The concavity and convexity of the Boros – Moll sequences

In their study of a quartic integral, Boros and Moll discovered a special class of sequences, which is called the Boros–Moll sequences. In this paper, we consider the concavity and convexity of the Boros–Moll sequences {di(m)}i=0. We show that for any integer m > 6, there exist two positive integers t0(m) and t1(m) such that di(m)+di+2(m) > 2di+1(m) for i ∈ [0, t0(m)] ⋃ [t1(m),m−2] and di(m)+di+2(m) < 2di+1(m) for i ∈ [t0(m)+1, t1(m)−1]. When m is a square, we find t0(m) = m− √ m−4 2 and t1(m) = m+ √ m−2 2 . As a corollary of our results, we show that lim m→+∞ card{i|di(m) + di+2(m) < 2di+1(m), 0 6 i 6 m− 2} √ m = 1.


Introduction and Main Results
The object of this paper is to study the concavity and convexity of the Boros-Moll sequences.Boros and Moll [4,5,6,7,8] explored a special class of Jacobi polynomials in their study of a quartic integral.They have shown that for any a > −1 and any nonnegative integer m, ∞ 0 1 (x 4 + 2ax 2 + 1) m+1 dx = π 2 m+3/2 (a + 1) m+1/2 P m (a), (1) where Using Ramanujan's Master Theorem, Boros and Moll [7,18] derived the following formula which indicates that the coefficients of a i in P m (a) are positive for 0 i m.Chen, Pang and Qu [12] gave a combinatorial proof to show that ( 2) is equal to (3).Let d i (m) be defined by The polynomials P m (a) will be called the Boros-Moll polynomials, and the sequences {d i (m)} m i=0 of the coefficients will be called the Boros-Moll sequences.It follows from ( 3) and ( 4) that The readers can find in [3] many proofs of this formula.Recall that P m (a) can be expressed as a hypergeometric function from which one sees that P m (a) can be viewed as the Jacobi polynomial P Some combinatorial properties of the Boros-Moll sequences have been established.Boros and Moll [5] proved that the sequence {d i (m)} m i=0 is unimodal and the maximum element appears in the middle, namely, They also established the unimodality of the sequence {d i (m)} m i=0 by taking a different approach [6].Amdeberhan, Dixit, Guan, Jiu and Moll [1] presented another proof of (8).Amdeberhan, Manna and Moll [2] analyzed properties of the 2-adic valuation of an the electronic journal of combinatorics 22(1) (2015), #P1.8 integer sequence and gave a combinatorial interpretation of the valuations of the integer sequence which is related to the Boros-Moll sequences.Moll [18] conjectured that the sequence {d i (m)} m i=0 is log-concave.Kauers and Paule [16] proved this conjecture based on the following four recurrence relations found by using the WZ-method [20]: and for 0 i m + 1, In fact, the recurrences ( 11) and ( 12) are also derived independently by Moll [19] by using the WZ-method [20].Chen and Gu [11] showed that the Boros-Moll sequences satisfy the reverse ultra log-concavity.Chen and Xia [13] proved that the Boros-Moll sequences satisfy the ratio monotone property which implies the log-concavity and the spiral property.They [14] also confirmed a conjecture given by Moll in [19].By constructing an intermediate function, Chen and Xia [15] proved the 2-log-concavity of the Boros-Moll sequences.
Chen, Dou and Yang [10] proved two conjectures of Brändén [9] on the real-rootedness of the polynomials Q n (x) and R n (x) which are related to the Boros-Moll polynomials P n (x).The first conjecture implies the 2-log-concavity of the Boros-Moll sequences, and the second conjecture implies the 3-log-concavity of the Boros-Moll sequences.
In this paper, we consider the concavity and convexity of the Boros-Moll sequences.Let {a i } n i=0 be a sequence of real numbers.Recall that the sequence {a i } n i=0 is said to be convex (resp.concave) if for 0 i n − 2. It is easy to see that for positive sequences, the log-convexity implies the convexity and the concavity implies the log-concavity.
The main results of this paper can be stated as follows.
Theorem 1.Let m, i be integers and m 6.We have where 2 ] when m is not a square.In order to prove Theorem 1, we establish the following two Theorems: Theorem 2. Let m, i be integers and m 6.We have Let m, i be integers and m 6.We have 2 is an integer if and only if m is a square.Therefore, from Theorems 2 and 3, we immediately prove Theorem 1.By Theorems 2 and 3, we can obtain the following corollary: Corollary 4. We have To conclude this section, we propose an open problem.Determine the signs of the differences

Proofs of the Main Results
In this section, we present proofs of the main results.We first represent d i (m)+d i+2 (m)− 2d i+1 (m) in terms of d i (m) and d i (m + 1).Lemma 5.For 1 i m − 2, we have where Proof.It follows from ( 10) and ( 12) that for 1 i m − 1, the electronic journal of combinatorics 22(1) (2015), #P1.8 and Lemma 5 follows from ( 18) and ( 19).This completes the proof.Now, we are ready to prove Theorem 2. Proof of Theorem 2. It is a routine to verify that Theorem 2 holds for 6 m 9. So, we can assume that m 10.It is easy to check that for 1 i m − 2, It is easy to verify that where In Section 3, we will prove that for Combining ( 20), ( 21) and (23), we deduce that It follows from ( 20) and ( 24) In order to establish the reverse ultra log-concavity of {d i (m)} m i=0 , Chen and Gu [11] gave an upper bound of the ratio d i (m + 1)/d i (m).They proved that for m 2 and 0 i m, the electronic journal of combinatorics 22(1) (2015), #P1.8 By ( 25) and (26), we see that for It should be noted that A(m, i) < 0 for 1 i m − 1.Thus, the above inequality can be rewritten as By Lemma 5 and (28), we see that . It remains to verify that d 0 (m) + d 2 (m) > 2d 1 (m) for m 10.In (12), let i = 2, we have Therefore, Employing ( 5), we deduce that and By (31), (32) and the ratio monotone property of the Boros-Moll sequences established by Chen and Xia in [13], we have which implies that the left hand side of (30) is positive for m 10.This completes the proof.Now we turn to prove Theorem 3. Proof of Theorem 3. It is easy to check that Theorem 3 is true for 6 m 931 by Maple.In the following, we assume that m 932.It is easy to verify that the electronic journal of combinatorics 22(1) (2015), #P1.8 where We can prove that for m 932 and i The proof of ( 36) is analogous to the proof of ( 23), and hence is omitted.In Section 4, we will prove that for m 55 and [ 2m 5 ] + 1 i m − 1, In view of (34), ( 36) and (37), we find that for i and m 932, which implies This is because A(m, i) < 0 for 1 i m − 1.In view of Lemma 5 and (39), we deduce that Theorem 3 implies that lim Corollary 4 follows from (40) and (41).The proof is complete.
3 Proof of (23) In this section, we present a proof of (23).
It is a routine to verify that for m 10, the electronic journal of combinatorics 22(1) (2015), #P1.8 Also, it is easy to check that for m 6, It follows from ( 42) and ( 43) that for m 10, It should be noted that 4m 2 − 16m + 1 > 0 for m 6.Therefore, for m 10 and It is easy to verify that for m 10, and Inequality (23) follows from (45), ( 46) and (47).This completes the proof.
4 Proof of (37) In this section, we provide a proof of (37).We are ready to prove (37) by induction on m.It is easy to check that (37) holds for m = 55.We assume that (37) is true for n 55, i.e., We aim to prove that (37) holds for n + 1, that is, It is easy to check that where F (n, i) is given by Let f (i) and g(i) be defined by In view of (51), ( 52) and (53), we deduce that for 2n+2 5 + 1 i n − 1 and n 55, It follows from ( 50) and (54) that for 2n+2 5 + 1 i n − 1 and n 55, .
By ( 48) and (55), we deduce that for 2n+2 5 It is a routine to verify that which yields (61).Hence the proof is complete by induction.