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Journal of Integer Sequences, Vol. 16 (2013),
Article 13.3.1
23 11
On a Conjecture of Andrica and Tomescu
Blair D. Sullivan1
Oak Ridge National Laboratory
1 Bethel Valley Road
Oak Ridge, TN 37831
USA
sullivanb@ornl.gov
Abstract
For positive integers n congruent to 0 or 3 (mod 4), let S(n) be the coefficient
of xn(n+1)/4 in the expansion of (1 + x)(1 + x2 ) · · · (1 + xn )q
We prove a conjecture of
Andrica and Tomescu that S(n) is asymptotically equal to
1
6
π
· 2n · n−3/2 .
Introduction
For positive integers n congruent to 0 or 3 (mod 4), let S(n) denote the coefficient of the
middle term of the expansion of the polynomial (1 + x)(1 + x2 ) · · · (1 + xn ). (In the case
n ≡ 1, 2 (mod 4), the quantity n(n + 1)/4 is not an integer, and the expansion has no
middle term.) This number also represents the number of partitions of Tn /2 = n(n + 1)/4
into distinct parts less than or equal to n, where Tn is the nth triangular number. Andrica
and
q Tomescu conjectured that as n approaches infinity, S(n) behaves asymptotically like
6
π
· 2n · n−3/2 . More formally, writing f (n) ∼ g(n) as usual to denote
lim
n→∞
f (n)
= 1,
g(n)
we have
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The submitted manuscript has been authored by a contractor of the U.S. Government under Contract
No. DE-AC05-00OR22725. Accordingly, the U.S. Government retains a non-exclusive, royalty-free license to
publish or reproduce the published form of this contribution, or allow others to do so, for U.S. Government
purposes.
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Conjecture 1. [Andrica, Tomescu [1]] S(n) ∼
q
6
π
· 2n · n−3/2 for n ≡ 0 or 3 (mod 4).
From [1], one can write S(n) in integral form via Cauchy’s formula as
Z
2n−1 2π
cos(t) cos(2t) · · · cos(nt) dt.
S(n) =
π 0
We will
to estimate this integral [2]. Rewriting, we have S(n) =
R use the Laplace method
Qn
2n−1 2π
fn (t) dt where fn (t) = k=1 cos(kt). In Section 2, we analyze the behavior of fn (t)
π
0
and note a technical lemma needed for the main proof of Conjecture 1, which is presented
in Section 3. We note that a similar approach was suggested in a review of [1] by Hwang [3]
published in Mathematical Reviews.
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Behavior of fn(t)
Lemma 2. Let 0 < ε < 1/4, and fn (t) =
o(n−3/2 ) as n → ∞.
Qn
k=1
cos(kt). Then
R
n−(3/2−ε) <|t|<π/2
|fn (t)| dt =
Proof. We break the integral into three pieces based on the value of |t|.
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Case 1. n−( 2 −ε) ≤ |t| ≤ n1 :
Since cos(x) = cos(−x), and cos is a monotone decreasing function on [0, π], fn (t) =
fn (−t) is also monotone decreasing for t ∈ [0, 1/n], and it suffices to give an appropriate
upper bound on fnR(n−(3/2−ε) ).
Since we need n−(3/2−ε) <|t|<π/2 fn (t) dt = o(n−3/2 ), given that 0 < ε < 1/4, it suffices to
show that for a constant c > 0,
fn (n−(3/2−ε) ) ≤ exp(−cn2ε (1 + o(1))).
2
4
Using the Taylor series expansion, we know cos(kt) ≤ 1 − (kt)
+ (kt)
. Substitution then
2!
4!
yields
n
n Y
Y
(kt)2 (kt)4
1−
,
fn (t) =
cos(kt) ≤
+
2!
4!
k=1
k=1
since k ≤ n and |t| ≤ 1/n implies kt ≤ 1. When t = n−(3/2−ε) , we have
fn (t) ≤
n Y
k=1
k 2 n−(3−2ε) k 4 n−(6−4ε)
1−
+
2
24
.
To evaluate, we note the terms of this product are all in [0, 1], and apply log(1 − x) ≤ −x:
log
n Y
k=1
1−
k 2 n−(3−2ε) k 4 n−(6−4ε)
−
2
24
2
n X
k 2 n−(3−2ε) k 4 n−(6−4ε)
−
≤
.
+
2
24
k=1
Writing
Pn
P
k 2 = (1/3 + o(1))n3 and nk=1 k 4 = (1/5 + o(1))n5 , we have
1
1
3 −(3−2ε)
+ o(1) n n
+
+ o(1) n5 n−(6−4ε) .
log fn = −
6
120
k=1
Letting c = 1/6, fn ≤ exp(−c(1 + o(1))n2ε + c(1 + o(1))n−1+4ε ), and recalling ε < 1/4,
fn ≤ exp(−c(1 + o(1))n2ε ) as desired.
Case 2. n1 ≤ |t| ≤ πn :
Qn Here we use will the monotonicity of fn (t) in n. It follows directly from fn (t) =
k=1 cos(kt) and 0 ≤ cos(x) ≤ 1 that |fn (t)| ≤ |fm (t)| for n ≥ m. Let hn = ⌊n/4⌋ be the
greatest integer in n/4. Then |fn (t)| ≤ |fhn (t)|. From Case 1, fhn (t) ≤ exp(−ch2ε
n (1 + o(1)))
−( 3 −ε)
−( 3 −ε)
−5/4
for hn 2
≤ |t| ≤ 1/hn . Since 1/n > hn
≥ hn 2
for n > 1050 and hn ≤ n/4 ≤ n/π
2ε
implies π/n ≤ 1/hn , we get |fn (t)| ≤ exp(−chn (1 + o(1))) for t ≤ π/n as n → ∞.
Case 3. πn ≤ |t| ≤ π2 :
Note that it suffices to show that |fn (t)| ≤ cn for a constant c < 1, since then
Z
|fn (t)| dt ≤ π · cn = o(n−3/2 ).
π/n≤|t|<π/2
To accomplish this, we first transform fn (t) from a product to a sum using the arithmeticgeometric mean inequality:
!1/n
n
n
Y
1X 2
2
2
1/n
≤
cos (kt).
(1)
cos (kt)
(fn (t)) =
n
k=1
k=1
The sum on the right-hand side can be simplified as
n
X
n
n 1X
n cos((n + 1)t) sin(nt)
cos (kt) = +
.
cos(2kt) = +
2 2 k=1
2
2
sin(t)
k=1
2
Combining equations 1 and 2, we can write
n/2
1 1
1
+
.
|fn (t)| ≤
2 2n sin(t)
(2)
(3)
for 0 ≤ |t| ≤ π/2. For
We will now apply the Jordan-style concavity inequality | sin(t)| ≥ 2|t|
π
π/n ≤ |t| ≤ π/2, substitution in equation 3 gives
n/2 n/2
1
1 π
π
1
|fn (t)| ≤
=
.
+
+
2 2n 2|t|
2 4n|t|
Observing that the right-hand side is monotonically decreasing in |t|, we have |fn (t)| ≤
fn (π/n). Evaluating, we see
n/2
1
1
−
|fn (t)| ≤
2 2n
p
proving |fn (t)| ≤ ( 7/16)n (since we may assume 2n ≥ 16 as n → ∞).
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We will also need the following straightforward lemma from analysis.
Lemma 3. Let c ∈ R and a(c), b(c) be real-valued functions such that
√
√
lim −a(c) c = lim b(c) c = ∞.
c→∞
c→∞
Then
Z
as c → ∞.
3
b(c)
e
−ct2
a(c)
dt ∼
Z
∞
2
e−ct dt
−∞
Main Result
We now prove Conjecture 1.
Theorem 4. For n ≡ 0 or 3 (mod 4), we have S(n) ∼
q
6
π
· 2n · n−3/2 .
Proof. When n ≡ 0 or 3 (mod 4), fn (t + mπ) = fn (t) for any integer m, so
2 · 2n−1
S(n) =
π
Z
π
2
− π2
fn (t) dt,
(4)
and we may assume
n (t).
R |t| ≤ π/2 when evaluating f−3/2
), so it suffices to consider |t| < n−(3/2−ε)
By Lemma 2, n−(3/2−ε) <|t|<π/2 |fn (t)| dt = o(n
when estimating fn (t) around t = 0. Recalling
fn (t) =
n
Y
eln(cos(kt)) ,
k=1
we first use Taylor series to approximate gk (t) = ln(cos(kt)) at t = 0. We have gk (t) =
−k 2 t2 /2 + R2 , where R2 is the Lagrange remainder. Then R2 is bounded by a constant times
(3)
(3)
t3 gk (t0 ) for some t0 near 0. Since gk (t) = −2k 3 sin(kt)/ cos3 (kt), and t0 is small (since
|t| < n−(3/2−ε) ), we have that R2 ≤ ak 3 t3 where a is constant. The absolute error for gk (t) is
thus bounded by ak 3 n−(9/2−3ε) .
Q
Q
k 2 t2
3 −(9/2−3ε)
Around t = 0, fn (t) can be approximated as δ nk=1 e− 2 with error δ ≤ nk=1 eak n
.
This simplifies to
Pn
2
2
2
fn (t) ≈ e−t /2 k=1 k = e−t n(n+1)(2n+1)/12 .
(5)
Our error bound simultaneously simplifies to
δ ≤ ean
−(9/2−3ε)
Pn
k=1
k3
= ean
−(9/2−3ε) n2 (n+1)2 /4
.
This proves that the error goes to one as n approaches infinity whenever ε < 16 .
Substituting (5) for fn (t) in equation 4, and applying Lemma 2, we find that
4
πS(n)
= (1 + o(1))
2n
By Lemma 3, this implies
Z
n−(3/2−ε)
e−n(n+1)(2n+1)t
2 /12
dt + o(n−3/2 ).
−n−(3/2−ε)
πS(n)
= (1 + o(1))
2n
Using
Z
Z
∞
e−n(n+1)(2n+1)t
2 /12
dt + o(n−3/2 ).
−∞
∞
e
−Ct2
dt =
−∞
r
π
C
for any constant C > 0 and n(n + 1)(2n + 1) ∼ 2n3 , we have
r
r
6 n −3/2
2n 12π
=
·2 ·n
,
S(n) ∼
π 2n3
π
as desired.
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Acknowledgments
This research was partially completed while the author was an intern at Microsoft Research,
Redmond, WA. We thank Henry Cohn for suggesting the problem and providing comments
which greatly improved the manuscript. Special thanks are due to Xavier Gourdon, who
pointed out a flaw in an earlier version of the paper, and suggested the arithmetic-geometric
mean inequality approach now used in the third case of Lemma 2.
References
[1] D. Andrica and I. Tomescu, On an integer sequence related to a product of trigonometric
functions, and its combinatorial relevance, J. Integer Sequences 5 (2002), Article 02.2.4.
[2] N. G. de Bruijn, Asymptotic Methods in Analysis, Dover, 1981.
[3] H.-K. Hwang, Review of [1], Math. Reviews, MR1938223 (2003j:05005).
2000 Mathematics Subject Classification: Primary 05A16. Secondary 11B75, 05A15, 05A18.
Keywords: triangular number, generating function, partition, asymptotic evaluation.
(Concerned with sequence A025591.)
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Received September 18 2008; revised versions received September 14 2012; February 14 2013.
Published in Journal of Integer Sequences, March 2 2013.
Return to Journal of Integer Sequences home page.
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