1
2
3
47
6
Journal of Integer Sequences, Vol. 10 (2007),
Article 07.1.3
23 11
Bounds for the Eventual Positivity
of Difference Functions of Partitions
into Prime Powers
Roger Woodford
Department of Mathematics
University of British Columbia
Vancouver, BC V6T 1Z2
Canada
rogerw@math.ubc.ca
Abstract
In this paper we specialize work done by Bateman and Erdős concerning difference
functions of partition functions. In particular we are concerned with partitions into
fixed powers of the primes. We show that any difference function of these partition
functions is eventually increasing, and derive explicit bounds for when it will attain
strictly positive values. From these bounds an asymptotic result is derived.
1
Introduction
Given an underlying set A ⊆ N, we denote the number of partitions of n with parts taken
(k)
from A by pA (n). The k-th difference function pA (n) is defined inductively as follows: for
(k)
k = 0, pA (n) = pA (n). If k > 0, then
(k)
(k−1)
pA (n) = pA
(k)
(k−1)
(n) − pA
(k)
(n − 1).
Let fA (x) be the generating function for pA (n). We have [5] the following power series
identity:
1
(k)
fA (x)
=
∞
X
(k)
pA (n)xn
(1)
n=0
= (1 − x)k
= (1 − x)k
∞
X
pA (n)xn
(2)
n=0
1
.
a
1
−
x
a∈A
Y
(3)
(k)
This may be used to define pA (n) for all k ∈ Z, including k < 0.
(k)
Bateman and Erdős [5] characterize the sets A for which pA (n) is ultimately nonnegative.
Note that if k < 0, then the power series representation of (1−x)k has nonnegative coefficients
(k)
so that pA (n) ≥ 0. For k ≥ 0, they prove the following: if A satisfies the property that
whenever k elements are removed from it, the remaining elements have greatest common
divisor 1, then
(k)
lim p (n)
n→∞ A
= ∞.
A simple consequence of this is the fact that pA (n) is eventually monotonic if k > 0.
(k)
No explicit bounds for when pA (n) becomes positive are included with the result of
Bateman and Erdős [5]. By following their approach but specializing to the case when
A = {pℓ : p is prime},
(4)
for some fixed ℓ ∈ N, we shall find bounds for n depending on k and ℓ which guarantee that
(k)
pA (n) > 0. For the remainder, A shall be as in (4), with ℓ ∈ N fixed.
In a subsequent paper, Bateman and Erdős [6] prove that in the special case when ℓ = 1,
(1)
pA (n) ≥ 0 for all n ≥ 2. That is, the sequence A000607 of partitions of n into primes is
increasing for n ≥ 1. Our result pertains to more general underlying sets; partitions into
squares of primes (A090677), cubes of primes, etc.
In a series of papers (cf. [10]- [13]), L. B. Richmond studies the asymptotic behaviour for
partition functions and their differences for sets satisfying certain stronger conditions. The
results none-the-less apply to the cases of interest to us, that is, where A is defined as above.
(k)
Richmond proves [12] an asymptotic formula for pA (n). Unfortunately, his formula is not
(k)
useful towards finding bounds for when pA (n) must be positive, since, as is customary, he
does not include explicit constants in the error term.
Furthermore, his asymptotic formula includes functions such as α = α(n) defined by
n=
X
a∈A
a
k
− α
.
−1 e −1
eαa
As we are seeking explicit constants, a direct approach will be cleaner than than attempting
to adapt the aforementioned formula.
2
Another result worthy of comment from Richmond [12] pertains to a conjecture of Bateman and Erdős [5]. His result applies to A as defined in (4), and states that
(k+1)
pA
(n)
(k)
pA (n)
= O(n−1/2 ), as n → ∞.
We omit the subscript and write p(k) (n) in case the underlying set is A, and omit the
superscript if k = 0. The letter B will always be used to denote a finite subset of A. We
shall also write ζn for the primitive n-th root of unity e2πi/n . The letters ζ and η shall always
denote roots of unity. We shall denote the m-th prime by pm .
Our main results are Theorems 1.1 and 1.2. The former is established in Sections 2 and 3.
q
2
Theorem 1.1. Let k be a nonnegative integer, let b0 = 2π 1 − π12 and let
%
$ µ ¶
3(k+1)
2
(k + 2)8ℓk+10ℓ+3 .
t= 6
b0
Then there are positive absolute constants a1 , . . . , a7 such that if
!t−1
Ã
2
¢t−1
¡
a2 aℓ3 log t
N = N (k, ℓ) = a1 t
,
+ a5 t3 a6 (a7 t)6ℓ
(k+3)ℓ
a4
then n ≥ N implies that p(k) (n) > 0.
Remark 1. The values of the constants are approximately
a1
a2
a3
a4
a5
a6
a7
≈ 1.000148266,
≈ 2757234.845,
≈ 1424.848799,
≈ 2.166322546,
≈ 1.082709333,
≈ .0193095561,
≈ 2.078207555.
For the remainder of the paper, b0 shall be as defined in Theorem 1.1. Note that b0 ≈
2.6474. Furthermore, Define
F (k, ℓ) = min {N ∈ N : n ≥ N implies that p(k) (n) > 0}.
We show in Section 4 that Theorem 1.1 yields the following asymptotic result:
Theorem 1.2. Fix ℓ ∈ N. Then as k → ∞,
¢
¡
log F (k, ℓ) = o (k + 2)8ℓk .
Following Bateman and Erdős [5], we first tackle the finite case.
3
Finite subsets of A
2
(k)
Lemma 2.1. Let B be a finite subset of A of size r, and suppose k < r. The function pB (n)
can be decomposed as follows:
(k)
(k)
(k)
pB (n) = gB (n) + ψB (n),
(k)
where gB (n) is a polynomial in n
(k)
and ψB (n) is periodic in n with
³
´−1
Q
of degree r−k−1 with leading coefficient (r − k − 1)! q∈B q
,
Q
period q∈B q.
(k)
Proof. We use partial fractions to decompose the generating function fB (x) as follows:
(k)
fB (x) =
q−1
YY
1
1
(1 − x)r−k q∈B j=1 1 − ζqj x
(5)
q−1
X X β(ζqj )
α2
αr−k
α1
+
+
.
.
.
+
+
=
,
1 − x (1 − x)2
(1 − x)r−k q∈B j=1 1 − ζqj x
(6)
where the αi , and β(ζqj ) are complex numbers that can be determined. Note that
αr−k =
Ã
Y
q
q∈B
!−1
.
The power series expansion for (1 − x)−h is given by
Hence, if
¶
∞ µ
X
n+h−1 n
1
=
x .
(1 − x)h
h−1
n=0
(k)
gB (n)
=
r−k
X
αh
h=1
and
(k)
ψB (n)
=
µ
¶
n+h−1
,
h−1
q−1
XX
β(ζqj )ζqjn ,
q∈B j=1
then the lemma is proved.
(k)
(k)
For the remainder of this paper, gB (n) and ψB (n) shall be as in Lemma 2.1, for a given
finite set B ⊆ A which shall be clear from the context.
Remark 2. Let B be as in Lemma 2.1. We wish to know the precise value of β(ζqj ). To
simplify notation a little, we will frequently write βζ instead, when ζ is clear from the context.
4
In particular, suppose η = ζqj , where q ∈ B, and 0 < j < q. Then
(k)
(1 − ηx)fB (x) = (1 − x)k
Y 1
1
1 + ηx + · · · + (ηx)q−1 p∈B 1 − xp
p6=q
= βη + (1 − ηx)
Ã
X βζ
α1
αr−k
+ ··· +
+
1−x
(1 − x)r−k ζ6=η 1 − ζx
!
,
hence,
βη =
(1 − η̄)k Y 1
.
p
q
1
−
η̄
p∈B
p6=q
We shall frequently make use of the inequality
θ2 θ4
θ2
≤ cos θ ≤ 1 −
+ ,
2
2
24
which holds for all values of θ. Note that
p
|eiθ − 1| = 2(1 − cos θ),
√
√
so for −2 3 ≤ θ ≤ 2 3,
r
θ2
≤ |eiθ − 1| ≤ |θ|.
|θ| 1 −
12
In particular, for 0 ≤ θ ≤ π,
r
θ2
θ 1−
≤ |eiθ − 1| ≤ θ.
12
1−
Lemma 2.2. Suppose that ζ 6= 1 is a q-th root of unity, q ≥ 2. Then
b0
≤ |1 − ζ| ≤ 2.
q
Proof. Clearly |1 − ζ| ≤ 2. On the other hand, by equation (7),
|1 − ζ| ≥ |1 − e2πi/q |
s
2π
4π 2
≥
1−
q
12q 2
b0
≥ .
q
5
(7)
Lemma 2.3. Suppose that k < r, and B ⊆ A, satisfies |B| = r. Suppose further that η = ζqj ,
for some q ∈ B, j ∈ {1, . . . , q − 1}. Then for k ≥ 0,
 k r−2
q
 2 r−1
, if k ≥ 0;
b0
|βη | ≤ qr−k−2
 r−k−1 , if k < 0.
b
0
Proof. Making use of Remark 2 and Lemma 2.2 we have that for k ≥ 0:
|βη | =
|1 − ζq−j |k Y
1
q
|1 − ζq−jp |
p∈B
p6=q
k
2 Y
1
≤
q p∈B |1 − ζq |
p6=q
¶r−1
q
2
≤
q b0
2k q r−2
= r−1 .
b0
k
µ
A similar arguments works for the case when k < 0.
Theorem 2.1. Suppose that k < r, and B ⊆ A, satisfies |B| = r. Then

P
2k
r−1
 r−1
,
if k ≥ 0;
q∈B q
(k)
b0
|ψB (n)| ≤
P
1
r−k−1
 r−k−1
, if k < 0.
q∈B q
b
0
Proof. First assume that k ≥ 0. By Lemmas 2.3 and 2.1,
¯
¯
q−1
¯
¯X X
¯
(k)
j jn ¯
|ψB (n)| = ¯
β(ζq )ζq ¯
¯
¯
q∈B j=1
≤
≤
=
≤
q−1
XX
q∈B j=1
|β(ζqj )|
q−1 k r−2
XX
2 q
q∈B j=1
br−1
0
2k X
(q − 1)q r−2
br−1
0
q∈B
2k X r−1
q .
br−1
0
q∈B
A similar argument works for k < 0.
6
To obtain bounds for the coefficients αh , we will use Laurent series.
Lemma 2.4. Suppose that k < r, and B ⊆ A, satisfies |B| = r. Denote the largest element
of B by Q, and suppose further that 0 < r0 < |1 − ζQ |. Let
dB (r0 ) =
q−1
YY
q∈B j=1
Then
|αh | ≤
(|ζqj − 1| − r0 ).
1
r0r−k−h dB (r0 )
.
Proof. Let γ be the circle |z−1| = r0 . From equations (5) and (6), and the Laurent expansion
theorem, we have that
¯
¯
Z
¯ 1
¯
(k)
h−1
¯
|αh | = ¯
fB (z)(z − 1)
dz ¯¯
2πi γ
Z
q−1
YY
1
1
1
dz
≤
r−k−h+1
2π γ |z − 1|
|1 − ζqj z|
q∈B j=1
≤
1
r0r−k−h dB (r0 )
.
For the following proposition, we define several new constants:
¶
∞ µ
Y
π2
,
c1 =
1−
k
3
·
4
k=3
¶ 1k
∞ µ
Y
2
π2
1−
c2 =
,
k
3
·
4
k=3
c3 = 4π −6 c1 ,
2 log π
− 1,
c4 =
log 2
³ π ´1/2
c5 =
c2 ,
2
¶1/4
³ π ´1/2 µ
π2
1−
,
c6 =
2
48
sin 4π/5
,
c7 =
4π/5
b1 = c5 c6 = 1.471843248 . . . ,
c3
b2 =
= .003278645140 . . . ,
c6
b3 = c4 = 2.302992260 . . . ,
b4 = c7 π/2 = .3673657828 . . . .
7
Proposition 2.1. Let B ⊆ A satisfy |B| = r, and suppose that all the elements of B are
odd. Let Q = max{B}, and T = ⌈log2 Q⌉. If
⌈q⌉
r0 = min {|ζq 2j − 1| − |ζ2j − 1| : q ∈ B, j = 2, . . . , T },
then
0<
b4
2π
≤
r
<
|1
−
ζ
|
≤
.
0
Q
Q2
Q
Furthermore, if dB = dB (r0 ), then
P
dB ≥ b1
q∈B
q
µ
2
log Q
b3 − log 2
b2 Q e
¶r
.
Proof. Observe that T satisfies
2T −1 < Q < 2T .
Now, let
q−1
Y
(|ζqj − 1| − r0 ),
δq =
j=1
so that
dB =
Y
δq .
q∈B
Note that since each element of B is odd, we have that
q−1
δq =
2
Y
j=1
(|ζqj − 1| − r0 )2 .
Consequently,
δq ≥
T
Y
k=2 ⌈
Y
|ζ2k − 1|2
q
⌉≤j<⌈ 2k−1
⌉
q
µ
µ
¶¶⌈ k−1
T
⌉−⌈ 2qk ⌉
Y
2
π2
π2
≥
1−
4k−1
3 · 4k
k=2
q
µ
¶¶⌈ k−1
µ 2µ
¶¶⌈ 2q ⌉−⌈ 4q ⌉ Y
T µ
⌉−⌈ 2qk ⌉
2
π2
π2
π
π2
1−
=
1−
×
k−1
4
48
4
3 · 4k
k=3
¶¶ q−1
µ
¶¶ qk +1
µ 2µ
T µ
Y
4
2
π2
π2
π
π2
1−
1
−
≥
.
×
k−1
k
4
48
4
3
·
4
k=3
q
2k
Now
8
à 1 1 !q
µ
¶¶ qk +1
T µ
Y
2
π2
π 2(T −2) π 2( 4 − 2T )
π2
c1 cq2
≥ T
1−
T +1
3
k−1
k
−
−1
(
)
4
3
·
4
4 4 2T
4 2
k=3
à 1 1 !q
2 log π
−4
log
2
4π Q
π 2( 4 − Q )
≥
c1 cq2
1
log2 Q
44
Qe log 2
µ 1/2 ¶q
log2 Q
2 log π
π c2
−1
−
−4
≥ 4π c1 Q log 2 e log 2
π −2
21/2
= c3 Qc4 e−
log2 Q
log 2
cq5
So we have that
Y µµ c3 ¶
q
(c5 c6 ) Q e
c6
¶r
µ
log2 Q
q∈B q
b3 − log 2
b2 Q e
= b1
dB ≥
q∈B
P
2
log Q
c4 − log 2
¶
Our next task is to bound r0 from below. For q ∈ B, j ∈ {2, . . . , T }, let
p
f (x) = 2(1 − cos x),
2π
θ = θ(j) = j , and
2
2π l q m 2π
− j.
ε = ε(q, j) =
q 2j
2
Then by the mean value theorem,
⌈q⌉
|ζq 2j − 1| − |ζ2j − 1| = f (θ + ε) − f (θ)
ε sin c
=p
,
2(1 − cos c)
for some c ∈ (θ, θ + ε).
It is easily seen that 2π⌈q/2j ⌉/q can be no greater than 4π/5. On the interval [0, 4π/5],
we have sin x ≥ c7 x. Therefore
εc7 c
= εc7 .
(8)
c
Choose a ∈ N such that (a − 1)2j + 1 ≤ q ≤ a2j − 1. Then clearly a ≤ 2T −j . Furthermore,
f (θ + ε) − f (θ) ≥
9
ε≥
=
≥
≥
≥
2πa
2π
− j
j
a2 − 1
2
2π
2π
− j
j
2 − 1/a
2
2π
2π
− j
j
j−T
2 −2
2
2π
2π
−
2T − 1 2T
π
.
2Q2
Hence, by (8) we conclude that
b4
.
Q2
Bounding r0 from above is a far simpler matter. By definition,
r0 ≥
r0 < |ζQ − 1| ≤
3
2π
.
Q
(9)
Infinite subsets of N and A
(k)
(k)
Proposition 3.1. For k ≤ 0, and D1 ⊆ D2 ⊆ N, we have pD2 (n) ≥ pD1 (n) ≥ 0.
Proof. This follows immediately from equation 2 and the fact that for k ≤ 0, the power
series expansion for (1 − x)k has nonnegative coefficients.
For the sake of clarity and the comprehensiveness of this exposition we include the following theorem of Bateman and Erdős [5] suitably adapted to our needs.
Theorem 3.1. Let D ⊆ N be an infinite set. For any t ∈ N, we have that
pD (n)
(−1)
pD (n)
≤
(t − 1)2 n2t−3
1
+
.
t+1
t + 1 p(−1)
(n)
D
Proof. Denote by Pq (n), the number of partitions of n into parts from D such that there are
exactly q distinct parts. Pq (n) has generating function
∞
X
Pq (n)xn =
n=0
X
xaq
xa1
.
·
·
·
1 − xa 1
1 − xaq
X
xaq
xa1
,
·
·
·
1 − xa 1
1 − xaq
{a1 ,...aq }⊆D
If Rq (n) is defined by
∞
X
n=0
Rq (n)xn =
{a1 ,...aq }⊆[n]
10
where [n] = {1,
¡ .¢. . , n}, then Pq (n) ≤ Rq (n).
There are nq subsets of [n] of size q. Also, the coefficient of xn in
(xa1 + x2a1 + · · · ) · · · (xaq + x2aq + · · · )
is less than or equal to the coefficient of xn in
2
q
(x + x + · · · ) =
so
¶
∞ µ
X
m−1
m=q
q−1
xm ,
¶
µ ¶µ
n n−1
≤ n2q−1 .
Pq (n) ≤
q−1
q
Any partition n = n1 a1 + · · · nq aq , where a1 , . . . , aq ∈ A, gives rise to a partition of n − ai
for i = 1, . . . , q, namely
n − a1 = (n1 − 1)a1 + n2 a2 + · · · + nq aq ,
n − a2 = n1 a1 + (n2 − 1)a2 + · · · + nq aq ,
..
.
n − aq = n1 a1 + n2 a2 + · · · + (nq − 1)aq .
Note that no two distinct partitions of n can give rise to the same partition of any m < n in
this way, and so
n
n−1
X
X
qPq (n) ≤
pD (m).
q=1
m=0
Now if t ∈ N, then
(−1)
pD (n)
=
n
X
pD (m)
m=0
≥ pD (n) +
n
X
qPq (n)
q=1
= (t + 1)pD (n) +
n
X
q=1
(q − t)Pq (n)
≥ (t + 1)pD (n) − (t − 1)
t−1
X
Pq (n)
q=1
2 2t−3
≥ (t + 1)pD (n) − (t − 1) n
,
and the theorem is proved.
For the remainder of this section, we follow the approach of Bateman and Erdős [5]
and simultaneously make the results explicit by applying them to the special case under
consideration. To be consistent, we shall assume k ≥ 0, and use the following notation:
11
Notation 1. Let B be the least k+2 elements of A, and let C = A\B. B1 = {pℓk+3 , pℓk+4 , . . . , pℓk+2t }
be the least 2t − 2 elements of C, where t is determined as in the statement of Theorem 1.1
from the values of k and ℓ in question. Furthermore, let
g=
µ
2
b0
¶k+1
(k + 2)2ℓ(k+1)+1 , and
2ℓ(k+2)
h = 3(k + 2)
g=3
µ
2
b0
¶k+1
(10)
(k + 2)4ℓk+6ℓ+1 .
Finally, let us remark that numerous constants shall be defined in the subsequent argument.
Their definitions shall remain consistent throughout.
Note that the right hand side of (10) is increasing in k and ℓ, so g ≥ 16/b0 = 6.04 . . ..
Lemma 3.1. For all n ≥ 0, we have
(k)
pB (n) ≥ 1 − g.
(k)
(11)
(k+1)
Proof. Since |B| = k + 2, gB (n) is linear in n, and gB
Lemma 2.1,
(k)
gB (n)
(n) is a constant function. By
2
X
¶
µ
n+h−1
=
αh
h−1
h=1
where α2 = (p1 · · · pk+2 )−ℓ . Substituting x = 0 into (5) and (6) gives
α1 + α2 +
q−1
XX
β(ζqj ) = 1.
q∈B j=1
Hence,
(k)
gB (n)
−ℓ
= (p1 · · · pk+2 ) n + 1 −
so
12
q−1
XX
q∈B j=1
β(ζqj ),
(k)
(k)
(k)
pB (n) = gB (n) + ψB (n)
= (p1 · · · pk+2 )−ℓ n + 1 −
−ℓ
≥ (p1 · · · pk+2 ) n + 1 −
q−1
XX
(1 − ζqjn )β(ζqj )
q∈B j=1
q−1
XX
q∈B j=1
−ℓ
≥ (p1 · · · pk+2 ) n + 1 − 2
−ℓ
≥ (p1 · · · pk+2 ) n + 1 −
≥1−
Now, it is easy to see that
X
q∈B
µ
2
b0
¶k+1 X
|1 − ζqjn ||β(ζqj )|
q−1
XX
q∈B j=1
µ
2
b0
|β(ζqj )|
¶k+1 X
q k+1
q∈B
q k+1 .
q∈B
2ℓ(k+1)+1
q k+1 ≤ Qk+1
,
0 (k + 2) ≤ (k + 2)
where Q0 = max (B). From this, the Lemma follows.
Lemma 3.2. For all n ≥ 0, we have
(k+1)
1 + |pB
(n)| < g − 1.
(12)
Proof. Note that
(k+1)
gB
(n) = (p1 · · · pk+2 )−ℓ .
Making use of Theorem 2.1, we see that
(k+1)
g − 2 − |pB
(k+1)
(n)| ≥g − 2 − (p1 · · · pk+2 )−ℓ − |ψB
k
(n)|
X
q k+1
2
bk+1
0
q∈B
õ
µ
¶
¶k+1 !
k+2
k+1
X
2(k + 2)2ℓ
1 2pℓi
≥
−
b
2 b0
0
i=1
≥g − 2 − (p1 · · · pk+2 )−ℓ −
− 2 − (p1 · · · pk+2 )−1 .
Observe that for a fixed i, 1 ≤ i ≤ k + 2, the expression
µ
2(k + 2)2ℓ
b0
¶k+1
1
−
2
13
µ
2pℓi
b0
¶k+1
,
is positive and increasing in ℓ, for ℓ ≥ 1, so,
µ
2(k + 2)2ℓ
b0
¶k+1
1
−
2
µ
2pℓi
b0
¶k+1
≥
≥
≥
≥
≥
µ
2(k + 2)2
b0
¶k+1
1
−
2
µ
2pi
b0
¶k+1
µ
¶k+1
¶k+1
1 2pk+2
2(k + 2)2
−
b0
2
b0
2
2(k + 2)
pk+2
−
b0
b0
2
4
(k + 2) − pk+2
+
b0
b0
4
.
b0
µ
Hence,
(k+1)
g − 2 − |pB
4
− 2 − (p1 · · · pk+2 )−1
b0
8
13
≥
−
= 0.855167405 . . . > 0,
b0
6
(n)| ≥ (k + 2)
that is,
(k+1)
1 + |pB
(n)| < g − 1.
Corollary 3.1.
(k)
pB (n)
(k+1)
1 + |pB
(n)|
≥ 2 if n ≥ h.
(13)
Proof. It follows from the proof of Lemma 3.1 that
(k)
pB (n) ≥ (p1 · · · pk+2 )−ℓ n + 1 − g
> (p1 · · · pk+2 )−ℓ n − g.
The Corollary follows from this, together with Lemma 3.2 and the fact that
p1 · · · pk+2 ≤ (k + 2)2(k+2) .
Lemma 3.3. There is an h1 ∈ N such that such that n ≥ h1 implies
(t − 1)2 n2t−3
1
(t − 1)2 n2t−3
.
≤
≤
(−1)
t + 1 p(−1)
t
+
1
t
+
1
(n)
p
(n)
C
B
1
14
(14)
Proof. The first inequality is a consequence of Proposition 3.1. Note that
t≥6
≥6
µ
µ
2
b0
2
b0
¶3(k+1)
¶3k+3
211
b30
3 · 216
≥
b30
µ
log t ≥ log
µ
Hence
(k + 2)8ℓk+10ℓ+3 − 1
28k+13 − 1
¶k
3 · 216
b30
.
¶
+ k log
since log t ≤ t/e, if we let M = e log (211 /b30 ), then
k≤
µ
¶
211
.
b30
t
.
M
Choose r0 and dB1 with respect to the set B1 as in Proposition 2.1, and let Q = pℓk+2t =
max (B1 ). It is clear that t ≥ ⌊6(2/b0 )3 213 ⌋ = 21192, which we denote by t0 . By equation (9),
we have that
r0 ≤
2π
2π
π
≤
≤ .
ℓ
Q
(2t)
t0
r0 ≥
b4
b4
≥
,
Q2
(k + 2t)4ℓ
We also have that
and
P
dB1 ≥ b1
≥
=
q∈B1
q
¶2t−2
µ
log2 Q
b3 − log 2
b2 Q e
(k+3)ℓ (2t−2)
b1
Ã
³
b3 ℓ
b2 (k + 2t) (k + 2t)
!2t−2
(k+3)ℓ
(k + 2t)b3 ℓ
b2 · b1
(k + 2t)
−
2ℓ log (k+2t)
log 2
´2t−2
2ℓ log (k+2t)
log 2
(−1)
Let us now bound pB1 (n) from below. Assume that n ≥ 2t, and let b5 = 1/M + 2 =
2.078207555 . . ., and b7 = t0 /(t0 − π). Making use of Theorem 2.1, we have
15
(−1)
(−1)
(−1)
pB1 (n) =gB1 (n) + ψB1 (n)
µ
¶ 2t−2
µ
¶
X
n + 2t − 2
n+h−1
≥α2t−1
−
|αh |
2t − 2
h−1
h=1
(−1)
− |ψB1 (n)|
¶ 2t−2
µ
n + 2t − 3 X
(pk+3 · · · pk+2t )−ℓ n2t−2
|αh |
−
≥
2t − 3
(2t − 2)!
h=1
−
1
k+2t
X
ℓ(2t−2)
pm
2t−2
b0 m=k+3
2t−2
(k + 2t)−ℓ(2t−2) n2t−2 (n + 2t − 3)2t−3 X
r0h
≥
−
(2t − 2)!
(2t − 3)!
r2t−1 dB1
h=1 0
−
1
(2t − 2)(k
b02t−2
−ℓ(2t−2) 2t−2
+ 2t)2ℓ(2t−2)
(b5 t)
n
n2t−3 22t−3
1
−
(2t − 2)!
(2t − 3)! (1 − r0 )r02t−2 dB1
(2t − 2)(k + 2t)2ℓ(2t−2)
−
b02t−2
Ã
2ℓ log (k+2t) !2t−2
(b5 t)−ℓ(2t−2) n2t−2 b7 n2t−3 22t−3 (k + 2t)(4−b3 )ℓ+ log 2
−
≥
(k+3)ℓ
(2t − 2)!
(2t − 3)!
b4 b2 b1
≥
−
2t(b5 t)2ℓ(2t−2)
.
b02t−2
Observe that
(k + 2t)(4−b3 )ℓ+
2ℓ log (k+2t)
log 2
2ℓ log Ct
≤ (b5 t)(4−b3 )ℓ+ log 2
2 log C
2 log t
= eℓ(log b5 +log t)((4−b3 + log 2 )+ log 2 )
= eℓ( log 2 log
2
2
t+(4−b3 +
4 log b5
log 2
log b5
log b5 )
) log t+(4−b3 + 4 log
2 )
2
≤ eb6 ℓ log t ,
where b6 is a constant determined as follows. Let x0 = log t0 . Then we may take
¶
¶
µ
µ
4 log b5 1
4 log b5 log b5
2
+ 4 − b3 +
+ 4 − b3 +
b6 =
log 2
log 2
x0
log 2
x20
= 3.523150893 . . . .
16
It suffices to select h1 such that for n ≥ h1 ,
(−1)
pB1 (n)n−(2t−3) ≥ (t − 1)2 .
(15)
Observe that (15) is implied by
(b5 t)−ℓ(2t−2) n
b7
−
(2t − 2)!
2(2t − 3)!
Ã
2eb6 ℓ log
2
t
(k+3)ℓ
b4 b2 b1
!2t−2
−
2t(b5 t)2ℓ(2t−2)
≥ (t − 1)2 ,
2t−2
2t−3
n
b0
which is equivalent to
n ≥ (b5 t)ℓ(2t−2) ×


Ã
!2t−2
b6 ℓ log2 t
2ℓ(2t−2)
2t(b5 t)
(2t − 2)!
b7 (t − 1) 2e
+
+ (t − 1)2 (2t − 2)! .
2t−2
(k+3)ℓ
2t−3
n
b
b4 b2 b1
0
(16)
The inequality
en n! ≤ nn+1 ,
holds for n ≥ 7. This implies that
(2t − 2)!
(2t)3
≤
.
(2t)2t−3
e2t (2t − 1)
Thus, since n ≥ 2t by assumption, (16) is implied by
n ≥ (b5 t)ℓ(2t−2) (A1 + A2 + A3 ),
where
A1 = b7 t
Ã
2eb6 ℓ log
2
t
(k+3)ℓ
b4 b2 b1
!2t−2
16t4 (b5 t)2ℓ(2t−2)
(2t − 1)e2t b02t−2
(t − 1)2 (2t)2t
A3 =
.
(2t − 1)e2t
A2 =
The term A3 is negligible relative to A2 , yet A1 and A2 are not easily compared since one
or the other may dominate depending on the values chosen for k, and ℓ. None the less, we
may simplify matters a little by absorbing A3 into A2 in the following way:
17
16t
A2 + A3
b02t−2 (2t)2t
≤
+
2t − 1 t(2t − 1)(b5 t)2(2t−2)
t3 (b5 t)2ℓ(2t−2) /(e2t b02t−2 )
¶µ
¶2t−2
µ
2b0
4t
16t
+
=
2t − 1
2t − 1
b25 t
¶µ
¶2t0 −2
µ
2b0
16t0
4t0
≤
+
2t0 − 1
2t0 − 1
b25 t0
≤ 8.000188756.
So, let
A′2 = 8.000188756
t3 (b5 t)2ℓ(2t−2)
.
e2t b02t−2
Then we may take
h1 = (b5 t)ℓ(2t−2) (A1 + A′2 ).
Remark 3. Note that
$ µ ¶
%
3(k+1)
2
(k + 2)8ℓk+10ℓ+3 = ⌊2g 2 h⌋,
t= 6
b0
and so
1
1
≤
.
t+1
2(g + 1)(g − 1)h
Lemma 3.4. There exists an N = N (k, ℓ) > 0, such that if n ≥ N , then
(−1)
p(k) (n) ≥ pC (n) > 0.
Proof. The second inequality is obvious. For the first, by Proposition 3.1, Theorem 3.1,
Lemma 3.3 and Remark 3, we have that for n ≥ h1 ,
pC (n)
(−1)
pC (n)
≤
1
1
1
(t − 1)2
≤
+
.
(−1)
t+1
t + 1 pB (n)n−(2t−3)
(g + 1)(g − 1)h
1
Now, using, (11), (12), (13), (17), and the identity
(k)
p (n) =
n
X
m=0
we have that for n ≥ h + h1 − 1,
(k)
pB (n − m)pC (m),
18
(17)
p(k) (n) ≥2
(k+1)
X
0≤m≤n−h
(1 + |pB
− (g − 1)
≥2
≥2
n
X
m=0
n
X
m=0
X
(k+1)
(k+1)
(1 + |pB
2
− (g − 1)
Ã
Ã
≥
≥
m=0
n
X
(1 + |pB
(n − m)|)pC (m)
(n − m)|)pC (m) − (g + 1)(g − 1)
n
X
X
2
pC (m)
!
pC (m)
(−1)
n−h<m≤n pC (m)
(k+1)
(1 + |pB
X
pC (m)
n−h<m≤n
(n − m)|)pC (m)
(−1)
n−h<m≤n pC (m)
≥ 2 − (g − 1)
n
X
(k+1)
n−h<m≤n
(1 + |pB
(n − m)|)pC (m)
(−1)
pC (n)
!
×
n
X
m=0
(k+1)
(1 + |pB
(n − m)|)pC (m)
(n − m)|)pC (m)
pC (m)
m=0
(−1)
=pC (n).
Proof of Theorem 1.1.
By the proofs of Lemmas 3.3 and 3.4, it suffices us to choose N ≥ h+h1 = h+(b5 t)ℓ(2t−2) (A1 +
A′2 ). First observe that A′2 ≥ 2t. The quantity (b5 t)ℓ(2t−2) A1 may be simplified further with
an upper bound. Let
b9 =
1
log b5
+
+ b6 = 3.630910490 . . . ,
2
x0
x0
and
b10 =
2
.
b2 b4
Then
19
!2t−2
(b5 t)ℓ(2t−2) A1 = b7 t
Ã
b10 (b5 t)ℓ eb6 ℓ log
= b7 t
Ã
b10 eℓ(log b5 +log t+b6 log
≤ b7 t
Ã
b10 eb9 ℓ log
A′1 = b7 t
Ã
2
t
(k+3)ℓ
b1
(k+3)ℓ
b1
2
t
(k+3)ℓ
b1
2
t)
!2t−2
!2t−2
Denote
b9 ℓ log2 t
b10 e
(k+3)ℓ
b1
!2t−2
.
We will use the fact that h ≤ t to absorb h into (b5 t)ℓ(2t−2) A′2 in the following way:
e2t b02t−2
h + (b5 t)ℓ(2t−2) A′2
≤
+ 8.000188756
t2 (b5 t)3(2t−2)
t3 (b5 t)3ℓ(2t−2) /(e2t b02t−2 )
µ
¶2t−2
e2 eb0
+ 8.000188756
= 2
t
b35 t3
µ
¶2t0 −2
e2 eb0
≤ 2
+ 8.000188756
t0 b35 t30
≤ 8.000188757
Letting b8 = 8.0002, we may take
b8 t3 (b5 t)3ℓ(2t−2)
e2t b02t−2
Ã
!2t−2
2
b10 eb9 ℓ log t
N = A′1 +
= b7 t
(k+3)ℓ
b1
+
b8 t3 (b5 t)3ℓ(2t−2)
.
e2t b02t−2
We conclude the proof by restructuring the constants as follows:
a1 = b 7
a2 = b210
a3
a4
a5
a6
a7
= e2b9
= b21
= e−2 b8
= e−2 b−2
0
= b5 .
2
20
4
Asymptotic results
Observe that as k → ∞, log t ≍ k log k, when ℓ remains fixed. This implies that if ℓ > 2,
then
2
a3ℓ log t
→ 0, as k → ∞,
(k+2)ℓ
a4
and so in this situation, the second term dominates in the expression for N (k, ℓ) in Theorem 1.1.
If ℓ = 1 or 2, then
2
a3ℓ log t
= eλ1 (k) ,
(k+2)ℓ
a4
where λ1 (k) ≍ k 2 log2 k, and
(a7 t)6ℓ = eλ2 (k) ,
where λ2 (k) ≍ k log k. In this case the first term dominates in the expression for N (k, ℓ).
Thus we have proved the following corollary to Theorem 1.1:
Corollary 4.1. Let ℓ ∈ N be fixed. Then as k → ∞,
 Ã
!t−1 
ℓ log2 t
a2 a3
 , if ℓ = 1, 2;
F (k, ℓ) = O t
(k+3)ℓ
a4
³ ¡
¢ ´
3
6ℓ t−1
F (k, ℓ) = O t a6 (a7 t)
, if ℓ > 2.
Theorem 1.2 follows from Corollary 4.1 simply by taking logarithms.
5
Acknowledgements
The author is supported by an NSERC CGS D research grant and wishes to acknowledge
the Natural Sciences and Engineering Research Council for their generous funding, as well
as the support of his supervisor Dr. Izabella Laba.
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2000 Mathematics Subject Classification: Primary 05A17; Secondary 11P81.
Keywords: partition function, difference function.
(Concerned with sequences A000607 and A090677.)
Received April 10 2006; revised version received November 18 2006. Published in Journal of
Integer Sequences, December 29 2006.
Return to Journal of Integer Sequences home page.
22