Higher order Turán inequalities for combinatorial sequences

Advances in Applied Mathematics 110 (2019) 180–196

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Higher order Turán inequalities for combinatorial sequences Larry X.W. Wang Center for Combinatorics, LPMC, Nankai University, Tianjin, 300071, PR China

a r t i c l e

i n f o

Article history: Received 15 February 2019 Received in revised form 20 June 2019 Accepted 20 June 2019 Available online xxxx MSC: 05A20 Keywords: Higher order Turán inequality Associated Jensen polynomial Combinatorial sequence Motzkin number Fine number Franel number Domb number

a b s t r a c t The Turán inequality and its higher order analog arise in the study of Maclaurin coefficients of an entire function in the Laguerre-Pólya class. It is well known that if a real entire function ψ(x) is in the LP class, the Maclaurin coefficients satisfy both the Turán inequality and the higher order Turán inequality. Chen, Jia and Wang proved that for n ≥ 95, the higher order Turán inequality holds for the partition function p(n) and the 3-rd associated Jensen polynomials p(n) +3p(n +1)x +3p(n +2)x2 +p(n +3)x3 have only real zeros. Recently, Griffin, Ono, Rolen and Zagier showed that Jensen polynomials for a large family of functions, including those associated to ξ(s) and the partition function, are hyperbolic for sufficiently large n. This result gave evidence for Riemann hypothesis. In this paper, we give a unified approach to investigate the higher order Turán inequality for the sequences {an /n!}n≥0 , where an satisfy a three-term recurrence relation. In particular, we prove higher order Turán inequality for the sequences {an /n!}n≥0 , where an are the Motzkin numbers, the Fine number, the Franel numbers of order 3 and the Domb numbers. As a consequence, for these combinatorial sequences, the 3-rd associated Jensen polynomials an an+3 3 3an+1 3an+2 2 + x+ x + x n! (n + 1)! (n + 2)! (n + 3)! have only real zeros. Furthermore, for these combinatorial sequences we conjecture that for any given integer m ≥ 4, there exists an integer N (m) such that for n > N (m), the m-th associated Jensen polynomials

E-mail address: [email protected]. https://doi.org/10.1016/j.aam.2019.06.005 0196-8858/© 2019 Elsevier Inc. All rights reserved.

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181

m    m an+i i x i (n + i)! i=0

have only real zeros. © 2019 Elsevier Inc. All rights reserved.

1. Introduction The Turán inequality and the higher order Turán inequality are related to the Laguerre-Pólya class of real entire functions; see [15] and [28]. A sequence {an } of real numbers is said to satisfy the Turán inequality, if for n ≥ 1, a2n ≥ an−1 an+1 . The sequence is said to satisfy the higher order Turán inequality or cubic Newton inequality [24] if for n ≥ 1, 4(a2n − an−1 an+1 )(a2n+1 − an an+2 ) − (an an+1 − an−1 an+2 )2 ≥ 0.

(1.1)

A real entire function ψ(x) =

∞ 

γk

k=0

xk k!

(1.2)

is said to be in the Laguerre-Pólya class, denoted ψ(x) ∈ LP, if it can be represented in the form ψ(x) = cxm e−αx

2

+βx

∞ 

(1 + x/xk ) e−x/xk ,

k=1

 −2 where c, β, xk are real numbers, α ≥ 0, m is a nonnegative integer, and xk < ∞. These functions are the only ones that are uniform limits of polynomials whose zeros are real. For more background on the theory of the LP class, see [8], [21] and [26]. Jensen [18] proved that a real entire function ψ(x) belongs to LP class if and only if for any positive integer n, the n-th associated Jensen polynomial

gn (x) =

n    n k=0

k

γ k xk

(1.3)

has only real zeros. More properties of the Jensen polynomials can be found in [7,9–11].

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Pólya and Schur [25] proved that for a real entire function ψ(x) ∈ LP and any nonnegative integer m, the m-th derivative ψ (m) of ψ(x) also belongs to the LP class. It means that the n-th Jensen polynomial associated with ψ (m) , gn,m (x) =

n    n k=0

k

γk+m xk ,

(1.4)

has only real zeros for any nonnegative integers n and m. Dimitrov [15] observed that for a real entire function ψ(x) in the LP class, the Maclaurin coefficients satisfy the higher order Turán inequality 2 4(γk2 − γk−1 γk+1 )(γk+1 − γk γk+2 ) − (γk γk+1 − γk−1 γk+2 )2 ≥ 0

(1.5)

for k ≥ 1. This fact follows from a theorem of Mařík [23] stating that if a real polynomial n    n k=0

k

a k xk

(1.6)

of degree n ≥ 3 has only real zeros, then a0 , a1 , . . . , an satisfy the higher order Turán inequality. In [6], Chen, Jia and Wang proved the higher order Turán inequality for the partition function p(n) for n ≥ 95, as conjectured by Chen [4]. As a result, the polynomials p(n) + 3p(n + 1)x + 3p(n + 2)x2 + p(n + 3)x3 have only real zeros for n ≥ 95. Chen, Jia and Wang also conjectured that for any given positive integer m > 3, there exists a positive integer N (m) such that for n > N (m), m

 m i the polynomials i p(n + i)x have only real zeros. Recently, Griffin, Ono, Rolen i=0

and Zagier [17] showed that Jensen polynomials for a large family of functions, including those associated to ξ(s) and the partition function, are hyperbolic for sufficiently large n. This result gave evidence for Riemann hypothesis. Larson and Wagner [20] refined the approach of Griffin, Ono, Rolen and Zagier and gave the exact values of N (3), N (4) and N (5) and an upper bound for N (m) for any positive integer m. More results on the log-behavior of partition functions and the higher order Turán inequality can be found in [12], [13], [19] and [22]. Motivated by these results, it is natural to consider whether other combinatorial sequences also satisfy the higher order Turán inequality. Unfortunately, many celebrated combinatorial sequences are log-convex, not log-concave, such as the Motzkin numbers, the Fine numbers, the Franel numbers of order 3 and the Domb numbers. Thus they definitely do not satisfy the higher order Turán inequality. However, as we are intrigued by the properties of γn!n , it is interesting to consider the log-behavior of the sequence { an!n }n≥0 . Došlić [16] introduced log-balanced sequences as follows: A sequence {an }n≥0 of

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positive real numbers is log-balanced if {an }n≥0 is log-convex and the sequence { an!n }n≥0 is log-concave. In that paper, Došlić showed that the Motzkin numbers, the Fine numbers, the Franel numbers of order 3 and 4, the Apéry numbers, the central Delannoy numbers and the large Schöder numbers are log-balanced. These results inspired us that while the sequence {an }n≥0 is not log-concave, the sequence { an!n }n≥0 might satisfy higher order Turán inequality. In this paper, we will give a unified approach to investigate the higher order Turán inequality for combinatorial sequences { an!n }n≥0 , where an satisfy the following three-term recurrence relation an+1 = un an + vn an−1 ,

(1.7)

where un and vn are rational functions in n and un > 0 for n ≥ 1. We proved that the sequence { an!n }n≥0 satisfies the higher order Turán inequality when an are the Motzkin numbers, the Fine numbers, the Franel numbers of order 3 and the Domb numbers. Consequently, we obtain that the 3-rd associated Jensen polynomial 3    n an+k k x g3,n (x) = k (n + k)! k=0

has only real zeros for the Motzkin numbers, the Fine numbers, the Franel numbers of order 3 and the Domb numbers. In conclusion, we propose the following conjecture. Conjecture 1.1. Let {an }n≥0 be the sequences of the Motzkin numbers, the Fine numbers, the Franel number of order 3 and the Domb numbers. For any given integer m ≥ 4, there exists a positive integer N (m) such that for n > N (m), the m-th associated Jensen polynomial m    m an+i i x i (n + i)! i=0

has only real zeros. 2. A unified approach In this section, for an satisfy the recurrence relations (1.7), we shall give a unified approach to prove the sequence { an!n }n≥0 possess the higher order Turán inequality. To prove the sequence { an!n }n≥0 satisfies the higher order Turán inequality, it suffices to prove  4

  a2n+1 an+1 an−1 an an+2 a2n − − n!2 (n − 1)! (n + 1)! (n + 1)!2 n! (n + 2)! 2  an−1 an+2 an an+1 − > 0. − n! (n + 1)! (n − 1)! (n + 2)!

(2.1)

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Applying the recurrence relations an−1 =

an+1 − un an vn

and an+2 = un+1 an+1 + vn+1 an in the above inequality, we get

 4 n n2 u2n+1 + 4n2 vn + nu2n+1 + 16nvn + 16vn an+1 − (n + 1)3 (n + 2)2 (n!)4 vn2 an  3  2 2nr1 r2 an+1 an+1 + − (n + 1)3 (n + 2)2 (n!)4 vn2 an (n + 1)2 (n + 2)2 (n!)4 vn2 an 2 2

2 2 vn+1 n un vn+1 + 4n vn + 12nvn2 + 8vn2 2r3 an+1 − − > 0, (n + 1)2 (n + 2)2 (n!)4 vn2 an (n + 1)2 (n + 2)2 (n!)4 vn2 where r1 = 2n2 un vn + 3n2 un+1 vn − n2 un+1 vn+1 + n2 un u2n+1 + 8nun vn + 9nun+1 vn −nun+1 vn+1 + 8un vn + 6un+1 vn + nun u2n+1 , 2 r2 = 6n2 un un+1 vn − 4n2 un un+1 vn+1 + n2 u2n u2n+1 − 3n2 vn2 + n2 vn+1

−6n2 vn vn+1 + 12nun un+1 vn − 12nvn2 − 12nvn vn+1 − 12vn2 , 2 + n2 u2n un+1 vn+1 + 3n2 un vn vn+1 r3 = 2n2 un+1 vn2 − n2 un vn+1

+6nun+1 vn2 + 6nun vn vn+1 + 4un+1 vn2 . Let

n n2 u2n+1 + 4n2 vn + nu2n+1 + 16nvn + 16vn 4 f (x) = − x (n + 1)3 (n + 2)2 (n!)4 vn2 2nr1 r2 + x3 − x2 (n + 1)3 (n + 2)2 (n!)4 vn2 (n + 1)2 (n + 2)2 (n!)4 vn2

vn+1 n2 u2n vn+1 + 4n2 vn2 + 12nvn2 + 8vn2 2r3 − x− . (n + 1)2 (n + 2)2 (n!)4 vn2 (n + 1)2 (n + 2)2 (n!)4 vn2 (2.2)   To prove (2.1), we aim to show that f aan+1 > 0. It is easily verified that if we can find n appropriate g(n) and h(n) satisfying that there exists a positive N such that for n > N , (1) g(n) <

an+1 an

< h(n),

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(2) f (g(n)) > 0 and f (h(n)) > 0, (3) f (x) is monotone on the interval [g(n), h(n)], then  we immediately get that f (x) > 0 on the interval [g(n), h(n)]. It implies that f aan+1 > 0, as desired. n Furthermore, we can deduce the following criterion for the higher order Turán inequality for the sequence { an!n }n≥0 . Theorem 2.1. If a sequence {an }n≥0 satisfying the recurrence relation (1.7), and we can find g(n) and h(n) so that there exists a positive integer N such that (1) g(n) < aan+1 < h(n), n (2) f (g(n)) > 0 and f (h(n)) > 0, (3) f (i) (g(n)) and f (i) (h(n)) have the same signs for i = 1, 2, 3, then the sequence { an!n }n≥0 satisfies the higher order Turán inequality. Proof. Since f (x) is a polynomial of degree 4, we have f (3) (x) is monotone. From the condition f (3) (g(n)) and f (3) (h(n)) have the same signs, we deduce that f (2) (x) is monotone on the interval [g(n), h(n)]. Since f (2) (g(n)) and f (2) (h(n)) have the same signs, f  (x) is monotone on the interval [g(n), h(n)]. In view of f  (g(n)) and f  (h(n)) having the same signs, we conclude that f (x) is monotone on the interval [g(n), h(n)]. This completes the proof. 2 In Section 3, we will demonstrate how to find appropriate lower bound g(n) and upper bound h(n) for aan+1 for celebrated combinatorial sequences, such as the Motzkin n numbers, the Fine numbers, the Franel numbers of order 3 and the Domb numbers. 3. Applications 3.1. Motzkin numbers Let us consider the Motzkin numbers Mn , defined by the recurrence relation Mn+1 =

2n + 3 3n Mn + Mn−1 , n+3 n+3

where n ≥ 1 and M0 = M1 = 1, see Aigner [1]. 3n Setting un = 2n+3 n+3 and vn = n+3 in the definition (2.2) of f (x) leads to f (x) = −

16n5 + 228n4 + 1281n3 + 3535n2 + 4779n + 2529 4 x 9(n + 1)3 (n + 2)2 (n + 4)2 (n!)4

(3.1)

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+

64n5 + 816n4 + 4020n3 + 9520n2 + 10788n + 4644 3 x 9(n + 1)3 (n + 2)2 (n + 4)2 (n!)4

+

32n4 + 328n3 + 1238n2 + 1968n + 1098 2 x 9(n + 1)2 (n + 2)2 (n + 4)2 (n!)4

−

948 + 1212n + 496n2 + 64n3 16n2 + 84n + 105 x − . 3(n + 1)(n + 2)2 (n + 4)2 (n!)4 (n + 2)2 (n + 4)2 (n!)4

As mentioned in Section 2, we need to find appropriate lower bound g(n) and upper bound h(n) for MMn+1 . First, let us consider the upper bound h(n). n Lemma 3.1. For n ≥ 0, we have 3(2n + 5) Mn+1 . < Mn 2(n + 4)

(3.2)

Proof. By the recurrence relation (3.1) for the Motzkin numbers, we have for n ≥ 0 Mn+2 =

2n + 5 3(n + 1) Mn+1 + Mn . n+4 n+4

Dividing both sides of the above equality by Mn+1 gives 2n + 5 3(n + 1) Mn Mn+2 + = . Mn+1 n+4 n + 4 Mn+1

(3.3)

It is well known that Mn is log-convex for n ≥ 1. Thus we have Mn+1 Mn+2 > . Mn+1 Mn

(3.4)

Using (3.4) into (3.3), we get 

Mn+1 Mn

2 −

2n + 5 Mn+1 3(n + 1) < 0. − n + 4 Mn n+4

It follows that  2n + 5 + (4n + 10)2 − 27 3(2n + 5) Mn+1 < . < Mn 2(n + 4) 2(n + 4) This completes the proof. 2 Setting h(n) =

3(2n + 5) , 2(n + 4)

(3.5)

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we get f (h(n)) =



9 64n6 − 384n5 − 12420n4 − 79129n3 − 223935n2 − 299979n − 154865 . 16(n + 1)3 (n + 2)2 (n + 4)6 (n!)4

It can be checked that for n ≥ 100, f (h(n)) > 0. On the other hand, f  (h(n)) = −

608n6 + 9936n5 + 67464n4 + 243469n3 + 492057n2 + 527265n + 233663 2(n + 1)3 (n + 2)2 (n + 4)5 (n!)4

f  (h(n)) = −

1)3 (n

r(n) + 2)2 (n + 4)4 (n!)4

9(n +

4 64n6 + 1072n5 + 7464n4 + 27625n3 + 57265n2 + 62961n + 28647 (3) f (h(n)) = − , 3(n + 1)3 (n + 2)2 (n + 4)3 (n!)4 where r(n) = 512n7 + 9856n6 + 83000n5 + 395852n4 +1150543n3 + 2026685n2 + 1990617n + 836019. It is easily seen that f  (h(n)), f  (h(n)) and f (3) (h(n)) are negative for n ≥ 0. Now let us turn to search an appropriate lower bound g(n) for MMn+1 . Fortunately, the n lower bound g(n) =

3(4n + 3)(4n + 7) 16(n + 1)(n + 3)

(3.6)

given in our previous paper [5] is available. In fact, we have f (g(n)) =

65536(n +

1)7 (n

9t(n) , + 2)2 (n + 3)3 (n + 4)2 (n!)4

where t(n) = 262144n9 + 3538944n8 + 21037056n7 + 71883776n6 + 154887168n5 +217141872n4 + 196826284n3 + 110421291n2 + 34543872n + 4561893. Obviously, f (g(n)) > 0 for n ≥ 0. On the other hand, f  (g(n)) = −

t1 (n) 1024(n + 1)6 (n + 2)2 (n + 3)2 (n + 4)2 (n!)4

(3.7)

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f  (g(n)) = − f (3) (g(n)) = −

t2 (n) 576(n + 1)5 (n + 2)2 (n + 3)(n + 4)2 (n!)4 512n6 + 7040n5 + 38832n4 + 109076n3 + 163355n2 + 122088n + 34533 6(n + 1)4 (n + 2)2 (n + 4)2 (n!)4

where t1 (n) = 216384n8 + 147456n7 + 436224n6 − 51328n5 −3349776n4 − 8562300n3 − 9910589n2 − 5526984n − 1194147 t2 (n) = 32768n8 + 532480n7 + 3662336n6 + 13841408n5 +31293040n4 + 43058540n3 + 34867695n2 + 14991408n + 2594241. One can see that f  (g(n)) < 0 for n ≥ 5, f  (g(n)) < 0 and f (3) (g(n)) < 0 for n ≥ 0. We have proved that for n ≥ 100, h(n) defined by (3.5) and g(n) defined by (3.6) satisfy all the conditions in Theorem 2.1. Hence, by Theorem 2.1, Mn satisfies the higher order Turán inequality for n ≥ 100. On the other hand, it is readily checked that for 1 ≤ n ≤ 99, Mn also satisfies the higher order Turán inequality. Hence, we reach the following result. n Theorem 3.2. Let {Mn } be the Motzkin numbers. We have that the sequence { M n! }n≥0 satisfies the higher order Turán inequality, i.e., for n ≥ 1,

 4

Mn−1 Mn+1 Mn2 − 2 n! (n − 1)! (n + 1)!  −



2 Mn+1 Mn Mn+2 − 2 (n + 1)! n! (n + 2)!

Mn−1 Mn+2 Mn Mn+1 − n! (n + 1)! (n − 1)! (n + 2)!



2 > 0.

Since (3.8) is equivalent to the positivity of the discriminant of the equation 3Mn+1 3Mn+2 2 Mn+3 3 Mn + x+ x + x = 0, n! (n + 1)! (n + 2)! (n + 3)! by Theorem 3.2 we get the following result. Theorem 3.3. For n ≥ 1, the polynomials 3Mn+1 3Mn+2 2 Mn Mn+3 3 + x+ x + x n! (n + 1)! (n + 2)! (n + 3)! have only real zeros.

(3.8)

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3.2. Fine numbers We proceed to consider the Fine numbers Fn , defined by the recurrence relation Fn+1 =

7n + 2 2(2n + 1) Fn + Fn−1 , 2n + 4 2n + 4

(3.9)

where n ≥ 1 and F0 = 1, F1 = 0, see Deutsch and Shapiro [14]. 3n Setting un = 2n+3 n+3 and vn = n+3 in the definition of f (x) gives f (x) = −

k0 + k1 x + k2 x2 + k3 x3 + k4 x4 16(1 + n)3 (2 + n)2 (3 + n)2 (1 + 2n)2 (n!)4

where k0 = 1152 + 9024n + 29072n2 + 50560n3 + 52500n4 + 33108n5 + 11776n6 + 1808n7 k1 = 1728 + 14880n + 54256n2 + 105728n3 + 118892n4 + 78328n5 + 28596n6 + 4536n7 k2 = −1728 − 12384n − 27084n2 − 16512n3 + 15177n4 + 26367n5 + 13779n6 + 2673n7 k3 = −7488n − 42576n2 − 93760n3 − 100684n4 − 58084n5 − 17700n6 − 2268n7 k4 = 4608n + 20496n2 + 33152n3 + 27236n4 + 12476n5 + 3084n6 + 324n7 . Now let us find appropriate upper bound for

Fn+1 Fn .

Lemma 3.4. For n ≥ 2, we have 4n + 6 Fn+1 . < Fn n+3

(3.10)

Proof. By the recurrence relation (3.9) for Fn , we have for n ≥ 2 Fn+2 =

7n + 9 2(2n + 3) Fn+1 + Fn . 2n + 6 2n + 6

Dividing both sides of the above equality by Fn+1 gives Fn+2 7n + 9 2(2n + 3) Fn + = . Fn+1 2n + 6 2n + 6 Fn+1

(3.11)

Since Fn is log-convex for n ≥ 2, we have Fn+2 Fn+1 > . Fn+1 Fn

(3.12)

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Using (3.12) into (3.11), we get 

Fn+1 Fn

2 −

7n + 9 Fn+1 2(2n + 3) < 0. − 2n + 6 Fn 2n + 6

It follows that 7n + 9 + Fn+1 < Fn



(7n + 9)2 + 4(2n + 6)(4n + 6) 4n + 6 = . 4(n + 3) n+3

This completes the proof. 2 Let h(n) =

4n + 6 , n+3

(3.13)

we have that for n ≥ 50, f (h(n)) > 0 and for n ≥ 1, f  (h(n)) < 0, f (2) (h(n)) < 0 and f (3) (h(n)) < 0. On the other hand, the lower bound g(n) =

4(n + 1)2 − 2(n + 1) + (n + 1)(n + 2)

2 3

(3.14)

given in [5] is still available. In fact, it is easy to see that for n ≥ 10, f (g(n)) > 0, f  (g(n)) < 0, f (2) (g(n)) < 0 and f (3) (g(n)) < 0. We have verified that for n ≥ 50, h(n) given by (3.13) and g(n) given by (3.14) satisfy all the conditions in Theorem 2.1. Hence, we deduce that for n ≥ 50, Fn satisfies the higher order Turán inequality. Meanwhile, for 1 ≤ n ≤ 49, Fn satisfies the higher order Turán inequality. Hence, we obtain the following result. n Theorem 3.5. Let {Fn } be the Fine numbers. We have that the sequence { M n! }n≥3 satisfies the higher order Turán inequality, i.e., for n ≥ 1,

 4

  2 Fn+1 Fn2 Fn−1 Fn+1 Fn Fn+2 − − n!2 (n − 1)! (n + 1)! (n + 1)!2 n! (n + 2)!  2 Fn Fn+1 Fn−1 Fn+2 − − > 0. n! (n + 1)! (n − 1)! (n + 2)!

An immediate consequence of Theorem 3.5 is as follows. Theorem 3.6. For n ≥ 3, the polynomials 3Fn+1 3Fn+2 2 Fn+3 3 Fn + x+ x + x n! (n + 1)! (n + 2)! (n + 3)! have only real zeros.

(3.15)

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3.3. Domb numbers Now we consider the sequence of the Domb numbers Dn given by the recurrence relation n3 Dn = 2(2n − 1)(5n2 − 5n + 2)Dn−1 − 64(n − 1)3 Dn−2 ,

(3.16)

where n ≥ 2 and D0 = 1 and D1 = 4. The n-th Domb number Dn is the number of 2n-step polygons on the diamond lattice. For more background of the Domb numbers, see [2] and [3]. The upper bound h(n) =

2(2n + 1)3 (n + 1)3

(3.17)

given in [5] is appropriate. For details, we have for n ≥ 10, f (h(n)) > 0, f  (h(n)) > 0, f  (h(n)) < 0 and f (3) (h(n)) < 0. Now we need a lower bound for g(n) for DDn+1 such that f (g(n)) > 0, f  (g(n)) > 0, n  (3) f (g(n)) < 0 and f (g(n)) < 0. We found that a lower bound g(n) directly deduced from the log-behavior of the Domb numbers does not satisfy these inequalities. It means that the method in finding the lower bound for the Motzkin numbers and the Fine numbers does not work on finding a desired lower bound for the Domb numbers. Thus, we try to get a lower bound by adjusting the upper bound (3.17). Let g(n) =

16n3 + 24n2 − 15n − 32 , (n + 1)3

(3.18)

we have for n ≥ 30, f (g(n)) > 0, f  (g(n)) > 0, f  (g(n)) < 0 and f (3) (g(n)) < 0. The remainder is to prove that g(n) is indeed a lower bound for DDn+1 by induction. n Lemma 3.7. For n ≥ 1, we have Dn+1 16n3 + 24n2 − 15n − 32 > . Dn (n + 1)3

(3.19)

Proof. We use induction on n. Assume that 16(n − 1)3 + 24(n − 1)2 − 15(n − 1) − 32 Dn > Dn−1 n3 is true, we proceed to show (3.19). By recurrence relation (3.16), we have 2(2n + 1)(5n2 + 5n + 2) 64n3 Dn−1 Dn = − . 3 Dn−1 n (n + 1)3 Dn

(3.20)

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Applying inductive assumption (3.20) to the above equality gives Dn 2(2n + 1)(5n2 + 5n + 2) > Dn−1 n3 −

64n6 . (n + 1)3 (16(n − 1)3 + 24(n − 1)2 − 15(n − 1) − 32)

(3.21)

It can be checked that for n ≥ 1, the right hand side of (3.21) is larger than 16n3 +24n2 −15n−32 . We complete the induction. 2 (n+1)3 By Theorem 2.1, the higher order Turán inequality for the Domb numbers holds for n ≥ 10. For 1 ≤ n ≤ 10, we can immediately verify that the Domb numbers satisfies the higher order Turán inequality. Hence, we arrive at the following two results. Theorem 3.8. Let {Dn } be the Domb numbers. We have that the sequence { Dn!n }n≥1 satisfies the higher order Turán inequality, i.e., for n ≥ 1,  4

  2 Dn+1 Dn2 Dn−1 Dn+1 Dn Dn+2 − − n!2 (n − 1)! (n + 1)! (n + 1)!2 n! (n + 2)! 2  Dn−1 Dn+2 Dn Dn+1 − − > 0. n! (n + 1)! (n − 1)! (n + 2)!

(3.22)

Theorem 3.9. For n ≥ 1, the polynomials 3Dn+1 3Dn+2 2 Dn+3 3 Dn + x+ x + x n! (n + 1)! (n + 2)! (n + 3)! have only real zeros. 3.4. Franel numbers of order 3 The Franel numbers of order r are defined by Fn(r) =

n  r  n k=0

k

.

It is known that the Franel numbers of order 3 satisfy a homogeneous linear recurrence relation (3)

Fn+1 = see [27].

7n2 + 7n + 2 (3) 8n2 (3) Fn + F , 2 (n + 1) (n + 1)2 n−1

(3.23)

L.X.W. Wang / Advances in Applied Mathematics 110 (2019) 180–196

Substitute un =

7n2 +7n+2 (n+1)2

and vn =

f (x) = −

8n2 (n+1)2

in the formula (2.2) of f (x) and generate

r 4 x4 + r 3 x3 + r 2 x2 + r 1 x + r 0 , 64n2 (n + 1)3 (n + 2)6 (n!)4

where r0 = 256 + 3072n + 32832n2 + 141504n3 + 311872n4 + 401088n5 + 317120n6 +153664n7 + 42432n8 + 5184n9 ; r1 = 768 + 15040n + 127552n2 + 461968n3 + 897536n4 + 1042448n5 + 752448n6 +333104n7 + 83328n8 + 9072n9 ; r2 = 64 + 10624n + 67332n2 + 208824n3 + 377541n4 + 417549n5 + 285918n6 +118410n7 + 27225n8 + 2673n9 ; r3 = −768 − 17456n − 92036n2 − 232258n3 − 337476n4 − 304410n5 − 174476n6 −62246n7 − 12684n8 − 1134n9 ; r4 = 256 + 4000n + 16825n2 + 34915n3 + 42793n4 + 33291n5 + 16771n6 +5345n7 + 987n8 + 81n9 . (3)

First, we concern with the lower bound for

Fn+1 (3)

Fn

.

Lemma 3.10. For n ≥ 3, we have (3)

Fn+1 (3)

>

Fn

8n2 + 8n + (n + 1)2

16 9

.

(3)

Proof. From the log-convexity of Fn+1 and the recurrence relation (3.23), we have (3)

Fn+1 (3)

(3)

>

Fn

7n2 + 7n + 2 8n2 Fn + . 2 (n + 1) (n + 1)2 F (3) n+1

(3)

It follows that

Fn+1 (3)

Fn

must be larger than the large root of the equation x2 −

7n2 + 7n + 2 8n2 x − = 0. (n + 1)2 (n + 1)2

Hence, we get (3)

Fn+1 (3)

Fn as desired. 2

>

193

8n2 + 8n + (n + 1)2

16 9

,

194

L.X.W. Wang / Advances in Applied Mathematics 110 (2019) 180–196

Let g(n) =

8n2 + 8n + (n + 1)2

16 9

,

(3.24)

we have that for n ≥ 100, f (h(n)) > 0, f  (g(n)) < 0, f  (g(n)) < 0 and f (3) (g(n)) < 0. To satisfy the conditions in Theorem 2.1, we aim to find an upper bound h(n) satisfying that f (h(n)) > 0, f  (h(n)), f  (h(n)) and f (3) (h(n)) are all negative. By slightly adjusting the constant in the numerator of the lower bound (3.24), we find the following bound h(n) =

8n2 + 8n + 3 (n + 1)2

(3.25)

is available. It can be seen that for n ≥ 2, f (h(n)) > 0, f  (h(n)) < 0, f  (h(n)) < 0 and f (3) (h(n)) < 0. (3)

The remainder is to prove (3.25) is indeed an upper bound for

Fn+1 (3)

Fn

Lemma 3.11. For n ≥ 4, (3)

Fn+1 (3) Fn

<

8n2 + 8n + 3 . (n + 1)2

Proof. By Lemma 3.10, we have that for n ≥ 2, (3)

Fn

(3)

Fn−1

>

8n2 − 8n + n2

16 9

.

It follows from the recurrence relation (3.23) that (3)

Fn+1 (3)

Fn

(3)

7n2 + 7n + 2 8n2 Fn−1 = + (n + 1)2 (n + 1)2 Fn(3) <

7n2 + 7n + 2 8n4 + 2 (n + 1) (n + 1)2 8n2 − 8n +

=

72n4 − 4n − 31n2 + 4 . (n + 1)2 (3n − 2)(3n − 1)

It can be checked that for n ≥ 4, 8n2 + 8n + 3 72n4 − 4n − 31n2 + 4 < . (n + 1)2 (3n − 2)(3n − 1) (n + 1)2 This completes the proof. 2

16 9



.

L.X.W. Wang / Advances in Applied Mathematics 110 (2019) 180–196

195

Combining the above results, we can now deduce that for n ≥ 100, the Franel numbers of order 3 satisfies the higher order Turán inequality. On the other hand, it can be (3) checked that for 2 ≤ n ≤ 100, the Franel numbers Fn of order 3 also satisfies the higher order Turán inequality. Hence, we have the following theorems. (3) Fn

(3)

Theorem 3.12. For n ≥ 2, the Franel numbers Fn inequality,

of order 3 satisfy the following

⎛

⎞⎛ ⎞ (3) (3) (3) 2 (3) (3) 2 (3) Fn+2 Fn−1 Fn+1 Fn+1 F F n n ⎠⎝ ⎠ 4⎝ 2 − − n! (n − 1)! (n + 1)! (n + 1)!2 n! (n + 2)!  −

(3)

(3)

(3)

(3) Fn+1 Fn−1 Fn+2 Fn − n! (n + 1)! (n − 1)! (n + 2)!

2 > 0.

(3.26)

Theorem 3.13. For n ≥ 1, the polynomials (3)

(3)

(3)

(3) 3Fn+1 3Fn+2 2 Fn+3 3 Fn + x+ x + x n! (n + 1)! (n + 2)! (n + 3)!

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