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Journal of Integer Sequences, Vol. 7 (2004),
Article 04.2.1
23 11
An Asymptotic Expansion for the
Catalan-Larcombe-French Sequence
Lane Clark
Department of Mathematics
Southern Illinois University, Carbondale
Carbondale, IL 62901–4408
USA
lclark@math.siu.edu
Abstract
We give an elementary development of a complete asymptotic expansion for the
Catalan-Larcombe-French sequence.
1
Introduction
In their delightful paper, Larcombe and French [3] developed a number of properties of the
sequence (A053175) P0 = 1, P1 = 8, P2 = 80, P3 = 896, P4 = 10816, . . . originally discussed
by Catalan [1]. In addition to a generating function, the following formula for P n was derived
µ ¶µ ¶
2q (2p)! (2q)!
1 X 2p
(n ∈ N) .
(1)
Pn =
q
n! p+q=n p
p! q!
¡ ¢2
Recently, Larcombe et al. [4] showed that Pn /2 2n
→ 1 as n → ∞ by a rather lengthy
n
analysis. In this short paper, we give an elementary development of a complete asymptotic
¡ ¢2
. We conclude with a table of numerical calculations as a companion
expansion for Pn /2 2n
n
of the theoretical results.
2
Main result
The positive integers are denoted by P; the nonnegative integers by N; the nonnegative
rational numbers by Q0 ; and the complex numbers by C. Let z 0 = (z)0 ≡ 1 and (z)p =
1
P
(z) · · · (z − p + 1) when z ∈ C and p ∈ P. For z ∈ C and p ∈ P, (z)p = pq=0 s(p, q)z q
where the s(p, q) are the Stirling numbers of the first kind (see [2; pp. 212–214]). We write
f (z) = O(g(z)) provided there exist real constants C, D with |f (z)| ≤ C|g(z)| for all |z| ≥ D.
For 0 ≤ p ≤ n, let
µ ¶2 µ
¶2
2p
2n − 2p
ap =
p! (n − p)! ∈ P
p
n−p
hence, ap = an−p and, let
hence, bp = bn−p
¶
µ ¶µ
2n − 2p
2p
∈P
bp =
n−p
p
¡ ¢−1
and ap = n! b2p np . For 0 ≤ p ≤ n − 1,
bp+1
(2p + 1)(n − p)
=
bp
(p + 1)(2n − 2p + 1)
then, for 0 ≤ p ≤ (n − 1)/2,
0<
bp+1
≤1
bp
hence, for 1 ≤ p ≤ (n − 1)/2
bp
b1
bp
=
···
≤1
b0
bp−1
b0
0<
(2)
which is correct for p = 0 also.
Fix s ≥ 1. For n = 2m + 1 ≥ 2s + 3, (1) and symmetry give
¶
n µ ¶µ
1 X 2p
2n − 2p (2p)! (2n − 2p)!
Pn =
n! p=0 p
p! (n − p)!
n−p
µ ¶−1
m
n
m
X
2 X
1 X
2 n
bp
ap =
ap = 2
=
p
n! p=0
n! p=0
p=0
=
Now
¡n¢
p
2b20
(
¶2 µ ¶−1
s µ
X
bp
n
p=0
b0
p
¶2 µ ¶−1 )
m µ
X
bp
n
+
.
p
b0
p=s+1
(3)
is an increasing sequence of positive integers for 0 ≤ p ≤ (n − 1)/2, so (2) gives
¶2 µ ¶−1
µ
¶−1
m µ
X
n
n
bp
≤n
0<
= O(n−s ) as n → ∞.
p
b
s
+
1
0
p=s+1
For 0 ≤ p ≤ (n − 1)/2,
bp
=
b0
µ ¶
2p (n)2p
p (2n)2p
2
(4)
hence,
¶2 µ ¶−1
s µ
X
n
bp
p=0
p
b0
s
X
µ ¶2
(n)3p
2p
=
p!
.
2
(2n)
p
2p
p=0
(5)
For z ∈ C, let f0 (z) ≡ 1 and, for 1 ≤ p ≤ s, let
¡ ¢2
¶−2
µ ¶2
p µ
p! 2p
(z)p Y
(z)3p
2j − 1
2p
p
1−
=
fp (z) = p!
p
(2z)22p
24p z 2p
2z
j=1
=
=
p!
¡2p¢2
p
24p
∞
X
p
X
s(p, q)z q−2p
q=0
b(p, r)z −r
(
̰
p
X
Y
j=1
(r + 1)(j − 0.5)r z −r
r=0
!)
(z ∈ C; |z| ≥ s)
(6)
r=p
where b(p, r) ∈ Q0 for r ≥ p and b(p, p) = p!
|(z)p | ≤ (|z| + s)p ≤
¡3
|z|
2
¢p
¡2p¢2
p
/24p . For 1 ≤ p ≤ s,
(|z| ≥ 2s)
|(2z)2p | ≥ (2|z| − 2s)2p ≥ |z|2p ,
hence,
¯
¯ (z)3p
¯
¯ (2z)2
2p
¯ µ ¶p
¯
¯≤ 4
¯
|z|
(|z| ≥ 2s).
(7)
For r, s ≥ p ≥ 1, Laurent’s Theorem (see [5; V. 2, p. 6]), standard estimates for the integral
and (7) give
¯
¯
I
¯ ¯ 1
¯
¯
f
(z)
p
¯b(p, r)¯ = ¯
¯
dz
¯ 2πi
¯
r+1
|z|=2s z
µ ¶2
µ ¶2 µ ¶p
2
1
2s
2p
.
≤
s!
≤ p!
s
p
s
(2s)r
(8)
Then (8) gives
hence, (6,9) give
¯∞
¯
µ ¶2 X
∞
¯X
¯
¡
¢
2s
¯
−r ¯
−s
b(p, r)z ¯ ≤ |z| s!
|z|−t = O |z|−s ,
¯
¯ r=s
¯
s
t=0
fp (z) =
s−1
X
r=p
¡
¢
b(p, r)z −r + O |z|−s
3
(9)
(10)
¢
¡
where fs (z) = O |z|−s . For s ≥ 1, (10) gives
gs (z) :=
s−1
X
(z ∈ C; |z| ≥ s)
fp (z)
p=0
=
s−1
X
c(s, r)z −r + O(|z|−s )
(11)
r=0
P
where c(s, 0) = 1 for s ≥ 1 and c(s, r) = rp=1 b(p, r) ∈ Q0 for 1 ≤ r ≤ s − 1. Observe
that c(s + 1, r) = c(s, r) for 0 ≤ r ≤ s − 1. The analysis for n = 2m ≥ 2s + 2 is identical
¡ n ¢−1
¡ n ¢−1
. Then (3–5,11) give the following complete
not 2b2m m
except that Pn includes b2m m
¡2n¢2
asymptotic expansion for Pn /2 n .
Theorem. Fix s ≥ 1. There exist effectively calculable nonnegative rational numbers
c(s, 0) = 1, c(s, 1), . . . , c(s, s − 1) so that
µ
2n
Pn /2
n
¶2
=
s−1
X
c(s, r)n−r + O(n−s ) as n → ∞.
¥
r=0
Let
hs (n) =
s−1
X
c(s, r)n−r
r=0
hence,
1
,
4n
h1 (n) = 1,
h2 (n) = 1 +
h4 (n) = 1 +
17
207
1
+
+
2
4n
32n
128n3
h3 (n) = 1 +
1
17
+
,
4n
32n2
and
1
207
14875
17
+
+
.
+
2
3
4n
32n
128n
2048n4
¤
£
¡ ¢2
− hs (n) ns . Our theorem shows Qs (n) = O(1) as n → ∞. The table
Let Qs (n) = Pn /2 2n
n
below which gives the first 10 digits of Q2 (n) and Q3 (n) for several values of n was found
using Mathematica. This provides numerical evidence for our theorem.
h5 (n) = 1 +
4
n
Q2 (n)
100 .5481946735
200 .5395231003
300 .5367229603
400 .5353390483
500 .5345137769
600 .5339656896
700 .5335752174
800 .5332829178
900 .5330559013
1000 .5328744940
2000 .5320604149
3000 .5317898711
Q3 (n)
1.6944673581
1.6546200668
1.6418880975
1.6356193435
1.6318884961
1.6294137740
1.6276521934
1.6263343131
1.6253112456
1.6244940166
1.6208298850
1.6196133523
Acknowledgement. I wish to thank the referee for comments and suggestions which have
resulted in an improved version of the paper.
References
1. E. Catalan, Sur les Nombres de Segner, Rend. Circ. Mat. Pal. 1 (1887), 190–201.
2. L. Comtet, Advanced Combinatorics, D. Reidel, Boston, MA, 1974.
3. P.J. Larcombe and D.R. French, On the ‘Other’ Catalan Numbers: A Historical Formulation Re-examined, Congr. Numer. 143 (2000), 33–64.
4. P.J. Larcombe, D.R. French and E.J. Fennessey, The Asymptotic Behaviour of the
Catalan-Larcombe-French Sequence {1, 8, 80, 896, 10816, . . .}, Util. Math. 60 (2001),
67–77.
5. A.I. Markushevich, Theory of Functions of a Complex Variable, Chelsea Publishing
Co., New York, NY, 1985.
2000 Mathematics Subject Classification: Primary 05A10; Secondary 05A16, 41A60.
Keywords: asymptotic expansion, Catalan-Larcombe-French sequence.
(Concerned with sequence A053175.)
Received October 28 2002; revised version received February 2 2004. Published in Journal
of Integer Sequences, March 23 2004.
Return to Journal of Integer Sequences home page.
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