The Log-Behavior of Some Sequences Related to a Linear Recurrence Sequence

#A100 INTEGERS 25 (2025)
THE LOG-BEHAVIOR OF SOME SEQUENCES RELATED TO A
LINEAR RECURRENCE SEQUENCE
Feng-Zhen Zhao
Depa men o Ma hema ics, Shanghai Uni e si y, Shanghai, China
[email p o ec ed]
Recei ed: 3/10/25, Accep ed: 10/24/25, Published: 11/5/25
Abs ac
Le {An}n≥0be a sequence sa is ying he ecu ence ela ion
(n+ 1)An+1 = 2(n+ 1)An+ 3(n−1)An−1,
wi h he ini ial condi ions A0= 1 and A1= 1. Le ∆ be he ( o wa d) di e ence
ope a o . In his pape , we a e in e es ed in he log-beha io o some sequences
in ol ing An. We mainly show ha {∆2An}n≥3and {∆3An}n≥3a e log-balanced.
In addi ion, we p o e ha some sequences such as {nAn}n≥1,{∆(nAn)}n≥1, and
{n·∆An}n≥1a e log-conca e.
1. In oduc ion
Le {An}n≥0be a sequence sa is ying he ecu ence ela ion
(n+ 1)An+1 = 2(n+ 1)An+ 3(n−1)An−1,(1)
wi h he ini ial condi ions A0= 1 and A1= 1. The sequence {An}n≥0is sequence
A005773 in he OEIS [9] and {An}n≥1appea ed in Exe cise 6.46 o S anley [11].
The alue o Anis equal o he numbe o symme ic Dyck pa hs o semileng h
2n−1 wi h no peaks a e en le el. Fo mo e p ope ies o {An}n≥0, see sequence
A005773 in he OEIS [9]. Some sequences in ol ing Anplay an impo an ole in
combina o ial enume a ion p oblems. Le ∆ be he ( o wa d) di e ence ope a o .
The i s di e ence sequence o {An}n≥0is {∆An}n≥0={An+1 −An}n≥0and
{∆An}n≥0is ela ed o sequence A025566 in he OEIS [9]. Le {an}n≥0deno e
sequence A025566 in he OEIS [9]. The alue o an+1 is he numbe o Mo zkin
(2n)-pa hs whose las weak alley occu s immedia ely a e s ep n. Fo he de ini ion
o he weak alley and mo e p ope ies o {an}n≥0, see sequence A025566 in he
DOI: 10.5281/zenodo.17535297
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OEIS [9]. Fo n≥2, an= ∆An−1=An−An−1. The second di e ence sequence
o {An}n≥0is
∆2An= ∆(∆An) = ∆(An+1 −An) = An+2 −2An+1 +An.
The sequence {∆2An}n≥0is sequence A026135 in he OEIS [9] and he alue o
∆2Anis he o al numbe o ows o consecu i e peaks in all Mo zkin (n+2)-pa hs.
The sequence {nAn}n≥0is sequence A132894 in he OEIS [9] and he alue o nAn
is he numbe o peaks in all pa hs o leng h n+1 wi h s eps U= (1,1), D= (1,−1),
and H= (1,0), which s a a (0,0) and s ay weakly abo e he x-axis. Hence he
p ope ies o sequences ela ed o Andese e o be s udied.
The main pu pose o his pape is o in es iga e he log-beha io o some se-
quences in ol ing An. Now we ecall some de ini ions ha we will need in his pa-
pe . A posi i e sequence {zn}n≥0is called log-con ex (log-conca e) i z2
n≤zn−1zn+1
(z2
n≥zn−1zn+1) o each n≥1. A log-con ex sequence {zn}n≥0is called log-
balanced i {zn
n!}n≥0is log-conca e (Doˇsli´c [3] ga e his de ini ion). I is clea ha
a sequence {zn}n≥0is log-con ex (log-conca e) i and only i i s quo ien sequence
{zn+1
zn}n≥0is nondec easing (noninc easing) and a log-con ex sequence {zn}n≥0is
log-balanced i and only i {zn+1
(n+1)zn}n≥0is noninc easing. Log-con exi y (Log-
conca i y) is a e ile sou ce o combina o ial inequali ies and plays an impo an
ole in many subjec s (see o ins ance [1, 2, 4, 5, 6, 8, 10]). In his pape , we a e
in e es ed in he log-beha io o some sequences in ol ing An. Fo ins ance, we
show ha {∆2An}n≥3and {∆3An}n≥3a e log-balanced, {nAn}n≥1is log-conca e,
and {nnAn
n}n≥1is log-con ex.
2. Main Resul s
We i s gi e a lemma.
Lemma 1. Fo he sequence {An}n≥0sa is ying he ecu ence ela ion (1), le
xn=An+1
An(n≥0). Fo n≥0, we ha e
xn≤φn,(2)
and
xn+1
xn
≥2(3n2+ 6n+ 2)
(n+ 2)(2n+ 1)φn
,(3)
whe e φn=3(2n+1)
2(n+1) .
P oo . By using (1), we ge
xn= 2 + 3(n−1)
(n+ 1)xn−1
o n≥1.(4)
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We p o e by induc ion ha (2) holds. I is clea ha xj≤φj o 0 ≤j≤3.
Assume ha xk≤φk o k≥3. I ollows om (4) ha
xk+1 −φk+1 = 2 + 3k
(k+ 2)xk
−φk+1.
Since Liu and Wang [7] p o ed ha
xn≥µn o n≥0,(5)
whe e µn=6n
2n+1 , i ollows om (5) ha
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≤φn o n≥0. I ollows om (2) and (4) ha
xn+1
xn
≥2
φn
+3n
(n+ 2)φ2
n
=2(3n2+ 6n+ 2)
(n+ 2)(2n+ 1)φn
.
Fo he sequence {An}n≥0sa is ying he ecu ence ela ion (1), Liu and Wang
[7] p o ed ha {An}n≥0is log-con ex, and Zhao [12] showed ha {An+1 −An}n≥3
is log-con ex. Now we discuss he log-beha io o some sequences in ol ing Ansuch
as {∆2An}n≥0,{∆3An}n≥0, and {nAn}n≥1.
Theo em 1. Fo he sequence {An}n≥0sa is ying he ecu ence ela ion (1), he
sequences {∆An}n≥3and {∆2An}n≥3a e log-balanced.
P oo . Fo n≥0, pu xn=An+1
An,xn=An+2−An+1
(n+1)(An+1−An)(n≥1), and yn=∆2An+1
∆2An.
I is clea ha xn=xn(xn+1−1)
(n+1)(xn−1) . I ollows om (1) ha
∆2An=2(2n+ 1)
n+ 2 An o n≥0.(6)
The alues o {∆2Aj}0≤j≤10 a e
1,2,5,14,39,110,312,890,2550,7334,21161,61226,177575.
By applying (4), we de i e
xn=xn
(n+ 1)(xn−1) +3n
(n+ 1)(n+ 2)(xn−1) (n≥1).
Since he wo sequences {xn
(n+1)(xn−1) }n≥1and {n
(n+1)(n+2)(xn−1) }n≥2a e dec eas-
ing, i ollows ha {xn}n≥2is dec easing. We ha e ha {An+1−An
n!}n≥2is log-
conca e. No e ha Zhao [12] showed ha {An+1 −An}n≥3is log-con ex, and
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hence {∆An}n≥3is log-balanced. In o de o p o e ha he sequence {∆2An}n≥3
is log-con ex, we need o show ha {yn}n≥3is inc easing. By means o (6), we ge
yn=(n+ 2)(2n+ 3)
(n+ 3)(2n+ 1)xn o n≥0.(7)
I ollows om (7) ha
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) .
I ollows om (4) ha
(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 ha e
(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)22 + 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 inc easing. By (4), we ha e
xn
n+ 1 =2
n+ 1 +3(n−1)
(n+ 1)2xn−1
o n≥1.
Since {xn}n≥0is inc easing and {n−1
(n+1)2}n≥3is dec easing, i ollows ha he se-
quence {xn
n+1 }n≥3is dec easing. No ing ha
yn
n+ 1 =1 + 3
2n2+ 7n+ 3xn
n+ 1,
we ha e ha {yn
n+1 }n≥3is dec easing. This indica es ha {∆2An
n!}n≥3is log-conca e,
and hence {∆2An}n≥3is log-balanced.
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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 ini ial alues o {∆bn}n≥0
Theo em 2. Fo he sequence {An}n≥0sa is ying he ecu ence ela ion (1),
{∆3An}n≥2is log-con ex and {∆3An}n≥3is log-balanced.
P oo . Fo con enience, le bn= ∆2An(n≥0). I is clea ha ∆3An= ∆bn. I
ollows om (1) and (6) ha
bn+1 =2(n+ 2)(2n+ 3)
(n+ 3)(2n+ 1) bn+3(n−1)(2n+ 3)
(n+ 3)(2n−1) bn−1 o n≥1.(8)
Some ini ial alues o {∆bn}n≥0a e gi en in Table 1. By using (8), we ge
∆bn=1 + 6
2n2+ 7n+ 3bn+3(2n+ 3)(n−1)
(2n−1)(n+ 3) bn−1.
We ind ha {∆bj}2≤j≤23 is log-con ex. In o de o p o e ha he sequence
{∆bn}n≥2is log-con ex, we need o show ha {∆bn}n≥22 is log-con ex. Fo n≥1,

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le gn=(n−1)bn−1
n+3 . We i s p o e ha he sequence {gn}n≥22 is log-con ex. Fo
n≥0, pu xn=An+1
Anand hn=gn+1
gn(n≥1). I ollows om (6) ha
hn=n(n+ 1)(n+ 3)(2n+ 1)
(n−1)(n+ 2)(n+ 4)(2n−1)xn−1, n ≥1.
Due o (3), we ob ain
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.
De ine
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 compu a ion, we ha e
Γn= 10n6−145n5−1388n4−3406n3−2054n2+ 1031n+ 768 >0 (n≥22).
This implies ha {hn}n≥22 is inc easing, and hence {gn}n≥22 is log-con ex. No e
ha {1+ 6
2n2+7n+3 }n≥1,{bn}n≥3, and {2n+3
2n−1}n≥1a e log-con ex, and hence {∆bn}n≥22
is log-con ex. Fo n≥0, le yn=bn+1
bnand yn=∆bn+1
∆bn. I ollows om (8) ha
∆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 ha e
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 o show ha {∆bn
n!}n≥3is log-conca e. I su ices o p o e ha {yn
n+1 }n≥3
is dec easing. Since he 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≥2a e dec easing, i ollows ha {yn
n+1 }n≥3is dec easing.
Hence {∆bn}n≥3is log-balanced.
Theo em 3. Fo he sequence {An}n≥0sa is ying he ecu ence ela ion (1), we
ha e ha {nAn}n≥1,{∆(nAn)}n≥1,{n·∆An}n≥1, and {n·∆2An}n≥1a e log-
conca e.
P oo . Fo n≥0, le xn=An+1
An. Fo n≥1, se Wn=nAn,sn=Wn+1
Wn, n=
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). I is e iden ha
sn+1 −sn=n(n+ 2)xn+1 −(n+ 1)2xn
n(n+ 1) , n=sn(sn+1 −1)
sn−1,
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and
un=(n+ 1)xn(xn+1 −1)
n(xn−1) .
I ollows om (4) and (5) ha
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 o sn+1 −sn<0 o n≥1, and hence he sequence {nAn}n≥1is log-
conca e. I ollows om (1) ha
Wn+1 =2(n+ 1)
nWn+ 3Wn−1 o n≥2.(10)
Applying (10), we ha e
sn=2(n+ 1)
n+3
sn−1
o n≥1.
Thus we ob ain
n=sn
sn−1n+ 3
n+ 1 +3
sn=1 + 1
sn−1n+ 3
n+ 1 +3
sn−1.
I is ob ious ha { n}n≥1is dec easing. Thus, {∆(nAn)}n≥1is log-conca e. Now
we p o e ha he sequence {n·∆An}n≥1is log-conca e. I ollows om (4) ha
un=n+ 1
n+n+ 1
n(xn−1) +3(n+ 1)
(n+ 2)(xn−1) o n≥1.(11)
By using (11), we de i e
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).
No ing ha he sequence {xn}n≥0is inc easing, we ha e ha
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).
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I ollows om (5) ha
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 ha {un}n≥1is dec easing, and hence he sequence {n·∆An}n≥1is log-
conca e. I ollows om (6) ha n·∆2An= 2nAn·2n+1
n+2 . Since he wo sequences
{nAn}n≥1and {2n+1
n+2 }n≥1a e bo h log-conca e, i ollows ha {n·∆2An}n≥1is
also log-conca e.
Theo em 4. Fo he sequence {An}n≥0sa is ying he ecu ence ela ion (1),
{nnAn
n}n≥1is log-con ex.
P oo . Fo n≥0, pu xn=An+1
Anand n=(n+1)n+1An+1
n+1
nnAn
n
(n≥1). I su ices o
show ha { n}n≥1is inc easing. I is ob ious ha
n= (n+ 1)An+11 + 1
nn
xn
n.
Since he sequence {An}n≥0is log-con ex and x0= 1, i ollows ha {xn
n}n≥0is
inc easing. On he o he hand, he sequences {(1 + 1
n)n}n≥1and {(n+ 1)An+1}n≥1
a e inc easing. Thus, { n}n≥1is inc easing.
3. Concluding Rema ks
Fo he sequence {An}n≥0sa is ying he ecu ence ela ion (1), we ha e dis-
cussed he log-beha io o some sequences in ol ing An. We mainly p o ed ha
{∆2An}n≥3and {∆3An}n≥3a e log-balanced, whe e ∆ is he ( o wa d) di e ence
ope a o . Fo bn= ∆2An, he e is a conjec u e (see sequence A026135 in he OEIS
[9]) ha s a es he ollowing ecu ence ela ion
(n+ 2)bn−3(n+ 1)bn−1−(n+ 2)bn−2+ 3(n−3)bn−3= 0 o n≥3.
This conjec u e holds as a consequence o (1) and (6). Fo {An}n≥0, we now s a e
a conjec u e.
Conjec u e 1. Fo he sequence {An}n≥0sa is ying he ecu ence ela ion (1),
he e exis s a posi i e in ege Nsuch ha {∆4An}n≥Nis log-con ex.
The au ho plans o s udy he a ious p ope ies o {An}n≥0.
Acknowledgmen . The au ho is g a e ul o he anonymous e e ee o his/he
help ul commen s and sugges ions.
INTEGERS: 25 (2025) 9
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