Download SUMS OF SQUARES AND SUMS OF TRIANGULAR NUMBERS

SUMS OF SQUARES AND SUMS OF TRIANGULAR
NUMBERS INDUCED BY PARTITIONS OF 8
NAYANDEEP DEKA BARUAH, SHAUN COOPER AND MICHAEL HIRSCHHORN
Abstract. Let rk (n) and tk (n) denote the number of representations
of an integer n as a sum of k squares, and as a sum of k triangular
numbers, respectively. We prove that
1
t8 (n) = 10
(r8 (8n + 8) − 16r8 (2n + 2)) ,
2 × 32
and therefore the study of the sequence t8 (n) is reduced to the study of
subsequences of r8 (n). We give an additional 21 analogous results for
sums of squares and sums of triangular numbers induced by partitions
of 8. We give a brief indication of what happens for the case k ≥ 9.
Addresses:
Nayandeep Deka Baruah
Department of Mathematical Sciences, Tezpur University
Napaam-784028, Sonitpur, Assam, India
[email protected]
Shaun Cooper
Institute of Information and Mathematical Sciences, Massey University
Private Bag 102904, North Shore Mail Centre, Auckland, New Zealand
[email protected]
Michael Hirschhorn
School of Mathematics, University of New South Wales
Sydney 2052, Australia
[email protected]
1. Introduction
Let rk (n) and tk (n) denote the number of representations of an integer n
as a sum of k squares, and as a sum of k triangular numbers, respectively.
2000 Mathematics Subject Classification. Primary—11E25; Secondary—05A19, 11D85,
33D15.
Nayandeep Deka Baruah is partially supported by BOYSCAST Fellowship grant
SR/BY/M-03/05 from DST, Govt. of India.
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NAYANDEEP DEKA BARUAH, SHAUN COOPER AND MICHAEL HIRSCHHORN
That is, rk (n) is the number of solutions in integers of the equation
x21 + · · · + x2k = n,
and tk (n) is the number of solutions in non-negative integers of
xk (xk + 1)
x1 (x1 + 1)
+ ··· +
= n.
2
2
In [3], a remarkable relation between rk (n) and tk (n) was given:
Theorem 1.1. Suppose 1 ≤ k ≤ 7. Then for any non-negative integer n,
we have
k
k−1
2+
.
rk (8n + k) = ck tk (n), where ck = 2
4
Thus, the study of tk (n) for 1 ≤ k ≤ 7 is reduced to the study of the
subsequence rk (8n + k) of rk (n). Theorem 1.1 was rediscovered in [2], where
a simple proof was given using generating functions. Later, a combinatorial
proof of Theorem 1.1 was given in [7]. It has been proved [4] that Theorem
1.1 does not hold for any value k ≥ 8.
In [1], it was shown that a result similar to Theorem 1.1 holds for representations by other quadratic forms. In order to state the result, let us first
make some definitions.
Let λ = (λ1 , . . . , λm ) be a partition of k. That is, λ1 , . . . , λm are integers
satisfying λ1 ≥ · · · ≥ λm ≥ 1 and λ1 + · · · + λm = k. For any non-negative
integer n, define rλ (n) to be the number of solutions in integers of
λ1 x21 + · · · + λm x2m = n,
(1.1)
and define tλ (n) to be the number of solutions in non-negative integers of
xk (xk + 1)
x1 (x1 + 1)
+ · · · + λk
= n.
2
2
A solution of (1.1) (resp. (1.2)) is called a representation of n by squares
(resp. triangular numbers) induced by λ. For a partition λ = (λ1 , . . . , λm ),
let
i1
i1
i2
i1
i3
m
m−1
(1.3)
Cλ = 2 + 2
+
+
,
4
2
1
1
1
(1.2)
λ1
where ij denotes the number of parts in λ which are equal to j.
The main result in [1] is:
Theorem 1.2. Let λ = (λ1 , . . . , λm ) be a partition of k. Then, for 1 ≤ k ≤
7, we have
rλ (8n + k) = Cλ tλ (n),
for all non-negative integers n, where the value of Cλ is given by (1.3).
Theorem 1.1 is the special case of Theorem 1.2 for which λ is the partition
consisting of k ones.
SUMS OF SQUARES
3
The goal of this article is to extend Theorems 1.1 and 1.2 to the case
k = 8. By the non-existence result in [4], any result for k = 8 must have a
different form from the results for 1 ≤ k ≤ 7 in Theorems 1.1 and 1.2.
We will prove the following three theorems:
Theorem 1.3. Let λ = (λ1 , . . . , λm ) be a partition of 8. Let aλ (n) denote
the number of solutions in integers of
λ1 x21 + · · · + λm x2m = n,
for which x1 , . . . , xm , are not all even, and let bλ (n) denote the number of
solutions in integers for which x1 , . . . , xm , are all odd. Then for any nonnegative integer n, we have
Cλ
bλ (8n),
2m
where the value of Cλ is given by (1.3).
aλ (8n) =
Theorem 1.4. Let λ = (λ1 , . . . , λm ) be a partition of 8. Then for any
non-negative integer n, we have
rλ (8n + 8) − rλ (2n + 2) = Cλ tλ (n),
where the value of Cλ is given by (1.3).
Theorem 1.5. Let
λ ∈ {(1, 1, 1, 1, 1, 1, 1, 1), (2, 2, 1, 1, 1, 1), (4, 2, 1, 1), (3, 3, 1, 1), (7, 1)} .
Then for any non-negative integer n,

if n is even,
 Dλ rλ (n + 1),
tλ (n) =

Eλ (rλ (2n + 2) − rλ (n + 1)) , if n is odd,
where the values of Dλ and Eλ are given
λ
(1, 1, 1, 1, 1, 1, 1, 1)
(2, 2, 1, 1, 1, 1)
(3, 3, 1, 1)
(4, 2, 1, 1)
(7, 1)
by:
Dλ
Eλ
1/16 1/128
1/8 1/24
1/4
1/8
1/4
1/4
1/2
1/2
Theorem 1.4 is the analogue of Theorem 1.2 for partitions of 8. Theorems
1.3 and 1.4 are equivalent. Theorem 1.5 shows that for some partitions of 8,
tλ (n) may be obtained from subsequences of rλ (n) in another way. Theorem
1.5 does not hold for any of the other partitions of 8.
This work is organized as follows. In Section 2, we will show that Theorems 1.3 and 1.4 are equivalent. In Section 3, we will give a combinatorial
proof of Theorem 1.3. In Section 4, we will give a proof of Theorem 1.4
using generating functions. In Section 5, we give a proof of Theorem 1.5
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NAYANDEEP DEKA BARUAH, SHAUN COOPER AND MICHAEL HIRSCHHORN
using generating functions. Finally, in Section 6, we indicate what happens
for partitions of k when k ≥ 9.
For the remainder of this paper, unless otherwise stated it will be assumed
that λ = (λ1 , . . . , λm ) is a partition of 8.
2. Equivalence of Theorems 1.3 and 1.4
In this section, we show that Theorems 1.3 and 1.4 are equivalent.
First, observe that if we multiply the equation λ1 x21 + · · · + λm x2m = 2n by
4, it follows that rλ (2n) is the number of solutions of λ1 x21 +· · ·+λm x2m = 8n
for which x1 , . . . , xm , are all even. Therefore, by separating the solutions of
λ1 x21 + · · · + λm x2m = 8n into those for which x1 , . . . , xm , are all even, and
those for which x1 , . . . , xm , are not all even, it follows that
(2.1)
rλ (8n) = rλ (2n) + aλ (8n).
Second, observe that if
xm (xm + 1)
x1 (x1 + 1)
+ · · · + λm
, x1 , . . . , xm ≥ 0,
2
2
is a representation of n − 1 as a sum of triangular numbers induced by λ,
then multiplying by 8 and completing the square gives 2m representations
of 8n as a sum of odd squares induced by λ, namely
(2.2)
(2.3)
n − 1 = λ1
8n = λ1 (±(2x1 + 1))2 + · · · + λm (±(2xm + 1))2 ,
and conversely, each of the 2m representations in (2.3) arises from exactly
one representation in (2.2). Therefore
(2.4)
tλ (n − 1) = 2m bλ (8n).
The equivalence of Theorems 1.3 and 1.4 now follows from Eqs. (2.1) and
(2.4).
3. Combinatorial proof of Theorem 1.3
We shall require the following three lemmas from [1, Corollaries 6 – 8]:
Lemma 3.1. The number of representations of 8n + 4 as a sum of four odd
squares equals twice the number of representations of 8n + 4 as a sum of
four even squares.
Lemma 3.2. The number of representations of 8n+4 by the form 2x2 +y 2 +
z 2 in which x, y and z are all odd equals twice the number of representations
in which x, y and z are all even.
Lemma 3.3. The number of representations of 8n + 4 by the form 3x2 + y 2
in which x and y are both odd equals twice the number of representations in
which x and y are both even.
SUMS OF SQUARES
5
It will be useful to divide the 22 partitions of 8 into 7 classes. Let
S1
S2
S3
S4
S5
S6
S7
=
=
=
=
=
=
=
{(1, 1, 1, 1, 1, 1, 1, 1), (4, 1, 1, 1, 1)}
{(2, 2, 2, 1, 1), (4, 2, 1, 1)}
{(3, 2, 2, 1), (3, 3, 1, 1), (4, 3, 1)}
{(2, 1, 1, 1, 1, 1, 1), (2, 2, 1, 1, 1, 1)}
{(3, 1, 1, 1, 1, 1)}
{(3, 2, 1, 1, 1)}
the other 11 partitions of 8 that are not included
in any of S1 , · · · , S6 .
Lemma 3.4. Suppose
(3.1)
8n = λ1 x21 + · · · + λm x2m
is a representation of 8n as a sum of squares induced by λ, with x1 , . . . , xm ,
not all even. Let xA , xB , xC , xD denote any of the xi in (3.1) which occur
with coefficient 1, 2, 3 or 4, respectively. For example, if λ = (3, 2, 1, 1, 1)
and 8n = 3x21 + 2x22 + x23 + x24 + x25 , then xA ∈ {x3 , x4 , x5 }, xB ∈ {x2 },
xC ∈ {x1 } and xD ∈ ϕ. Then
(1) If λ ∈ S1 then either all of the xi are odd; or four of the xA are even
and the other xi are all odd.
(2) If λ ∈ S2 then either all of the xi are odd; or one of the xB and two
of the xA are even and the other xi are all odd.
(3) If λ ∈ S3 then either all of the xi are odd; or one of the xC and one
of the xA are even and the other xi are all odd.
(4) If λ ∈ S4 then either all of the xi are odd; or exactly 4 of the xA and
are even and the other xi are all odd; or one of the xB and two of
the xA are even and the other xi are all odd.
(5) If λ ∈ S5 , i.e., λ = (3, 1, 1, 1, 1, 1) and 8n = 3x21 +x22 +x23 +x24 +x25 +x26 ,
then either all of the xi are odd; or four of x2 , . . . , x6 are even and
the other xi are odd; or x1 and one of x2 , . . . , x6 are even and the
other xi are odd.
(6) If λ ∈ S6 , i.e., λ = (3, 2, 1, 1, 1) and 8n = 3x21 + 2x22 + x23 + x24 + x25 ,
then either all of the xi are odd; or x2 and two of x3 , x4 , x5 are
even and the other xi are odd; or x1 and one of x3 , x4 , x5 are even
and the other xi are odd.
(7) If λ ∈ S7 , then all of the xi are odd.
Proof. Analyse each of the 22 partitions of 8, one at a time. We give two
examples to illustrate the technique.
Example 1: λ = (5, 2, 1) ∈ S7 . Let 8n = 5x21 + 2x22 + x23 be a representation of 8n as a sum of squares induced by λ, with x1 , x2 , x3 not
all even. If x1 = 2j + 1 is odd, then 8n = 5(2j + 1)2 + x22 + x23 , and so
x22 + x23 ≡ 3 (mod 8). This implies x2 and x3 are also odd. If x1 = 2j is
6
NAYANDEEP DEKA BARUAH, SHAUN COOPER AND MICHAEL HIRSCHHORN
even, then 8n = 5(2j)2 + 2x22 + x23 , and so 2x22 + x23 ≡ 0 (mod 4). This
implies x2 and x3 are also even, but this case is excluded by hypothesis.
This proves Lemma 3.4 for the partition λ = (5, 2, 1).
Example 2: λ = (2, 2, 2, 1, 1) ∈ S2 . Let 8n = 2x21 + 2x22 + 2x23 + x24 + x25 be a
representation of 8n as a sum of squares induced by λ, with x1 , x2 , x3 , x4 , x5
not all even. If x1 , x2 , x3 are all odd, then x24 + x25 ≡ 2 (mod 8), and therefore x4 and x5 are also odd. If two of x1 , x2 , x3 are odd, then x24 + x25 ≡ 4
(mod 8), and therefore x4 and x5 are both even. If one of x1 , x2 , x3 is odd,
then x24 + x25 ≡ 6 (mod 8), which is impossible. If none of x1 , x2 , x3 are
odd, i.e., x1 , x2 , x3 are all even, then x24 + x25 ≡ 0 (mod 8), and so x4 and
x5 are also even; this is excluded by hypothesis. This proves Lemma 3.4 for
the partition λ = (2, 2, 2, 1, 1).
The other 20 partitions are handled similarly.
We can now prove Theorem 1.3.
Proof. First suppose λ ∈ S1 . Lemmas 3.1 and 3.4 imply that the number of
representations of 8n as a sum of squares induced by
λin which not all of
1 i1
the xi are odd, and not all of the xi are even, is
times the number
2 4
of representations in which all of the xi are odd. The binomial coefficient
arises from the number of orderings of the xi by parity in which four of the
xi are
even
and the others odd. Since i2 = i3 = 0 for λ ∈ S1 , we have
i2
i3
=
= 0. Therefore we have shown that
1
1
1
i1
i1
i2
i1
i3
aλ (8n) =
2+
+
+
bλ (8n)
2
4
2
1
1
1
for λ ∈ S1 . The other cases λ ∈ S2 , . . . , S7 follow similarly from Lemmas
3.1–3.4.
4. Generating function proof of Theorem 1.4
Let
ϕ(q) =
∞
X
n2
q ,
ψ(q) =
n=−∞
∞
X
n=0
q
n(n+1)/2
,
a(q) =
∞
X
∞
X
2 +mn+n2
qm
m=−∞ n=−∞
The generating functions for rλ (n) and tλ (n), where λ = (λ1 , . . . , λm ), are
∞
X
n=0
∞
X
n=0
rλ (n)q n = ϕ(q λ1 ) · · · ϕ(q λm ),
tλ (n)q n = ψ(q λ1 ) · · · ψ(q λm ).
.
SUMS OF SQUARES
7
Theorem 1.4 may be proved by repeatedly dissecting the generating function
for rλ (n). We illustrate the procedure by giving complete proofs for λ =
(3, 2, 2, 1) and (3, 1, 1, 1, 1, 1). The proofs for the other partitions of 8 are
similar, and usually simpler.
We will require the following Lemma. Observe that (4.6)–(4.10) give
dissections of various theta functions into their even and odd parts.
Lemma 4.1.
(4.1)
ϕ(q) + ϕ(−q) = 2ϕ(q 4 ),
(4.2)
ϕ(q) − ϕ(−q) = 4qψ(q 8 ),
(4.3)
ϕ(q)ϕ(−q) = ϕ2 (−q 2 ),
ϕ(q)ψ(q 2 ) = ψ(q)2 ,
(4.4)
(4.5)
a(q) = ϕ(q)ϕ(q 3 ) + 4qψ(q 2 )ψ(q 6 ),
(4.6)
ϕ(q) = ϕ(q 4 ) + 2qψ(q 8 ),
(4.7)
ϕ(q)2 = ϕ(q 2 )2 + 4qψ(q 4 )2 ,
(4.8)
ψ(q)ψ(q 3 ) = ϕ(q 6 )ψ(q 4 ) + qϕ(q 2 )ψ(q 12 ),
(4.9)
ϕ(q)ϕ(q 3 ) = a(q 4 ) + 2qψ(q 2 )ψ(q 6 ),
a(q) = a(q 4 ) + 6qψ(q 2 )ψ(q 6 ).
(4.10)
Proof. Eqs. (4.1)–(4.4) are parts (i)–(iv) of Entry 25 in [5, pp. 40–41].
Equation (4.5) is discussed in [6, p. 93] and references to proofs are given
there. Eq. (4.6) follows by adding (4.1) and (4.2), and (4.7) may be obtained
by combining the results in [5, p. 40, Entry 25]. A proof of (4.8) may be
found in [8, Preliminary lemmas, part (xxxiii)]. Another proof of (4.8),
using Schröter’s formulas, can be obtained by setting µ = 2, ν = 1 in [5, p.
69, (36.8)]. We will prove (4.9) and (4.10). By (4.6),
ϕ(q)ϕ(q 3 ) = ϕ(q 4 ) + 2qψ(q 8 ) ϕ(q 12 ) + 2q 3 ψ(q 24 )
= ϕ(q 4 )ϕ(q 12 ) + 4q 4 ψ(q 8 )ψ(q 24 ) + 2q ϕ(q 12 ψ(q 8 ) + q 2 ϕ(q 4 )ψ(q 24 )
= a(q 4 ) + 2qψ(q 2 )ψ(q 6 ),
where we have used (4.5) with q 4 for q and (4.8) with q 2 for q. Thus, (4.9)
is proved. Finally, (4.10) follows by substituting (4.9) into (4.5).
4.1. Proof of Theorem 1.4 for the partition (3, 2, 2, 1). Apply (4.9) to
the generating function to get
∞
X
r(3,2,2,1) (n)q n = ϕ(q 3 )ϕ(q 2 )2 ϕ(q)
n=0
=
a(q 4 ) + 2qψ(q 6 )ψ(q 2 ) ϕ(q 2 )2 .
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NAYANDEEP DEKA BARUAH, SHAUN COOPER AND MICHAEL HIRSCHHORN
Extract the even powers of q, then replace q 2 with q. The result is
∞
X
(4.11)
r(3,2,2,1) (2n)q n = a(q 2 )ϕ(q)2 .
n=0
Apply (4.7) to this to obtain
∞
X
r(3,2,2,1) (2n)q n = a(q 2 ) ϕ(q 2 )2 + 4qψ(q 4 )2 ,
n=0
and therefore,
∞
X
r(3,2,2,1) (4n)q n = a(q)ϕ(q)2 .
n=0
Apply (4.7) and (4.10) to this, to obtain
∞
X
r(3,2,2,1) (4n)q n = a(q 4 ) + 6qψ(q 2 )ψ(q 6 ) ϕ(q 2 )2 + 4qψ(q 4 )2 ,
n=0
from which we deduce
∞
X
r(3,2,2,1) (8n)q n = a(q 2 )ϕ(q)2 + 24qψ(q)ψ(q 2 )2 ψ(q 3 ).
n=0
Now use (4.11) to obtain
∞
X
n=0
n
r(3,2,2,1) (8n)q =
∞
X
n
r(3,2,2,1) (2n)q + 24
n=0
q n+1 ,
If we compare coefficients of
1.4 for the partition (3, 2, 2, 1).
∞
X
t(3,2,2,1) (n)q n+1 .
n=0
then we complete the proof of Theorem
4.2. Proof of Theorem 1.4 for the partition (3, 1, 1, 1, 1, 1). The procedure is similar to the proof for the partition (3, 2, 2, 1), therefore we will
be brief and use properties of Lemma 4.1 without comment.
∞
X
r(3,1,1,1,1,1) (n)q n = ϕ(q 3 )ϕ(q)5
n=0
= ϕ(q 2 )2 + 4qψ(q 4 )2
∞
X
2
a(q 4 ) + 2qψ(q 6 )ψ(q 2 ) ,
r(3,1,1,1,1,1) (2n)q n = a(q 2 )ϕ(q)4 + 16qa(q 2 )ψ(q 2 )4 + 16qϕ(q)2 ψ(q 3 )ψ(q 2 )2 ψ(q)
n=0
= a(q 2 )ϕ(q)4 + 16qa(q 2 )ψ(q 2 )4 + 16qψ(q 3 )ψ(q)5
2
= a(q 2 ) ϕ(q 2 )2 + 4qψ(q 4 )2 + 16qa(q 2 )ψ(q 2 )4
+ 16qψ(q 2 )2 ϕ(q 2 )2 + 4qψ(q 4 )2 ϕ(q 6 )ψ(q 4 ) + qϕ(q 2 )ψ(q 12 ) ,
∞
X
n=0
r(3,1,1,1,1,1) (4n)q n = a(q)ϕ(q)4 + 16qa(q)ψ(q 2 )4 + 16qϕ(q)3 ψ(q 6 )ψ(q)2
SUMS OF SQUARES
9
+ 64qϕ(q 3 )ψ(q 2 )3 ψ(q)2
= a(q)ϕ(q)4 + 16qa(q)ψ(q 2 )4 + 16qϕ(q)4 ψ(q 6 )ψ(q 2 )
+ 64qϕ(q)ϕ(q 3 )ψ(q 2 )4
2
= a(q 4 ) + 6qψ(q 6 )ψ(q 2 ) ϕ(q 2 )2 + 4qψ(q 4 )2
+ 16q a(q 4 ) + 6qψ(q 6 )ψ(q 2 ) ψ(q 2 )4
2
+ 16qψ(q 6 )ψ(q 2 ) ϕ(q 2 )2 + 4qψ(q 4 )2
+ 64qψ(q 2 )4 a(q 4 ) + 2qψ(q 6 )ψ(q 2 ) ,
∞
X
r(3,1,1,1,1,1) (8n)q n = a(q 2 )ϕ(q)4 + 16qa(q 2 )ψ(q 2 )4
n=0
+ 48qϕ(q)2 ψ(q 3 )ψ(q 2 )2 ψ(q) + 96qψ(q 3 )ψ(q)5
+ 128qϕ(q)2 ψ(q 3 )ψ(q 2 )2 ψ(q) + 128qψ(q 3 )ψ(q)5
= a(q 2 )ϕ(q)4 + 16qa(q 2 )ψ(q 2 )4 + 400qψ(q 3 )ψ(q)5
=
=
∞
X
n=0
∞
X
n=0
r(3,1,1,1,1,1) (2n)q n + 384qψ(q 3 )ψ(q)5
r(3,1,1,1,1,1) (2n)q n + 384
∞
X
t(3,1,1,1,1,1) (n)q n+1 .
n=0
If we equate coefficients of q n+1 , we complete the proof of Theorem 1.4 for
the partition (3, 1, 1, 1, 1, 1).
5. Proof of Theorem 1.5
In this section we use generating functions to prove Theorem 1.5. As
in the previous section, we will be brief and use properties of Lemma 4.1
without comment.
5.1. Proof of Theorem 1.5 for λ = (1, 1, 1, 1, 1, 1, 1, 1). We have
(5.1)
∞
X
r(1,1,1,1,1,1,1,1) (n)q n = ϕ(q)8 = (ϕ(q 2 )2 + 4qψ(q 4 )2 )4 ,
n=0
and
(5.2)
∞
X
t(1,1,1,1,1,1,1,1) (n)q n = ψ(q)8 = ϕ(q)4 ψ(q 2 )4 = ψ(q 2 )4 (ϕ(q 2 )2 +4qψ(q 4 )2 )2 .
n=0
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NAYANDEEP DEKA BARUAH, SHAUN COOPER AND MICHAEL HIRSCHHORN
If we extract odd powers of q from (5.1), extract even powers of q from (5.2),
and replace q 2 with q in each case, then we obtain
∞
X
r(1,1,1,1,1,1,1,1) (2n + 1)q n = 16ϕ(q)6 ψ(q 2 )2 + 256qϕ(q)2 ψ(q 2 )6
n=0
= 16ϕ(q)4 ψ(q)4 + 256qψ(q 2 )4 ψ(q)4 ,
and
∞
X
t(1,1,1,1,1,1,1,1) (2n)q n = ϕ(q)4 ψ(q)4 + 16qψ(q 2 )4 ψ(q 4 ).
n=0
It follows that
r(1,1,1,1,1,1,1,1) (2n + 1) = 16t(1,1,1,1,1,1,1,1) (2n).
This proves the first part of Theorem 1.5 for the partition (1, 1, 1, 1, 1, 1, 1, 1).
Next, extract the even powers of q from (5.1) and replace q 2 with q, to
get
∞
X
r(1,1,1,1,1,1,1,1) (2n)q n = ϕ(q)8 + 96qϕ(q)4 ψ(q 2 )4 + 256q 2 ψ(q 2 )8 .
n=0
Therefore
∞
X
2
r(1,1,1,1,1,1,1,1) (2n) − r(1,1,1,1,1,1,1,1) (n) q n = 96q ϕ(q 2 )2 + 4qψ(q 4 )2 ψ(q 2 )4 +256q 2 ψ(q 2 )8 .
n=0
If we extract even powers of q from this and replace q 2 with q, we obtain
∞
X
r(1,1,1,1,1,1,1,1) (4n) − r(1,1,1,1,1,1,1,1) (2n) q n = 768qϕ(q)2 ψ(q 2 )2 ψ(q)4 + 256qψ(q)8
n=0
= 1024qψ(q)8 .
Now extract odd powers of q from (5.2) and replace q 2 with q to get
∞
X
t(1,1,1,1,1,1,1,1) (2n − 1)q n = 8qψ(q)4 ϕ(q)2 ψ(q 2 )2 = 8qψ(q)8 .
n=1
If we compare the last two equations, we find that
r(1,1,1,1,1,1,1,1) (4n) − r(1,1,1,1,1,1,1,1) (2n) = 128t(1,1,1,1,1,1,1,1) (2n − 1).
This proves the second part of Theorem 1.5 for the partition (1, 1, 1, 1, 1, 1, 1, 1).
5.2. Proof of Theorem 1.5 for λ = (2, 2, 1, 1, 1, 1). The procedure is
similar to that for the partition (1, 1, 1, 1, 1, 1, 1, 1). We omit the details.
SUMS OF SQUARES
11
5.3. Proof of Theorem 1.5 for λ = (3, 3, 1, 1). By (4.8) we have
ψ(q)ψ(q 3 ) = ϕ(q 6 )ψ(q 4 ) + qψ(q 12 )ϕ(q 2 )
and replacing q by −q we obtain
ψ(−q)ψ(−q 3 ) = ϕ(q 6 )ψ(q 4 ) − qψ(q 12 )ϕ(q 2 ).
If we square and subtract, we obtain
ψ 2 (q)ψ 2 (q 3 ) − ψ 2 (−q)ψ 2 (−q 3 ) = 4qϕ(q 2 )ψ(q 4 )ϕ(q 6 )ψ(q 12 ).
Employing (4.4), we deduce that
ψ 2 (q)ψ 2 (q 3 ) − ψ 2 (−q)ψ 2 (−q 3 ) = 4qψ 2 (q 2 )ψ 2 (q 6 ).
(5.3)
Multiplying both sides by
ψ 2 (q)ψ 2 (q 3 ) + ψ 2 (−q)ψ 2 (−q 3 ) /ψ 2 (q 2 )ψ 2 (q 6 ),
we deduce that
ψ 4 (q)ψ 4 (q 3 ) − ψ 4 (−q)ψ 4 (−q 3 )
= 4q{ψ 2 (q)ψ 2 (q 3 ) + ψ 2 (−q)ψ 2 (−q 3 )}.
ψ 2 (q 2 )ψ 2 (q 6 )
With the help of (4.4), this becomes
ϕ2 (q)ϕ2 (q 3 ) − ϕ2 (−q)ϕ2 (−q 3 ) = 4q{ψ 2 (q)ψ 2 (q 3 ) + ψ 2 (−q)ψ 2 (−q 3 )},
which is equivalent to
∞
X
r(3,3,1,1) (n)(q n − (−q)n ) = 4q
n=0
∞
X
t(3,3,1,1) (n)(q n + (−q)n ).
n=0
q 2n+1
Equating the coefficients of
on both sides, we obtain the first part of
Theorem 1.5 for the partition (3, 3, 1, 1).
Next, replace q with −q in (4.9) and subtract the result from (4.9) to get
ϕ(q)ϕ(q 3 ) − ϕ(−q)ϕ(−q 3 ) = 4qψ(q 2 )ψ(q 6 ).
Squaring both sides and applying (4.3), we obtain
ϕ2 (q)ϕ2 (q 3 ) + ϕ2 (−q)ϕ2 (−q 3 ) − 2ϕ2 (−q 2 )ϕ2 (−q 6 ) = 16q 2 ψ 2 (q 2 )ψ 2 (q 6 ),
which is equivalent to
∞
X
n
n
r(3,3,1,1) (n)(q + (−q) ) −
n=0
∞
X
n=0
n 2n
r(3,3,1,1) (n)(−1) q
= 16q
2
∞
X
t(3,3,1,1) (n)q 2n .
n=0
q 4n
Equating coefficients of
on both sides, we complete the proof of the
second part of Theorem 1.5 for the partition (3, 3, 1, 1).
5.4. Proof of Theorem 1.5 for λ = (4, 2, 1, 1). The procedure is similar
to that for the partition (1, 1, 1, 1, 1, 1, 1, 1). We omit the details.
12
NAYANDEEP DEKA BARUAH, SHAUN COOPER AND MICHAEL HIRSCHHORN
5.5. Proof of Theorem 1.5 for λ = (7, 1). We set µ = 4, ν = 3 in [5, p.
69, (36.8)] to deduce that
ψ(q)ψ(q 7 ) = ϕ(q 28 )ψ(q 8 ) + qψ(q 14 )ψ(q 2 ) + q 6 ψ(q 56 )ϕ(q 4 ).
Replacing q by −q, we obtain
ψ(−q)ψ(−q 7 ) = ϕ(q 28 )ψ(q 8 ) − qψ(q 14 )ψ(q 2 ) + q 6 ψ(q 56 )ϕ(q 4 ).
Adding, we obtain
ψ(q)ψ(q 7 ) + ψ(−q)ψ(−q 7 ) = 2ϕ(q 28 )ψ(q 8 ) + 2q 6 ψ(q 56 )ϕ(q 4 ).
With the help of (4.1) and (4.2), we obtain
ϕ(q)ϕ(q 7 ) − ϕ(−q)ϕ(−q 7 ) = 2q{ψ(q)ψ(q 7 ) + ψ(−q)ψ(−q 7 )},
which is equivalent to
∞
∞
X
X
n
n
r(7,1) (n)(q − (−q) ) = 2q
t(7,1) (n)(q n + (−q)n ).
n=0
n=0
q 2n+1
Comparing coefficients of
on both sides, we complete the proof of the
first part of Theorem 1.5 for the partition (7, 1).
Next, setting µ = 4 and ν = 3 in [5, p. 68, (36.3)], we can deduce that
ϕ(q)ϕ(q 7 ) + ϕ(−q)ϕ(−q 7 ) = 2ϕ(q 8 )ϕ(q 56 ) + 8q 16 ψ(q 16 )ψ(q 112 ) + 4q 4 ψ(q 4 )ψ(q 28 ).
With the aid of (4.1) and (4.2) we can rewrite this as
ϕ(q)ϕ(q 7 ) + ϕ(−q)ϕ(−q 7 ) = ϕ(q 2 )ϕ(q 14 ) + ϕ(−q 2 )ϕ(−q 14 ) + 4q 4 ψ(q 4 )ψ(q 28 ),
which is equivalent to
∞
∞
∞
X
X
X
n
n
2n
2 n
4
rλ (n)(q + (−q) ) =
rλ (n)(q + (−q ) ) + 4q
tλ (n)q 4n .
n=0
n=0
n=0
q 4n
Equating the coefficients of
on both sides, we complete the proof of the
second part of Theorem 1.5 for the partition (7, 1).
6. Partitions of k for k ≥ 9
In this section, we indicate the results for partitions of k when k ≥ 9. For
simplicity, we restrict the discussion to the case of sums of k squares and
sums of k triangular numbers; i.e., when the partition λ consists of k ones.
As an example, consider the representations of 8n+9 as a sum of 9 squares.
Either: all of the squares are odd, and the number of such representations is 29 t9 (n);
Or: four of the squares are
even, and Lemma 3.1 implies the number of
such representations is 28 94 t9 (n);
Or: eight of the squares are even, and the number of such representations
SUMS OF SQUARES
is the coefficient of q 8n+9 in
9
8
13
ϕ(q 4 )8 × 2qψ(q 8 ).
Therefore
r9 (8n + 9) =
9
t9 (n) + 18[q 8n+9 ] qϕ(q 4 )8 ψ(q 8 ) ,
2 +2
4
9
8
where [q k ]f (q) denotes the coefficient of q k in the Taylor expansion of f (q)
about q = 0. Thus we have proved
1
(6.1)
t9 (n) = 15 [q 8n+9 ] ϕ(q)9 − 18qϕ(q 4 )8 ψ(q 8 ) .
2
Let us contrast this with the result for sums of 8 triangular numbers:
1
8n+8
8
4 8
[q
]
ϕ(q)
−
16ϕ(q
)
.
(6.2)
t8 (n) = 10
2 × 32
Thus while (6.2) enables members of the sequence t8 (n) to be expressed in
terms of members of the sequence r8 (n), namely
1
t8 (n) = 10
(r8 (8n + 8) − 16r8 (2n + 2)) ,
2 × 32
eq. (6.1) does not allow us to express members of the sequence t9 (n) solely
in terms of the sequence r9 (n). For this reason, the results for partitions of
k are less interesting when k ≥ 9, and we simply state the following result
for the record:
Proposition 6.1. Suppose 0 ≤ j ≤ 7 and k is a non-negative integer. Then
(
)
k
X
t8k+j (n) = [q 8k+j ] c0 ϕ(q)8k+j +
ci ϕ(q 4 )8i (qψ(q 8 ))8(k−i)+j ,
i=1
where c0 , . . . , ck are (real) constants which depend on j and k.
References
[1] C. Adiga, S. Cooper and J. H. Han, A general relation between sums of squares and
sums of triangular numbers, Int. Journal of Number Theory, 1 (2005), 175–182.
[2] P. Barrucand, S. Cooper and M. D. Hirschhorn, Relations between squares and triangles, Discrete Math. 248 (2002), no. 1-3, 245–247.
[3] P. T. Bateman and M. I. Knopp, Some new old-fashioned modular identities, Ramanujan J. 2 (1998), no. 1-2, 247–269.
[4] P. T. Bateman, B. A. Datskovsky and M. I. Knopp, Sums of squares and the preservation of modularity under congruence restrictions, Symbolic computation, number
theory, special functions, physics and combinatorics (Gainesville, FL, 1999), 59–71,
Dev. Math., 4, Kluwer Acad. Publ., Dordrecht, 2001.
[5] B. C. Berndt, Ramanujan’s Notebooks, Part III, Springer-Verlag, New York, 1991.
[6] B. C. Berndt, Ramanujan’s Notebooks, Part V, Springer-Verlag, New York, 1998.
[7] S. Cooper and M. D. Hirschhorn, A combinatorial proof of a result from number
theory, Integers 4 (2004), A9, 4 pp. (electronic).
[8] S. Cooper and M. D. Hirschhorn, Results of Hurwitz type for three squares, Discrete
Math. 274, (2004), 9–24.