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Combinatorial sequences arising from a rational integral

Victor H.Moll

Department of Mathematics

Tulane University

New Orleans,LA 70118

vhm@https://www.wendangku.net/doc/4c17040816.html,

Submitted:;Accepted:January 23,2007;Published:March 14,2007

Abstract

We present analytical properties of a sequence of integers related to the evaluation of a rational integral.We also discuss an algorithm for the evaluation of the 2-adic valuation of these integers that has a combinatorial interpretation.

1Introduction

The sequence of positive integers b l,m =m

k =l 2k 2m ?2k m ?k m +k m k l (1.1)

for m ∈N and 0≤l ≤m appeared in the process of evaluating the de?nite integral N 0,4(a ;m )= ∞0

dx (x 4+2ax 2+1)m +1.(1.2)The author has shown in [7]that the polynomial

P m (a ):=2

?2m m l =0b l,m a l (1.3)satis?es P m (a ):=

1π

2m +3/2(a +1)m +1/2N 0,4(a ;m ).(1.4)The coe?cients b l,m do not have a natural combinatorial interpretation,but they have some combinatorial ?avor.The goal of this work is to present several conjectures that illustrate this.For instance,in Section 3we discuss a criteria developed in order to establish the unimodality of {b l,m :0≤l ≤m }.We have conjectured that b l,m are logconcave,that is,b 2l,m ≥b l ?1,m b l +1,m .We present several of our attempts to establish this conjecture.In the last section we discuss arithmetical properties of b l,m .In particular we describe an algorithm to evaluate their 2-adic valuation.Based on extensive numerical data,we have conjectured that this valuation can be determined in terms of two natural operators acting on sequences:the ?rst one simply repeats the initial element of a sequence,that is,F ({a 1,a 2,a 3,···}):={a 1,a 1,a 2,a 3,···},

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and the second one picks every other term:

T({a1,a2,a3,···}):={a1,a3,a5,a7,···}.(1.6) The algorithm involves the operator

L l:=ω(l)

j=1

F[T n j?c](1.7)

whereω(l)∈N,c is de?ned by

c:={ν2(m):m≥1}={0,1,0,2,0,1,0,3,0,···},(1.8) and the exponents n j are(conjecturally)related to the distinct compositions of a binary sequence of?xed length.Conjecture4.8presents all the details.We conclude that the combinatorics of b l,m is hidden in their arithmetic properties.

Section2presents a hypergeometric evaluation of(1.2).Section3discusses the unimodal-ity of the sequence b l,m and describes our work on the conjectured logconcavity.Section4 presents an alternative expression for b l,m that is used to discuss their divisibility properties and to state our main conjecture.

2A hypergeometric evaluation of the integral

The integral(1.2)is now evaluated by standard methods in terms of the hypergeometric

function

2F1[a,b;c;z]=

∞

k=0

(a)k(b)k

(c)k k!

z k.(2.1)

Here a∈R,k∈N,and(a)k=a(a+1)(a+2)···(a+k?1)is the Pochhammer symbol, with the usual convention(a)0=1.The reader will?nd in[3]detailed information about this function.In particular,the integral representations

2F1[a,b;c;z]=

Γ(c)

Γ(b)Γ(c?b)

t b?1(1?t)c?b?1(1?tz)?a dt(2.2)

and

2F1[a,b;c;z]=

Γ(c)

Γ(b)Γ(c?b)

∞

s b?1(1+s)a?c(1+sz)?a ds(2.3)

appear there.The gamma function in(2.2)and(2.3)is given by

Γ(z)=

∞

t z?1e?t dt(2.4)

for z>0.This function has a meromorphic extension to C with simple poles at the negative integers.The special values

Γ(n)=(n?1)!andΓ(n+1

2)=

√

22n

(2n)!

(2.5)

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for n ∈N ,will be used throughout.

The reader will ?nd in [7,9]alternative proofs of the value of N 0,4(a ;m ).The latter establishes a connection between N 0,4(a ;m ),the Taylor expansion around c =0of the function a +√1+c and some results of Ramanujan.This was quite a tour de force .Theorem 2.1.Let a >?1and m ∈N .De?ne

P m (a )=1π

2m +3/2(a +1)m +1/2N 0,4(a ;m ).(2.6)Then P m (a )is a polynomial in a given by

P m (a )=2?2m m k =02k 2m ?2k m ?k m +k m (a +1)k .(2.7)

Proof.The change of variable t =x 2yields N 0,4(a ;m )=12 ∞0t ?1/2(t +t +)?m ?1(t +t ?)?m ?1dt where t ±=?a ±√a 2?1are the roots of t 2+2at +1=0.The representation (2.2)and the identity t +t ?=1show that N 0,4(a ;m )is given by N 0,4(a ;m )=π24m +3/2 4m +22m +1 a ?√a 2?1×2F 1 m +1,12;2m +2;2 1?a 2+a √a 2?1

.This can now be simpli?ed using Kummer’s formula 2F 1 α,β;2β;4z (1+z )2

=(1+z )2α2F 1 α,α+12?β;β+12;z 2 ,described in [3].Apply it with z =√a ?1/√a +1,α=12and β=m +1and use a ?√a 2?1=1√2 √a +1?√a ?1 to obtain N 0,4(a ;m )=π24m +5/2 4m +22m +1 1√a +1

2F 1 12,?m ;m +32;a ?1a +1 .This can be simpli?ed further using the relation

2F 1[a,b ;c ;z ]=Γ(c )Γ(c ?a ?b )Γ(c ?a )Γ(c ?b )

2F 1[a,b ;a +b ?c +1;1?z ]++(1?z )c ?a ?b Γ(c )Γ(a +b ?c )Γ(a )Γ(b )

2F 1[c ?a,c ?b ;c ?a ?b +1;1?z ],that in the case b =?m ∈N reduces to 2F 1[a,?m ;c ;z ]=Γ(c )Γ(c ?a +m )Γ(c ?a )Γ(c +m )

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in view of1/Γ(?m)=0.The resulting expression for the quartic integral is

N0,4(a;m)=

22m+3/2

√

a+1

2F1

,?m;?2m;

a+1

.(2.8)

The second argument of the2F1is a negative integer so the hypergeometric series termi-nates.This proves that P m,de?ned by(2.6),is a polynomial given by

P m(a)=

k=0

2?m+k(a+1)k.

It is elementary to check that this form is equivalent to(2.7).The proof is complete.

We now de?ne d l(m)to be the coe?cient of a l in P m(a).

Corollary2.2.For m∈N and0≤l≤m,we have d l(m)=2?2m b l,m,that is,

P m(a)=2?2m

l=0

b l,m a l.(2.9)

Proof.Expand the term(a+1)k in(2.7).

3Unimodality and logconcavity

A?nite sequence of numbers{d0,d1,···,d m}is said to be unimodal if there exists an index 0≤j≤m such that d0≤d1≤···≤d j and d j≥d j+1≥···≥d m.The sequence {d0,d1,···,d m}with d j>0is said to be logarithmically concave(or logconcave for short)if d j?1d j+1≤d2j for1≤j≤m?1.It is easy to see that if a sequence is logconcave then it is unimodal[22].We say that a polynomial is unimodal(logconcave)if the sequences of its coe?cients is unimodal(logconcave).

Unimodal sequences arise often in combinatorics,geometry,and algebra,and have been the subject of considerable research.The reader is referred to[11,18]for surveys of the diverse techniques employed to prove that speci?c sequences are unimodal.Aside from establishing the unimodality(or logconcavity)of a speci?c sequence,it is desirable to produce a combinatorial proof.The reader will?nd in[23]an account of Kathy Ohara’s proof of the unimodality of gaussian polynomials.A combinatorial proof of the logconcavity of i(n,k), the number of permutations of n letters with k inversions,appears in[4].

We?rst established the unimodality of the sequence{b l,m}in[8]by a complicated argu-ment.The proof of Theorem3.1given in[6]is completely elementary.The identity(2.7) shows that the unimodality of b l,m follows from it.

Theorem3.1.If P(x)is a polynomial with positive nondecreasing coe?cients,then P(x+1) is unimodal.

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The theorem can be improved to conclude the unimodality of the polynomial P (x +d ),with arbitrary d >0.The case d ∈N appears in [2]and [20]treats arbitrary d ∈R +.We now turn to the question of logconcavity of the sequence {b l,m }.Based on extensive numerical evidence,we propose

Conjecture 3.2.For each m ∈N ,the sequence {b l,m :0≤l ≤m }is logconcave.We now describe our (failed)attemps to setteled this question.

(A).The ?rst attempt is based on a result of F.Brenti [11]that is in the same spirit as Theorem 3.1:

Theorem 3.3.Let Q (x )be a logconcave polynomial.Then so is Q (x +1).

The hypothesis of this theorem do not hold in our case.De?ne

Q (x )=2?2m

m k =02k 2m ?2k m ?k m +k m x k ≡m k =0

a k x k .(3.1)

Then P m (a )=Q (a +1).But the polynomial Q (x )is not unimodal.Indeed,

24m ?2k a 2k ?a k ?1a k +1 = 2m m ?k 2 m +k m 2× 1?k (m ?k )(2m ?2k +1)(m +k +1)(k +1)(m +k )(2m ?2k ?1)(m ?k +1)

could be negative.Symbolic experiments show that roughly √m terms are negative.

(B)The WZ-method developed by H.Wilf and D.Zeilberger can be used to produce a recurrence for the numbers b l,m .Details on this procedure can be found in [17].One ?nds that b l,m satis?es

b l +1(m )=2m +1l +1b l,m ?(m +l )(m +1?l )l (l +1)b l ?1,m .(3.2)Therefore the sequence {b l,m }is logconcave provided

(m +l )(m +1?l )b 2l ?1,m +l (l +1)b 2l,m ?l (2m +1)b l ?1,m b l,m ≥0.(3.3)

We have extensive numerical evidence to support the next conjecture:

Conjecture 3.4.The left-hand side of (3.3)attains its minimum at l =m with value 22m m (m +1) 2m m 2.

The inequality (3.3)can be written in terms of u =b l,m /b l ?1,m as

l (l +1)u 2?l (2m +1)u +(m +l )(m +1?l )≥0.

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Unfortunately,the discriminant of the quadratic form,is

disc=l(4l3?4m2?4m?3l),(3.5) that is not strictly negative.

(C)A useful criterion to establish the logconcavity of a sequence{a j}is provided by the zeros of the polynomial P(x)=a0+a1x+···+a m x m.

Theorem3.5.If the polynomial P has only real roots,then it is logconcave.

The reader will?nd in[5]a proof and several examples.

The analysis of the zeros of P m(a)was discussed in[7]and[8].It turns out that P m(a)is

part of the family of Jacobi polynomials P(α,β)

m(z)de?ned by

P(α,β) m (z)=

k=0

(?1)m?k

m+β

m?k

m+k+α+β

z+1

.(3.6)

The special valuesα=m+1

2,β=?(m+12)produce P m(a).The zeros of Jacobi polynomials

are studied in detail in[19](see page145?).We concluded that P m(a)has at most one real zero.

The zeros of P m(a)have interesting properties.Dimitrov[12]recently established our conjecture that,when scaled appropriately,the limit curve of these zeros is the left half of Bernoulli’s lemniscate

L={z∈C:|z2?1|=1,Re z<0}.(3.7) A generalization.The coe?cients{b l,m}seem to have a property much stronger than the logconcavity stated in Conjecture3.2.Introduce the operator L on the space of sequences, via

L(a l):=a2l?a l?1a l+1.(3.8) The?nite sequence{a1,···,a n}is replaced by{...,0,0,a1,···,a n,0,0,...}before apply-ing L.Thus,{a l}is logconcave if L(a l)is nonnegative.We say that{a l}is r-logconcave if L(k)(a l)≥0for0≤k≤r.The sequence{a l}is∞-logconcave if it is r-logconcave for every r∈N.

Conjecture3.6.For each m∈N,the sequence{b l,m:0≤l≤m}is∞-logconcave.

The binomial coe?cients is the canonical sequence on which these issues are tested.The solution of the next conjecture should provide guiding principles on how to approach Con-jecture3.6.

Conjecture 3.7.For m∈N?xed,the sequence of binomial coe?cients{

}is∞-

logconcave.

A direct calculation proves the existence of rational functions R r(m,l)such that

L(r)

R r(m,l).(3.9)

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Moreover R r satisfy the recurrence

R r +1(m,l )=R 2r (m,l )? l (m ?l )(l +1)(m ?l +1)

2r ×R r (m,l ?1)R r (m,l +1).Therefore we only need to prove that R r (m,l )≥0.This could be di?cult.

Note .Conjecture 3.2has been established by Manuel Kauers and Peter Paule.The preprint A computer proof of Moll’s logconcavity conjecture established the conjecture using the RISC package MultiSum.

4Divisibility properties of b l,m

The original expression for b l,m (1.1),written in the form

b l,m =2l

m k =l 2k ?l 2m ?2k m ?k m +k m k l (4.1)

shows that the power of 2that divides b l,m is at least l .A more detailed study of this power requires the alternative representation of b l,m discussed in this section.

The evaluation of b l,m using (4.1)is e?cient if l is close to m .Indeed,b m,m =2m 2m m

and b m ?1,m =2m ?1(2m +1) 2m m .The formulas described below are e?cient for l small.

A direct computation of the (?nite)Taylor series of the polynomial P m yields its coe?-cients in terms of de?nite integrals related to (1.2).The details of the next theorem appear in [10].

Theorem 4.1.There exist polynomials αl (m )and βl (m ),with positive integer coe?cients,such that b l,m =2m ?l l !m ! αl (m )m k =1(4k ?1)?βl (m )m k =1

(4k +1) .(4.2)The degrees of αl and βl are l and l ?1respectively.For instance

b 0,m

=2m m !m k =1(4k ?1)(4.3)b 1,m =2m ?1m ! (2m +1)m k =1(4k ?1)?m k =1(4k +1) (4.4)

Numerical calculations on the roots of these polynomials,lead us to conjecture the location of these roots.The next theorem was established by J.Little in [15].

Theorem 4.2.For every m ∈N ,all the zeros of the polynomials αl ,βl lie on the vertical

line Re m =?12

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The proof is based on the fact that the auxiliary polynomialsαl(u),βl(u),with u=

(s?1)/2,satisfy the three-term recurrence

p l+1(s)=2sp l(s)?(s2?(2l?1)2)p l?1(s).(4.5) A generalization of the classical Sturm separation theorem(see[16]for proofs)is then used to establish the result.

The valuations.Arithmetic properties of numbers appearing in combinatorics have always been of interest.The reader will?nd in[1]information about the prime decomposition of

Catalan numbers and[21]describes divisibility by2of the Stirling numbers of second kind.

We now describe divisibility properties of the sequence{b l,m}.We recall?rst some basic de?nitions on valuations.Given a prime p and a rational number r,there exist unique

integers a,b,m with a and b not divisible by p such that

r=a

p m(4.6)

The integer m is the p-adic valuation of r and we denote it byνp(m).Observe that we depart from the usual convention m=?νp(m).

A basic result of number theory states that

νp(m!)=

∞

k=1

p k

.(4.7)

Naturally the sum is?nite and we can end it at k= log2m .There is a famous result of Legendre[13,14]for the p-adic valuation of m!.It states that

νp(m!)=m?s p(m)

p?1

,(4.8)

where s p(m)is the sum of the base-p digits of m.In particular

ν2(m!)=m?s2(m).(4.9) The2-adic value of b0,m follows directly from(4.3).It follows that

ν2(b0,m)=m?ν2(m!)

and Legendre’s result(4.9)reduces this to

ν2(b0,m)=s2(m).(4.10) The next coe?cient b1,m given in(4.4)was analyzed in[10].The main result there is: Theorem4.3.The2-adic valuation of b1,m is given by

ν2(b1,m)=s2(m)+ν2(m(m+1)).(4.11) The key element of the proof is to express the products in the de?nition of b1,m in terms of the Stirling numbers of the?rst kind

x(x+1)(x+2)···(x+r?1)=

k=0

x k(4.12)

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and the representation

r?k

k?1

i=0

2k?i

C k,i(4.13)

for some integers C k,i.

We now present some conjectures about the functionsν2(b l,m)for l≥2.Introduce the

notation

A l,m=αl(m)

k=1

(4k?1)?βl(m)

k=1

(4k+1),(4.14)

so that Theorem4.1states that

A l,m=l!m!2?m+l b l,m,for m≥l.(4.15) For example,

A1,m=m!2?m+1b1,m(4.16) and Theorem4.3implies that

ν2(A1,m)=ν2(2m(m+1)).(4.17) The?rst few values ofν2(A1,m)are given by

ν2(A1,m)={2,2,3,3,2,2,4,4,2,2,···}(4.18) and we observe that this set consists of blocks of length2,starting at odd integers,on which the function A1,m has the same value.The explicit formula(4.17)can be used to chek this property.Indeed,for m odd,we have

ν2(A1,m)=ν2(2m(m+1))=1+ν2(m+1),

and

ν2(A1,m+1)=ν2(2(m+1)(m+2))=1+ν2(m+1).

This type of block structure it is conjectured to remains valid for l≥2.

De?nition4.4.Let s∈N,s≥2.We say that a sequence{a j:j∈N}is simple of length s,or just s-simple if,for each t∈{0,1,2,···},we have

a st+1=a st+2=···=a s(t+1).(4.19)

For example,the sequence{ν2(A1,m),m∈N}is2-simple.

The function A2,m is given by

A2,m=2(2m2+2m+1)

k=1

(4k?1)?2(2m+1)

k=1

(4k+1).(4.20)

and its2-adic values are

ν2(A2,m)={5,5,5,5,6,6,6,6,5,5,5,5,···}.

Similarly

ν2(A3,m)={7,7,9,9,8,8,9,9,7,7,10,10,···}.

Thereforeν2(A2,m)is4-simple andν2(A3,m)is2-simple.

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Conjecture 4.5.Let l ∈N be ?xed.Then the set {ν2(A l,m ):m ≥l }is an s -simple sequence with s =21+ν2(l ).

A recurrence .The recurrence (3.2)for the numbers b l,m and (4.15)yield

A l +1,m =2(2m +1)A l,m ?4(m +l )(m +1?l )A l ?1,m .

(4.21)Using the WZ-method,now in the variable m ,automatically produces the recurrence b l,m +2=2(8m 2?4l 2+24m +19)(m +2)(m ?l +2)b l,m +1?4(4m +5)(4m +3)(m +l +1)(m +2)(m +1)(m ?l +2)

b l,m that yields A l,m +2=8m 2?4l 2+24m +19m ?l +2A l,m +1?(4m +5)(4m +3)(m +l +1)m ?l +2

A l,m .In particular,for l =1we obtain

(4.22)

A 1,m +2=8m 2+24m +15m +1A l,m +1?(4m +5)(4m +3)(m +2)m +1A 1,m .Some partial results on Conjecture 4.5can be obtained from this recurrence.To illustrate this,let m =2k ?1and as induction hypothesis we assume

ν2(A 1,2k ?1)=ν2(A 1,2k ):=t.

(4.23)

To establish the conjecture we need to prove that

ν2(A 1,2k +1)=ν2(A 1,2k +2).(4.24)We write A 1,2k ?1=2t x and A 1,2k =2t y ,with x,y odd integers.Then m =2k ?1in (4.22)yields kA 1,2k +1=2

t ?1 (32k 2+16k ?1)y ?(128k 3+64k 2?2k ?1)x .Writing

A 1,2k +1=2w z

with z odd and using m =2k in (4.22)yields

(2k +1)A 1,2k +2=2w (32k 2+48k +15)z ?2t +1(k +1)(8k +3)(8k +5)y.

In particular,if w

A geometrical interpretation .The graphs of the function ν2(A l,m ),where we reduce the repeating blocks to a single value,are shown in the next ?gures.Online Journal of Analytic Combinatorics,Issue 2(2007),#410

51015202530

Figure1:The2-adic valuation of d1(m)

The main experimental result is that these graphs have an initial segment from which the rest is determined by adding a central piece followed by a folding rule.For example,in the case l=1,the?rst few values of the reduced table are

{2,3,2,4,2,4,2,3,2,5,...}.

The ingredients are:

initial segment:{2,3,2},

central piece:the value at the center of the initial segment,namely3.

rules of formation:start with the initial segment and add1to the central piece and re?ect. This produces the sequence

{2,3,2}→{2,3,2,4}→{2,3,2,4,2,3,2}→{2,3,2,4,2,3,2,5}→

→{2,3,2,4,2,3,2,5,2,3,2,4,2,3,2}.

The details are shown in Figure1.

The di?culty with this procedure is that,at the moment,we have no form of determining the initial segment nor the central piece.Figure2shows the beginning of the case l=9. From here one could be tempted to predict that this graph extends as in the case l=1. This is not correct as it can be seen in Figure3.The new pattern described seems to be the correct one as shown in Figure4.

The initial pattern could be quite elaborate.Figure5illustrates the case l=53.

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26.5

25.5

234567

Figure2:The beginning for l=9

An algorithm and the main conjecture.We now present an algorithm that we hope will lead to the derivation of an analytic expression forν2(A l,m).The algorithm requires two operators de?ned on sequences:

F({a1,a2,a3,···}):={a1,a1,a2,a3,···},

and

T({a1,a2,a3,···}):={a1,a3,a5,a7,···}.

Now recall the sequence c de?ned in(1.8):

c:={ν2(m):m≥1}={0,1,0,2,0,1,0,3,0,···}.

We begin the description of the algorithm with the example l=12.The sequence

X1(12):={ν2(A12(m)):m≥1}

begins with

X1(12)={34,34,34,34,34,34,34,34,36,36,36,36,36,36,36,36,

35,35,35,35,35,35,35,35,36,36,36,36,36,36,36,36,

34,34,34,34,34,34,34,34,37,37,37,37,37,37,37,37,···}.

This is8-simple,illustrating Conjecture4.5,in view of21+ν2(12)=8.

The?rst step in the algorithm is to replace X1(12)by Y1(12)=T3(X1(12)).This sequence contains every eight element of X1(12)and it begins with

Y1(12)={34,36,35,36,34,37,36,37,34,36,35,36,34,38,37,38,

34,36,35,36,34,37,36,37,34,36,35,36,34,39,38,39,

34,36,35,36,34,37,36,37,34,36,35,36,34,38,37,38,···}. Online Journal of Analytic Combinatorics,Issue2(2007),#412

5101520

Figure 3:The continuation of l =9

The next step is to de?ne Z 1(12):=Y 1(12)?c ,with c as above.This sequence begins with

Z 1(12)={30,31,31,30,30,33,33,31,31,32,32,31,31,33,33,30,

30,31,31,30,30,34,34,32,32,33,33,32,32,34,34,30,

30,31,31,30,30,33,33,31,31,32,32,31,31,33,33,30,···},

that is almost 2-simple,except that the ?rst element appears only once.This motivates the map F .The last step is to de?ne W 1(12):=F (Z 1(12)),that produces

W 1(12)={30,30,31,31,30,30,33,33,31,31,32,32,31,31,33,33,

30,30,31,31,30,30,34,34,32,32,33,33,32,32,34,34,

30,30,31,31,30,30,33,33,31,31,32,32,31,31,33,33,···}.

This new sequence is 2-simple and the ?rst loop is completed.

Algorithm .

1)Start with the sequence X 1(l ):={ν2(A l (m )):m ≥1}.

2)Find n 1∈N so that the sequence X 1(l )is 2n 1-simple.De?ne Y 1(l ):=T n 1(X 1(l )).We conjecture that n 1=1+ν2(l ).

3)Introduce the constant shift Z 1(l ):=Y 1(l )?c .

4)De?ne W 1(l ):=F (Z 1(l )).

The sequence W 1is 2n 2-simple.Then return to step 1)with W 1instead of X 1.

Conjecture 4.6.After a ?nite number of steps,the algorithm yields a constant sequence of the form {a,a,a,a,a,a,···}.The constant term will be denoted by a ∞(l ).

Online Journal of Analytic Combinatorics,Issue 2(2007),#413

1020304050

Figure4:The pattern for l=9persists

161

160

159

158

157

156

10203040506070

Figure5:The initial pattern for l=53

Online Journal of Analytic Combinatorics,Issue2(2007),#414

De?nition4.7.Letω(l)be the number of steps required in the previous conjecture.The sequence of integers

?(l):=

n1,n2,n3,···,nω(l)

(4.25)

is called the reduction sequence of l.

We now present the values of the limiting constant a∞(l)and the sets?(l)for1≤l≤32. The data presented below was checked with tables of size200and validated with size400.

l a∞(l)?(l) 121

252 371,1 4113 5131,2 6162,1 7181,1,1 8234 9251,3 10282,2 11301,1,2 12343,1 13361,2,1 14392,1,1 15411,1,1,1 16475

l a∞(l)?(l) 17491,4 18522,3 19541,1,3 20583,2 21601,2,2 22632,1,2 23651,1,1,2 24704,1 25721,4,1 26752,2,1 27771,1,2,1 28813,1,1 29831,2,1,1 30862,1,1,1 31881,1,1,1,1 32956

The main conjecture stated below provides a combinatorial interpretation of the sets ?(l).

Conjecture4.8.The sets?(l)associated to the integers l from2j+1to2j+1are the2j distinct compositions of j+1.

To obtain the order in which the compositions appear,write all the binary sequences of length j in lexicographic order and then preprend a1to each of these.For instance,for j=3we obtain

1000,1001,1010,1011,1100,1101,1110,1111.

Read these sequences from right to left.The?rst part of the set associated to l is the number of digits up to and including the?rst1read in the corresponding binary sequence;the second one is the number of additional digits up to and including the second1read,and so on.In the case j=3,this yields

4;1,3;2,2;1,1,2;3,1;1,2,1;2,1,1;1,1,1,1

as desired.

Online Journal of Analytic Combinatorics,Issue2(2007),#415

One last experimental conjecture.Based on observations of the values ofν2(A l,m), Dante Manna has conjectured an analytic expression for this function:

ν2(A l,m)=ν2((m?l+1)2l)+l.(4.26) This is a generalization of Theorem4.3.Expressing the Pochhammer symbol as quotients of factorials,we can write(4.26)as

ν2(A l,m)=3l+s2(m?l)?s2(m+l).(4.27) It follows that the asymptotic value in Conjecture4.6is

a∞(l)=3l?s2(l).(4.28) Acknowledgements.The author acknowledges the partial support of NSF-DMS0409968. The author wishes to thank T.Amdeberhan for providing WZ-expertise and insight,Aaron Jaggard for identifying the data that lead to the main conjecture and Dante Manna for comments that improved the manuscript.

This work was completed when the author was a Katrina refugee at the Courant Institute of New York University.Their hospitality is gratefully acknowledged.

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