The Log-Behavior of Some Sequences Related to a Linear Recurrence Sequence
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#A100 INTEGERS 25 (2025) THE LOG-BEHAVIOR OF SOME SEQUENCES RELATED TO A LINEAR RECURRENCE SEQUENCE Feng-Zhen Zhao Department of Mathematics, Shanghai University, Shanghai, China [email protected] Received: 3/10/25, Accepted: 10/24/25, Published: 11/5/25 Abstract Let {An}n≥0be a sequence satisfying the recurrence relation (n+ 1)An+1 = 2(n+ 1)An+ 3(n−1)An−1, with the initial conditions A0= 1 and A1= 1. Let ∆ be the (forward) difference operator. In this paper, we are interested in the log-behavior of some sequences involving An. We mainly show that {∆2An}n≥3and {∆3An}n≥3are log-balanced. In addition, we prove that some sequences such as {nAn}n≥1,{∆(nAn)}n≥1, and {n·∆An}n≥1are log-concave. 1. Introduction Let {An}n≥0be a sequence satisfying the recurrence relation (n+ 1)An+1 = 2(n+ 1)An+ 3(n−1)An−1,(1) with the initial conditions A0= 1 and A1= 1. The sequence {An}n≥0is sequence A005773 in the OEIS [9] and {An}n≥1appeared in Exercise 6.46 of Stanley [11]. The value of Anis equal to the number of symmetric Dyck paths of semilength 2n−1 with no peaks at even level. For more properties of {An}n≥0, see sequence A005773 in the OEIS [9]. Some sequences involving Anplay an important role in combinatorial enumeration problems. Let ∆ be the (forward) difference operator. The first difference sequence of {An}n≥0is {∆An}n≥0={An+1 −An}n≥0and {∆An}n≥0is related to sequence A025566 in the OEIS [9]. Let {an}n≥0denote sequence A025566 in the OEIS [9]. The value of an+1 is the number of Motzkin (2n)-paths whose last weak valley occurs immediately after step n. For the definition of the weak valley and more properties of {an}n≥0, see sequence A025566 in the DOI: 10.5281/zenodo.17535297
INTEGERS: 25 (2025) 2 OEIS [9]. For n≥2, an= ∆An−1=An−An−1. The second difference sequence of {An}n≥0is ∆2An= ∆(∆An) = ∆(An+1 −An) = An+2 −2An+1 +An. The sequence {∆2An}n≥0is sequence A026135 in the OEIS [9] and the value of ∆2Anis the total number of rows of consecutive peaks in all Motzkin (n+2)-paths. The sequence {nAn}n≥0is sequence A132894 in the OEIS [9] and the value of nAn is the number of peaks in all paths of length n+1 with steps U= (1,1), D= (1,−1), and H= (1,0), which start at (0,0) and stay weakly above the x-axis. Hence the properties of sequences related to Andeserve to be studied. The main purpose of this paper is to investigate the log-behavior of some sequences involving An. Now we recall some definitions that we will need in this paper. A positive sequence {zn}n≥0is called log-convex (log-concave) if z2 n≤zn−1zn+1 (z2 n≥zn−1zn+1) for each n≥1. A log-convex sequence {zn}n≥0is called logbalanced if {zn n!}n≥0is log-concave (Doˇsli´c [3] gave this definition). It is clear that a sequence {zn}n≥0is log-convex (log-concave) if and only if its quotient sequence {zn+1 zn}n≥0is nondecreasing (nonincreasing) and a log-convex sequence {zn}n≥0is log-balanced if and only if {zn+1 (n+1)zn}n≥0is nonincreasing. Log-convexity (Logconcavity) is a fertile source of combinatorial inequalities and plays an important role in many subjects (see for instance [1, 2, 4, 5, 6, 8, 10]). In this paper, we are interested in the log-behavior of some sequences involving An. For instance, we show that {∆2An}n≥3and {∆3An}n≥3are log-balanced, {nAn}n≥1is log-concave, and {nnAn n}n≥1is log-convex. 2. Main Results We first give a lemma. Lemma 1. For the sequence {An}n≥0satisfying the recurrence relation (1), let xn=An+1 An(n≥0). For n≥0, we have xn≤φn,(2) and xn+1 xn ≥2(3n2+ 6n+ 2) (n+ 2)(2n+ 1)φn ,(3) where φn=3(2n+1) 2(n+1) . Proof. By using (1), we get xn= 2 + 3(n−1) (n+ 1)xn−1 for n≥1.(4)
INTEGERS: 25 (2025) 3 We prove by induction that (2) holds. It is clear that xj≤φjfor 0 ≤j≤3. Assume that xk≤φkfor k≥3. It follows from (4) that xk+1 −φk+1 = 2 + 3k (k+ 2)xk −φk+1. Since Liu and Wang [7] proved that xn≥µnfor n≥0,(5) where µn=6n 2n+1 , it follows from (5) that xk+1 −φk+1 ≤2 + 3k (k+ 2)µk −φk+1 = 2 + 2k+ 1 2(k+ 2) −3(2k+ 3) 2(k+ 2) = 0. Thus xn≤φnfor n≥0. It follows from (2) and (4) that xn+1 xn ≥2 φn +3n (n+ 2)φ2 n =2(3n2+ 6n+ 2) (n+ 2)(2n+ 1)φn . For the sequence {An}n≥0satisfying the recurrence relation (1), Liu and Wang [7] proved that {An}n≥0is log-convex, and Zhao [12] showed that {An+1 −An}n≥3 is log-convex. Now we discuss the log-behavior of some sequences involving Ansuch as {∆2An}n≥0,{∆3An}n≥0, and {nAn}n≥1. Theorem 1. For the sequence {An}n≥0satisfying the recurrence relation (1), the sequences {∆An}n≥3and {∆2An}n≥3are log-balanced. Proof. For n≥0, put xn=An+1 An,xn=An+2−An+1 (n+1)(An+1−An)(n≥1), and yn=∆2An+1 ∆2An. It is clear that xn=xn(xn+1−1) (n+1)(xn−1) . It follows from (1) that ∆2An=2(2n+ 1) n+ 2 Anfor n≥0.(6) The values of {∆2Aj}0≤j≤10 are 1,2,5,14,39,110,312,890,2550,7334,21161,61226,177575. By applying (4), we derive xn=xn (n+ 1)(xn−1) +3n (n+ 1)(n+ 2)(xn−1) (n≥1). Since the two sequences {xn (n+1)(xn−1) }n≥1and {n (n+1)(n+2)(xn−1) }n≥2are decreasing, it follows that {xn}n≥2is decreasing. We have that {An+1−An n!}n≥2is logconcave. Note that Zhao [12] showed that {An+1 −An}n≥3is log-convex, and
INTEGERS: 25 (2025) 4 hence {∆An}n≥3is log-balanced. In order to prove that the sequence {∆2An}n≥3 is log-convex, we need to show that {yn}n≥3is increasing. By means of (6), we get yn=(n+ 2)(2n+ 3) (n+ 3)(2n+ 1)xnfor n≥0.(7) It follows from (7) that yn+1 −yn=(n+ 3)(2n+ 5) (n+ 4)(2n+ 3)xn+1 −(n+ 2)(2n+ 3) (n+ 3)(2n+ 1)xn =(n+ 3)2(2n+ 1)(2n+ 5)xn+1 −(n+ 2)(n+ 4)(2n+ 3)2xn (n+ 3)(n+ 4)(2n+ 1)(2n+ 3) . It follows from (4) that (n+ 3)2(2n+ 1)(2n+ 5)xn+1 −(n+ 2)(n+ 4)(2n+ 3)2xn = (n+ 3)2(2n+ 1)(2n+ 5)2 + 3n (n+ 2)xn−(n+ 2)(n+ 4)(2n+ 3)2xn. By using (2), we have (n+ 3)2(2n+ 1)(2n+ 5)xn+1 −(n+ 2)(n+ 4)(2n+ 3)2xn ≥(n+ 3)2(2n+ 1)(2n+ 5)2 + 3n (n+ 2)φn−(n+ 2)(n+ 4)(2n+ 3)2φn = 2(n+ 3)2(2n+ 1)(2n+ 5) + 2n(n+ 1)(n+ 3)22 + 1 n+ 2 −3(n+ 2)(n+ 4)(2n+ 1)(2n+ 3)2 2(n+ 1) = 5n2−10n−30 + 2n(n+ 1) n+ 2 +3(n+ 2)(n+ 4) 2(n+ 1) >0 (n≥3). Thus {yn}n≥3is increasing. By (4), we have xn n+ 1 =2 n+ 1 +3(n−1) (n+ 1)2xn−1 for n≥1. Since {xn}n≥0is increasing and {n−1 (n+1)2}n≥3is decreasing, it follows that the sequence {xn n+1 }n≥3is decreasing. Noting that yn n+ 1 =1 + 3 2n2+ 7n+ 3xn n+ 1, we have that {yn n+1 }n≥3is decreasing. This indicates that {∆2An n!}n≥3is log-concave, and hence {∆2An}n≥3is log-balanced.
INTEGERS: 25 (2025) 5 n∆bn 0 1 1 3 2 9 3 25 471 5 202 6 578 7 1660 8 4784 9 13827 10 40065 11 116349 12 338539 13 986753 14 2880595 15 8420967 16 24648441 17 72229449 18 211881339 19 622137483 20 1828359477 21 5377601112 22 15828541932 23 46622437134 Table 1: Some initial values of {∆bn}n≥0 Theorem 2. For the sequence {An}n≥0satisfying the recurrence relation (1), {∆3An}n≥2is log-convex and {∆3An}n≥3is log-balanced. Proof. For convenience, let bn= ∆2An(n≥0). It is clear that ∆3An= ∆bn. It follows from (1) and (6) that bn+1 =2(n+ 2)(2n+ 3) (n+ 3)(2n+ 1) bn+3(n−1)(2n+ 3) (n+ 3)(2n−1) bn−1for n≥1.(8) Some initial values of {∆bn}n≥0are given in Table 1. By using (8), we get ∆bn=1 + 6 2n2+ 7n+ 3bn+3(2n+ 3)(n−1) (2n−1)(n+ 3) bn−1. We find that {∆bj}2≤j≤23 is log-convex. In order to prove that the sequence {∆bn}n≥2is log-convex, we need to show that {∆bn}n≥22 is log-convex. For n≥1,
INTEGERS: 25 (2025) 6 let gn=(n−1)bn−1 n+3 . We first prove that the sequence {gn}n≥22 is log-convex. For n≥0, put xn=An+1 Anand hn=gn+1 gn(n≥1). It follows from (6) that hn=n(n+ 1)(n+ 3)(2n+ 1) (n−1)(n+ 2)(n+ 4)(2n−1)xn−1, n ≥1. Due to (3), we obtain hn+1 hn ≥4(n−1)(n+ 2)2(n+ 4)2(2n+ 3)(3n2−1) 3n(n+ 1)(n+ 3)2(n+ 5)(2n−1)(2n+ 1)2. Define Γn= 4(n−1)(n+ 2)2(n+ 4)2(2n+ 3)(3n2−1) −3n(n+ 1)(n+ 3)2(n+ 5)(2n−1)(2n+ 1)2. By computation, we have Γn= 10n6−145n5−1388n4−3406n3−2054n2+ 1031n+ 768 >0 (n≥22). This implies that {hn}n≥22 is increasing, and hence {gn}n≥22 is log-convex. Note that {1+ 6 2n2+7n+3 }n≥1,{bn}n≥3, and {2n+3 2n−1}n≥1are log-convex, and hence {∆bn}n≥22 is log-convex. For n≥0, let yn=bn+1 bnand yn=∆bn+1 ∆bn. It follows from (8) that ∆bn+1 =6 (n+ 4)(2n+ 3) ·∆bn+3n(2n+ 5) (n+ 4)(2n+ 1) +6 (n+ 4)(2n+ 3)bn.(9) By (9), we have yn n+ 1 =6 (n+ 1)(n+ 4)(2n+ 3) +3n(2n+ 3)(2n+ 5) + 6(2n+ 1) (n+ 1)(n+ 4)(2n+ 1)(2n+ 3)(yn−1). We need to show that {∆bn n!}n≥3is log-concave. It suffices to prove that {yn n+1 }n≥3 is decreasing. Since the sequences {6 (n+1)(n+4)(2n+3) }n≥2,{1 yn−1}n≥3, and {3n(2n+3)(2n+5)+6(2n+1) (n+1)(n+4)(2n+1)(2n+3) }n≥2are decreasing, it follows that {yn n+1 }n≥3is decreasing. Hence {∆bn}n≥3is log-balanced. Theorem 3. For the sequence {An}n≥0satisfying the recurrence relation (1), we have that {nAn}n≥1,{∆(nAn)}n≥1,{n·∆An}n≥1, and {n·∆2An}n≥1are logconcave. Proof. For n≥0, let xn=An+1 An. For n≥1, set Wn=nAn,sn=Wn+1 Wn,tn= Wn+2−Wn+1 Wn+1−Wn, and un=(n+1)·∆An+1 n·∆An=(n+1)(An+2−An+1) n(An+1−An). It is evident that sn+1 −sn=n(n+ 2)xn+1 −(n+ 1)2xn n(n+ 1) , tn=sn(sn+1 −1) sn−1,
INTEGERS: 25 (2025) 7 and un=(n+ 1)xn(xn+1 −1) n(xn−1) . It follows from (4) and (5) that n(n+ 2)xn+1 −(n+ 1)2xn=n(n+ 2)2 + 3n (n+ 2)xn−(n+ 1)2xn ≤n(n+ 2)2 + 3n (n+ 2)µn−(n+ 1)2µn =−3n 2(2n+ 1) <0 (n≥1). This leads to sn+1 −sn<0 for n≥1, and hence the sequence {nAn}n≥1is logconcave. It follows from (1) that Wn+1 =2(n+ 1) nWn+ 3Wn−1for n≥2.(10) Applying (10), we have sn=2(n+ 1) n+3 sn−1 for n≥1. Thus we obtain tn=sn sn−1n+ 3 n+ 1 +3 sn=1 + 1 sn−1n+ 3 n+ 1 +3 sn−1. It is obvious that {tn}n≥1is decreasing. Thus, {∆(nAn)}n≥1is log-concave. Now we prove that the sequence {n·∆An}n≥1is log-concave. It follows from (4) that un=n+ 1 n+n+ 1 n(xn−1) +3(n+ 1) (n+ 2)(xn−1) for n≥1.(11) By using (11), we derive un+1 −un=−1 n(n+ 1) +n+ 2 (n+ 1)(xn+1 −1) +3(n+ 2) (n+ 3)(xn+1 −1) −n+ 1 n(xn−1) −3(n+ 1) (n+ 2)(xn−1). Noting that the sequence {xn}n≥0is increasing, we have that un+1 −un≤ − 1 n(n+ 1) +n+ 2 (n+ 1)(xn−1) +3(n+ 2) (n+ 3)(xn−1) −n+ 1 n(xn−1) −3(n+ 1) (n+ 2)(xn−1) =3n(n+ 1) −(n+ 2)(n+ 3)xn n(n+ 1)(n+ 2)(n+ 3)(xn−1).
INTEGERS: 25 (2025) 8 It follows from (5) that 3n(n+ 1) −(n+ 2)(n+ 3)xn≤3n(n+ 1) −(n+ 2)(n+ 3)µn =−n(7n+ 11) 2n+ 1 (n≥1). This implies that {un}n≥1is decreasing, and hence the sequence {n·∆An}n≥1is logconcave. It follows from (6) that n·∆2An= 2nAn·2n+1 n+2 . Since the two sequences {nAn}n≥1and {2n+1 n+2 }n≥1are both log-concave, it follows that {n·∆2An}n≥1is also log-concave. Theorem 4. For the sequence {An}n≥0satisfying the recurrence relation (1), {nnAn n}n≥1is log-convex. Proof. For n≥0, put xn=An+1 Anand vn=(n+1)n+1An+1 n+1 nnAn n (n≥1). It suffices to show that {vn}n≥1is increasing. It is obvious that vn= (n+ 1)An+11 + 1 nn xn n. Since the sequence {An}n≥0is log-convex and x0= 1, it follows that {xn n}n≥0is increasing. On the other hand, the sequences {(1 + 1 n)n}n≥1and {(n+ 1)An+1}n≥1 are increasing. Thus, {vn}n≥1is increasing. 3. Concluding Remarks For the sequence {An}n≥0satisfying the recurrence relation (1), we have discussed the log-behavior of some sequences involving An. We mainly proved that {∆2An}n≥3and {∆3An}n≥3are log-balanced, where ∆ is the (forward) difference operator. For bn= ∆2An, there is a conjecture (see sequence A026135 in the OEIS [9]) that states the following recurrence relation (n+ 2)bn−3(n+ 1)bn−1−(n+ 2)bn−2+ 3(n−3)bn−3= 0 for n≥3. This conjecture holds as a consequence of (1) and (6). For {An}n≥0, we now state a conjecture. Conjecture 1. For the sequence {An}n≥0satisfying the recurrence relation (1), there exists a positive integer Nsuch that {∆4An}n≥Nis log-convex. The author plans to study the various properties of {An}n≥0. Acknowledgment. The author is grateful to the anonymous referee for his/her helpful comments and suggestions.
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