general case of a convolution of arbitrarily many sequences, provided that the given recurrences (with the possible exception of one of them) satisfy one additional hypothesis; namely, that the degree of the leading coefficient should be greater than the degrees of the other coefficients. 2. A SPECIAL CASE: FRANEL’S CONJECTURE Franel’s Conjecture states that, for a given positive integer r , the sequence (Sr (n ))n PN defined by n n r Sr (n ) 5 k 50 k

OS D

satisfies a linear recurrence relation of the form P0(n )Sr (n ) 1 P1(n )Sr (n 2 1) 1 ? ? ? 1 Ps (n )Sr (n 2 s ) 5 0 ,

for all n > 0 ,

707 0195-6698 / 97 / 060707 1 06 $25.00 / 0

ej960123

÷ 1997 Academic Press Limited

708

M. Stoll

where the Pj are polynomials with integer coefficients, P0 is not the zero polynomial, and (this is the crucial point) s 5  1–2 (r 1 1). The cases r 5 1 and r 5 2 are well known: S1(n ) 5 2n and S2(n ) 5 ( 2nn) , whence the recurrences S1(n ) 2 2S1(n 2 1) 5 0

nS2(n ) 2 2(2n 2 1)S2(n 2 1) 5 0 .

and

Just over 100 years ago, in 1894, Franel [2, 3] established the following recurrences for r 5 3 and r 5 4: n 2S3(n ) 2 (7n 2 2 7n 1 2)S3(n 2 1) 2 8(n 2 1)2S3(n 2 2) 5 0 , n 3S4(n ) 2 2(2n 2 1)(3n 2 2 3n 1 1)S4(n 2 1) 2 4(n 2 1)(4n 2 3)(4n 2 5)S4(n 2 2) 5 0 . This led him to conjecture the result cited above. (He also stated a conjecture on the form of the polynomials appearing as coefficients, but this turned out to be wrong for r 5 5 .) Perlstadt [5] confirmed this conjecture by producing recurrences of length 3 for r 5 5 and r 5 6 , and Cusick [1] nearly proved it by giving an ingenious algorithm for computing the recurrences. (In our opinion, his proof is incomplete: he does not rule out the possibility that the recurrence relation he obtains is trivial, i.e. that all of the coefficients vanish.) We shall now give a simple proof of Franel’s conjecture, which also has the advantage that it can be generalized to a large class of P-recursive sequences. We keep r fixed throughout our discussion. Instead of Sr (n ) , we consider the sequence (an ) defined by

O k! (n12 k)! 5 n1! S (n). n

an 5

k 50

r

r

r

r

It is obviously sufficient to prove existence of a recurrence relation of length s 5  1–2 (r 1 1) for (an ). We take B (X ) 5

O n1! X , `

n

r

n 50

then A(X ) 5

O a X 5 B(X ) . `

n

n 50

2

n

Furthermore, denoting by D the differential operator X d / dX , we have the relation D rB (X ) 5 XB (X ). Using this relation, we see that we may express any term X mD nA(X ) as a linear combination of terms X i (D jB (X ))(D kB (X )) , where 0 < j < k , r and i > 0. Let us define Vn 5

O

i .0 0
Q ? X i (D jB (X ))(D kB (X )).

Cony olutions of P -recursiy e sequences

709

Then A(X ) P V0 , DVn ’ Vn11 , XVn ’ Vn1r , and the dimension of Vn is dim Vn 5 4h(i , j , k ) 3 0 < i , 0 < j < k , r , ri 1 j 1 k < n j < 4hi 3 0 < i < n / r j ? 4h( j , k ) 3 0 < j < k , r j <

Snr 1 1DSr 12 1D

5n

r11 1 constant, 2

where ‘constant’ refers to a quantity that does not depend on n. The set of (m 1 1)(s 1 1) elements D iX jA(X ) with 0 < i < m , 0 < j < s (recall that s 5  1–2 (r 1 1)) therefore is contained in the vector space Vsr1m , which has dimension at most 1–2 m (r 1 1) 1 C , where C is a constant not depending on m. Since 1–2 (r 1 1) , s 1 1 5  1–2 (r 1 1) 1 1 , these elements have to be linearly dependent for m large enough (a closer look shows that m 5 O(r 3) will be sufficient). This linear dependence means that there is a non-trivial relation P0(D )A(X ) 1 P1(D )XA(X ) 1 ? ? ? 1 Ps (D )X sA(X ) 5 0 , which, when expressed in terms of the an , is exactly what we want: P0(n )an 1 P1(n )an 21 1 ? ? ? 1 Ps (n )an2s 5 0 .

3. THE GENERAL RESULT We now want to formalize the ideas used in the above example in order to obtain a more general result. First, we have to set up some notation. In the following, we will always use the complex numbers C, but any other field of characteristic zero would do as well. We keep the notation D for the differential operator X d / dX acting on CvX b. W 5 C[X , D ] will be the ring of differential operators generated by D and X (acting as multiplication by X ); it is some kind of Weyl algebra. In W , we have the fundamental relation DX 5 X (D 1 1). For f P CvX b , we let V ( f ) denote the W -module Wf generated by f. Recall that f is called D -finite , if C(X ) NC[X ] V ( f ) is a finitedimensional C(X )-vector space; see, for example, [6]. In other words, this means that there is some 0 ? L P W with Lf 5 0 , i.e. there are polynomials P0 ? 0 , P1 , . . . , Pr such that P0(D )f 1 P1(D )Xf 1 ? ? ? 1 Pr (D )X rf 5 0 , This is equivalent to the following relation for the coefficients ( fn ) of f 5 on fn X n : P0(n )fn 1 P1(n )fn21 1 ? ? ? 1 Pr (n )fn2r 5 0 ,

for all n.

A sequence satisfying a relation of this type is called P -recursiy e. Let l . 0 be a fixed real number. We define W (l ) to be the set of the L P W of the form L 5 Dn 2

O

i ,n i 1l j
aij D iX j

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M. Stoll

for some n > 0 . Furthermore, we define a degree function degl on W by giving D degree 1 and X degree l : degl

O a D X 5 maxhi 1 lj 3 a ? 0j. i

j

ij

ij

i ,j

Having established this notation, we are ready to formulate and prove our main result. ˜ P W , with THEOREM. Let f1 , . . . , fr , g P CvX b and let L1 , . . . , Lr P W (l ) , 0 ? L ˜ L1 f1 5 ? ? ? 5 Lr fr 5 L g 5 0. Let e 1 , . . . , er be positiy e integers and set f 5 f e11 ? ? ? f errg . Let

s5

PS

D

1 r degl Lj 1 ej 2 1 ˜, degl L l j51 ej

and s 5 s  . Then there are polynomials 0 ? P0 , P1 , . . . , Ps such that P0(D )f 1 P1(D )Xf 1 ? ? ? 1 Ps (D )X sf 5 0. PROOF. Using Leibniz’s rule, we see that the C-vector space spanned by the elements X i (D k11 f1)(D k12 f1) ? ? ? (D k1e1 f1) ? ? ? (D kr1 fr )(D kr2 fr ) ? ? ? (D krer fr )(D kg ) ,

(1)

˜. with 0 < kj 1 < kj 2 < ? ? ? < kjej for all j contains V ( f ). Let dj 5 degl Lj and d˜ 5 degl L j i ˜ Write L 5 oi ,j aij X D and let (a , b ) be the pair with maximal i amongst the (i , j ) with aij ? 0 and i 1 l j 5 d˜ . The relations Lj fj 5 0 imply that we can replace D dj fj by Lj fj , where Lj P W has degree degl Lj < dj and contains D at most to the power dj 2 1. ˜ g 5 0 means that we can replace X bD ag by L ˜ g , where again degl L ˜ < d˜ , and Similarly, L ˜ ˜ where the terms of degree d in L contain D at most to the power a 2 1 . By induction, we conclude from this that, for every L with degl L 5 n , Lf is contained in the C-vector space Vn generated by the elements in (1) with the additional restrictions kjej , dj ,

li 1

Ok 1k
(k , a

and

jl

or i , b ).

j ,l

Its dimension is bounded by

Sln 1 1D P Sd 1ee 2 1Da 1 b P Sd 1ee 2 1D(n 1 1) a 1 lb d 1e 21 < (n 1 maxh1, l j) P S D l e r

dim Vn <

r

j

j 51

j

j

j51

j

j

j

r

j

j 51

j

j

5 (n 1 maxh1, l j)s . We now consider the (m 1 1)(s 1 1) elements D iX jf with 0 < i < m , 0 < j < s. They are all contained in Vm1ls , which has dimension at most ms 1 (l s 1 maxh1, l j)s .

Cony olutions of P -recursiy e sequences

711

Since s 1 1 . s , this is smaller than (m 1 1)(s 1 1) when m is large enough. Then the elements considered must be linearly dependent over C, whence the theorem. h 4. SOME EXAMPLES One consequence of this theorem, which contains Franel’s conjecture as a special case, is the following. COROLLARY. Let r and m be positiy e integers. The sequence (S (rm)(n ))n >0 , defined by

O Sk , . .n. , k D 5 O r

S (rm )(n ) 5

k11???1km 5n

1

m

n !r r r, k11???1km 5n k1! ? ? ? km !

satisfies a linear recurrence relation of length at most s5

r S

1 r 1m 21 m

D  5  m1 Sr 1 mr 2 1D  5  (r 1r!mm2! 1)!  .

PROOF. Take an 5 1 / n !r S (rm )(n ). It suffices to prove the result for (an ) instead of (S (n )). Let A 5 on an X n be the generating function of (an ) and B 5 on X n / n !r. Then D rB 2 XB 5 0 and A 5 B m. (m ) r

We take l 5 r , f1 5 B , e1 5 m and g 5 1 in the theorem and obtain s5

r S

1 r 1m 21 m

D h

as claimed.

For example, choosing r 5 2 and m 5 3 , we have that an 5 S (3) 2 (n ) satisfies a relation of length 2. It can easily be computed by the algorithm that can be extracted from the proof of the theorem. One obtains n 2an 2 (10n 2 2 10n 1 3)an 21 1 9(n 2 1)2an 22 5 0 ,

for all n > 0.

(2)

Our theorem does not always give the minimal length of a recurrence relation. As an example, take f1(X ) 5 o X n / n ! and f2(X ) 5 o S3(n )X n / n !. We have the equations Df1 2 Xf1 5 0 , D f2 2 (7D 2 7D 1 2)Xf2 2 8(D 2 1)X 2f2 5 0 . 3

2

Taking l 5 1 (which is best possible here), the theorem predicts a relation of length 3 for the coefficients of f1 f2 . It turns out, however, that there is already one of length 2. Multiplying the coefficients by n ! , we obtain for bn 5 onk50 ( nk )S3(k ) , the relation n 2bn 2 (10n 2 2 10n 1 3)bn 21 1 9(n 2 1)2bn 22 5 0 ,

for all n > 0.

Now, this is exactly the same relation as in (2)! And since a0 5 b 0 5 1 , we can deduce the amazing identity

O SnkD O Skj D 5 O SnkD O Skj D 5 O SnkD S2kkD. n

k

k 50

j 50

3

n

k 50

2

k

j 50

2

n

k 50

2

712

M. Stoll

As an additional example, consider alternating sums of the type considered in Franel’s conjecture (r is a positive integer):

O (21) Sn 21nkD . n

Ar (n ) 5

r

k

k 52n

McIntosh [4] has applied Cusick’s method to this case, which gives the bound  1–2 r  for the minimal length of a recurrence relation for Ar . (Using asymptotics, he also gives lower bounds for some r.) However, his proof has the same gap as Cusick’s proof for Franel’s conjecture. Now, we can easily derive this result from our theorem. For this, define n (21)k 1 ar (n ) 5 , r (n 2 k )!r k 50 k !

O

then ar (2n 1 1) 5 0 (symmetry) and (21)n (2n )!r ar (2n ) 5 Ar (n ). Take A 5 on ar (n )X n , B 5 on (2X )n / n !r and C 5 on X n / n !r. Then A 5 BC , and we have the equations D rB 1 XB 5 0

D rC 2 XC 5 0.

and

We apply the theorem with l 5 r , f1 5 B , f2 5 C , e 1 5 e 2 5 1 and g 5 1 , and obtain s 5 r. Hence there are polynomials P0 , P1 , . . . , Pr with P0 ? 0 such that P0(n )ar (n ) 1 P1(n )ar (n 2 1) 1 ? ? ? 1 Pr (n )ar (n 2 r ) 5 0 ,

for all n > 0.

We now replace n by 2n , multiply by (2n )! , and use that ar (odd) 5 0. Writing s 5  1–2 r  , the result is r

P0(2n )Ar (n ) 2 (2n )r (2n 2 1)rP2(2n )Ar (n 2 1) 1 ? ? ? 1 (21)s (2n )r (2n 2 1)r ? ? ? (2n 2 2s 1 1)rP2s (2n )Ar (n 2 s ) 5 0 , a recurrence relation of length s 5  1–2 r  for Ar . REFERENCES 1. T. W. Cusick, Recurrences for sums of powers of binomial coefficients, J. Combin. Theory , Ser. A , 52 (1989), 77 – 83. 2. J. Franel, L’Interme´ diaire des Mathe´ maticiens , Vol. 1, Gaulthier-Villars et fils, Paris, pp. 45 – 47. 3. J. Franel, L’Interme´ diaire des Mathe´ maticiens , Vol. 2, Gaulthier-Villars et fils, Paris, 1895, pp. 33 – 35. 4. R. J. McIntosh, Recurrences for alternating sums of powers of binomial coefficients, J. Combin. Theory , Ser. A , 63 (1993), 223 – 233. 5. M. A. Perlstadt, Some recurrences for sums of powers of binomial coefficients, J. Number Theory , 27 (1987), 304 – 309. 6. R. P. Stanley, Differentiably finite power series, Europ. J. Combin . , 1 (1980), 175 – 188. Receiy ed 30 October 1994 and accepted 21 May 1996 MICHAEL STOLL Mathematisches Institut , Uniy ersita¨ t Du¨ sseldorf , Uniy ersita¨ tsstr.1 , D-Y0225 Du ¨ sseldorf , Germany E-mail: stollêmath.uni-duesseldorf.de