Jul 2, 2010 - Keywords: Representations, triangular numbers, squares, pentagonal num- bers, octagonal numbers. Abstract. In a recent paper, I began with ...
The number of representations of a number by various forms involving triangles, squares, pentagons and octagons Michael D. Hirschhorn July 2, 2010 Dedicated to the memory of K. Venkatachaliengar on the occasion of the Centenary of his birth. Keywords: Representations, triangular numbers, squares, pentagonal numbers, octagonal numbers. Abstract In a recent paper, I began with four classical results due to Jacobi, Dirichlet and Lorenz which give the number of representations of a number by various forms involving squares in terms of divisor functions. I deduced another sixteen similar representation theorems involving triangular numbers and/or squares, including celebrated results of Legendre and Ramanujan (and omitted a seventeenth, given here). In this paper I deduce a further twenty–nine representation theorems involving triangular numbers, squares, pentagonal numbers and octagonal numbers.
1
Introduction
In a recent paper[1], I began with four classical results due to Jacobi, Dirichlet and Lorenz which give the number of representations of a number by various forms involving squares in terms of divisor functions. I deduced another sixteen similar representation theorems involving triangular numbers and/or squares, including celebrated results of Legendre and Ramanujan (and omitted a seventeenth, given here as (3.21)). In this paper I deduce a further twenty–nine representation theorems involving triangular numbers, squares, pentagonal numbers (numbers of the form (3k 2 ± k)/2) and octagonal numbers (numbers of the form 3k 2 ± 2k). These formulas all follow easily from results in [1].
1
Since the appearance of [1], similar results have been given by H. Y. Lam [2] and R. S. Melham [3]. For more than 300 further results (most as yet unproven) the reader is referred to Melham [4]. For integral x, let dr,m (x) denote the number of divisors d of x with d ≡ r (mod m), σ(x) the sum of the divisors of x, both 0 if x is not an integer. In what follows, for example, r{Π + 2Π}(n) means the number of representations of n as the sum of a pentagonal number and twice a pentagonal number. We prove the following results: (Note that Π stands for ‘pentagonal number’ and Ω for ‘octagonal number’.) For n ≥ 0, r{△ + Π}(n) = d1,6 (6n + 1) − d5,6 (6n + 1), r{△ + 4Π}(n) = d1,24 (24n + 7) + d19,24 (24n + 7) − d5,24 (24n + 7) − d23,24 (24n + 7), r{3△ + Π}(n) = d1,12 (12n + 5) − d11,12 (12n + 5), r{3△ + 2Π}(n) = d1,8 (24n + 11) − d7,8 (24n + 11), r{4△ + Π}(n) = d1,24 (24n + 13) + d19,24 (24n + 13) − d5,24 (24n + 13) − d23,24 (24n + 13),
(1.1) (1.2) (1.3) (1.4) (1.5)
r{6△ + Π}(n) = d1,8 (24n + 19) − d7,8 (24n + 19), (1.6) r{△ + 2Ω}(n) = d1,24 (24n + 19) + d13,24 (24n + 19) − d11,24 (24n + 19) − d23,24 (24n + 19), r{2△ + Ω}(n) = d1,12 (12n + 7) − d11,12 (12n + 7),
(1.7) (1.8)
r{3△ + Ω}(n) = d1,24 (24n + 17) + d19,24 (24n + 17) − d5,24 (24n + 17) − d23,24 (24n + 17), r{12△ + Ω}(n) = (d1,8 (6n + 11) + d3,8 (6n + 11) − d5,8 (6n + 11)
(1.9)
− d7,8 (6n + 11))/2, r{ + 2Π}(n) = d1,3 (12n + 1) − d2,3 (12n + 1),
(1.10) (1.11)
r{2 + Π}(n) = d1,3 (24n + 1) − d2,3 (24n + 1), r{3 + Π}(n) = d1,8 (24n + 1) + d3,8 (24n + 1) − d5,8 (24n + 1)
(1.12)
− d7,8 (24n + 1), (1.13) r{3 + 4Π}(n) = d1,8 (6n + 1) + d3,8 (6n + 1) − d5,8 (6n + 1) − d7,8 (6n + 1), (1.14) r{ + Ω}(n) = d1,3 (3n + 1) − d2,3 (3n + 1) + 2(d1,3 ((3n + 1)/4) − d2,3 ((3n + 1)/4)),
(1.15)
r{3 + Ω}(n) = d1,4 (3n + 1) − d3,4 (3n + 1), (1.16) r{3 + 2Ω}(n) = d1,8 (3n + 2) + d3,8 (3n + 2) − d5,8 (3n + 2) − d7,8 (3n + 2), (1.17) r{6 + Ω}(n) = d1,8 (3n + 1) + d3,8 (3n + 1) − d5,8 (3n + 1) − d7,8 (3n + 1), (1.18) r{Π + Π}(n) = d1,4 (12n + 1) − d3,4 (12n + 1), 2
(1.19)
r{Π + 2Π}(n) = d1,8 (24n + 3) − d7,8 (24n + 3) − (d1,8 ((8n + 1)/3) − d7,8 ((8n + 1)/3)), r{Π + 4Π}(n) = d1,8 (24n + 5) − d7,8 (24n + 5),
(1.20) (1.21)
r{Π + Ω}(n) = (d1,8 (24n + 9) + d3,8 (24n + 9) − d5,8 (24n + 9) − d7,8 (24n + 9) − (d1,8 ((8n + 3)/3) + d3,8 ((8n + 3)/3) − d5,8 ((8n + 3)/3) − d7,8 ((8n + 3)/3)))/2, (1.22) r{Π + 2Ω}(n) = d1,24 (24n + 17) + d13,24 (24n + 17) − d11,24 (24n + 17) − d23,24 (24n + 17), (1.23) r{2Π + Ω}(n) = d1,12 (12n + 5) − d11,12 (12n + 5), (1.24) r{4Π + Ω}(n) = (d1,8 (6n + 3) + d3,8 (6n + 3) − d5,8 (6n + 3) − d7,8 (6n + 3) − (d1,8 ((2n + 1)/3) + d3,8 ((2n + 1)/3) − d5,8 ((2n + 1)/3) − d7,8 ((2n + 1)/3)))/2, (1.25) r{Ω + Ω}(n) = d1,4 (3n + 2) − d3,4 (3n + 2),
(1.26)
r{Ω + 2Ω}(n) = (d1,8 (3n + 3) + d3,8 (3n + 3) − d5,8 (3n + 3) − d7,8 (3n + 3) − (d1,8 ((n + 1)/3) + d3,8 ((n + 1)/3) − d5,8 ((n + 1)/3) − d7,8 ((n + 1)/3)))/2, r{3△ + 3△ + Π + Π}(n) = σ(6n + 5)/6,
(1.27) (1.28)
r{3 + 3 + Ω + Ω}(n) = (σ(3n + 2) − 4σ((3n + 2)/4))/3.
(1.29)
2
Preliminary results
As usual, let φ(q) =
∞ X
2
qn ,
ψ(q) =
−∞
X
2
q (n
+n)/2
.
n≥0
We need the easy lemmas φ(q) = φ(q 9 ) + 2qΩ(q 3 ), 3
(2.1)
9
ψ(q) = Π(q ) + qψ(q ) where Ω(q) =
∞ X
2
q 3n
+2n
, Π(q) =
∞ X
(2.2)
2
q (3n
+n)/2
.
−∞
−∞
Note that the powers appearing in Π(q) are the pentagonal numbers and the powers appearing in Ω(q) are the octagonal numbers.
3
3
Proofs of theorems
We start with the results from [1]. They all hold for n ≥ 0, with the exception that (3.15), (3.16), (3.17) and (3.20) hold only for n ≥ 1. r{△ + △}(n) = d1,4 (4n + 1) − d3,4 (4n + 1),
(3.1)
r{△ + 2△}(n) = d1,8 (8n + 3) − d7,8 (8n + 3), r{△ + 3△}(n) = d1,3 (2n + 1) − d2,3 (2n + 1),
(3.2) (3.3)
r{△ + 4△}(n) = d1,8 (8n + 5) − d7,8 (8n + 5), r{△ + 12△}(n) = (d1,3 (8n + 13) − d2,3 (8n + 13))/2,
(3.4) (3.5)
r{3△ + 4△}(n) = (d1,3 (8n + 7) − d2,3 (8n + 7))/2, (3.6) r{△ + }(n) = d1,8 (8n + 1) + d3,8 (8n + 1) − d5,8 (8n + 1) − d7,8 (8n + 1), (3.7) r{△ + 2}(n) = d1,4 (8n + 1) − d3,4 (8n + 1), r{△ + 6}(n) = d1,3 (8n + 1) − d2,3 (8n + 1),
(3.8) (3.9)
r{2△ + }(n) = d1,4 (4n + 1) − d3,4 (4n + 1), r{2△ + 3}(n) = d1,3 (4n + 1) − d2,3 (4n + 1),
(3.10) (3.11)
r{3△ + 2}(n) = d1,3 (8n + 3) − d2,3 (8n + 3), (3.12) r{4△ + }(n) = d1,8 (2n + 1) + d3,8 (2n + 1) − d5,8 (2n + 1) − d7,8 (2n + 1), (3.13) r{6△ + }(n) = d1,3 (4n + 3) − d2,3 (4n + 3), r{ + }(n) = 4(d1,4 (n) − d3,4 (n)),
(3.14) (3.15)
r{ + 2}(n) = 2(d1,8 (n) + d3,8 (n) − d5,8 (n) − d7,8 (n)),
(3.16)
r{ + 3}(n) = 2(d1,3 (n) − d2,3 (n)) + 4(d1,3 (n/4) − d2,3 (n/4)), r{△ + △ + △ + △}(n) = σ(2n + 1),
(3.17) (3.18)
r{△ + △ + + }(n) = σ(4n + 1), r{ + + + }(n) = 8(σ(n) − 4σ(n/4)),
(3.19) (3.20)
to which we should add r{△ + △ + 2△ + 2△} = σ(4n + 3)/4.
(3.21)
(3.1) is equivalent to ψ(q)2 =
X
(d1,4 (4n + 1) − d3,4 (4n + 1))q n ,
n≥0
or, Π(q 3 )2 + 2qΠ(q 3 )ψ(q 9 ) + q 2 ψ(q 9 )2 =
X
(d1,4 (4n + 1) − d3,4 (4n + 1))q n .
n≥0
4
If we extract those terms in which the power of q is congruent to 0 modulo 3 and replace q 3 by q we obtain X Π(q)2 = (d1,4 (12n + 1) − d3,4 (12n + 1))q n n≥0
which yields (1.19), while if we extract those terms in which the power of q is congruent to 1 modulo 3, divide by 2q and replace q 3 by q, we obtain X ψ(q 3 )Π(q) = (d1,4 (12n + 5) − d3,4 (12n + 5))/2 q n n≥0
which gives (1.3). (3.2) is equivalent to (Π(q 3 ) + qψ(q 9 ))(Π(q 6 ) + q 2 ψ(q 18 )) =
X
(d1,8 (8n + 3) − d7,8 (8n + 3))q n .
n≥0
If we extract those terms in which the power of q is congruent to 0 modulo 3, and replace q 3 by q, we obtain X Π(q)Π(q 2 ) + qψ(q 3 )ψ(q 6 ) = (d1,8 (24n + 3) − d7,8 (24n + 3))q n , n≥0
while qψ(q 3 )ψ(q 6 ) =
X
(d1,8 (8n + 3) − d7,8 (8n + 3))q 3n+1
n≥0
=
X
(d1,8 ((8n + 1)/3) − d7,8 ((8n + 1)/3))q n .
n≥0
By subtraction, we obtain Π(q)Π(q 2 ) X = (d1,8 (24n + 3) − d7.8 (24n + 3) − (d1,8 ((8n + 1)/3) − d7,8 ((8n + 1)/3)))q n , n≥0
which gives (1.20). If from the above we extract those terms in which the power of q is congruent to 1 modulo 3, divide by q and replace q 3 by q we obtain X ψ(q 3 )Π(q 2 ) = (d1,8 (24n + 11) − d7,8 (24n + 11))q n n≥0
which gives (1.4), while if we extract those terms in which the power of q is 2 modulo 3, divide by q 2 and replace q 3 by q, we obtain X ψ(q 6 )Π(q) = (d1,8 (24n + 19) − d7,8 (24n + 19))q n n≥0
5
which gives (1.6). (3.3) is equivalent to (Π(q 3 ) + qψ(q 9 ))ψ(q 3 ) =
X
(d1,3 (2n + 1) − d2,3 (2n + 1))q n .
n≥0
If we extract those terms in which the power of q is congruent to 0 modulo 3 and replace q 3 by q, we obtain X ψ(q)Π(q) = (d1,3 (6n + 1) − d2,3 (6n + 1))q n , n≥0
which gives (1.1). (3.4) is equivalent to (Π(q 3 ) + qψ(q 9 ))(Π(q 12 ) + q 4 ψ(q 36 )) =
X
(d1,8 (8n + 5) − d7,8 (8n + 5))q n .
n≥0
If we extract those terms in which the power of q is congruent to 0 modulo 3 and replace q 3 by q we obtain X Π(q)Π(q 4 ) = (d1,8 (24n + 5) − d7,8 (24n + 5))q n , n≥0
which gives (1.21). (3.5) is equivalent to (Π(q 3 ) + qψ(q 9 ))ψ(q 12 ) =
X
(d1,3 (8n + 13) − d2,3 (8n + 13))/2 q n .
n≥0
If we extract those terms in which the power of q is congruent to 0 modulo 3 and replace q 3 by q, we obtain X ψ(q 4 )Π(q) = (d1,3 (24n + 13) − d2,3 (24n + 13))/2 q n , n≥0
from which (1.5) follows. (3.6) is equivalent to ψ(q 3 )(Π(q 12 ) + q 4 ψ(q 36 )) =
X
(d1,3 (8n + 7) − d2,3 (8n + 7))/2 q n .
n≥0
If we extract those terms in which the power of q is 0 modulo 3 and replace q 3 by q, we obtain X ψ(q)Π(q 4 ) = (d1,3 (24n + 7) − d2,3 (24n + 7))/2 q n , n≥0
6
which gives (1.2). (3.7) is equivalent to (Π(q 3 ) + qψ(q 9 ))(φ(q 9 ) + 2qΩ(q 3 )) X = (d1,8 (8n + 1) + d3,8 (8n + 1) − d5,8 (8n + 1) − d7,8 (8n + 1))q n , n≥0
If we extract those terms in which the power of q is congruent to 0 modulo 3 and replace q 3 by q, we obtain X φ(q 3 )Π(q) = (d1,8 (24n + 1) + d3,8 (24n + 1) − d5,8 (24n + 1) − d7,8 (24n + 1))q n , n≥0
which gives (1.13), while if we extract those terms in which the power of q is congruent to 2 modulo 3, divide by 2q 2 and replace q 3 by q, we obtain ψ(q 3 )Ω(q) X = (d1,8 (24n + 17) + d3,8 (24n + 17) − d5,8 (24n + 17) − d7,8 (24n + 17))/2 q n , n≥0
which gives (1.9). If we extract those terms in which the power of q is congruent to 1 modulo 3, divide by q and replace q 3 by q, we obtain 2Π(q)Ω(q) + ψ(q 3 )φ(q 3 ) X = (d1,8 (24n + 9) + d3,8 (24n + 9) − d5,8 (24n + 9) − d7,8 (24n + 9))q n n≥0
while ψ(q 3 )φ(q 3 ) =
X
(d1,8 (8n + 1) + d3,8 (8n + 1) − d5,8 (8n + 1) − d7,8 (8n + 1))q 3n
n≥0
=
X
(d1,8 ((8n + 3)/3) + d3,8 ((8n + 3)/3) − d5,8 ((8n + 3)/3)
n≥0
− d7,8 ((8n + 3)/3))q n . If we subtract the second from the first and divide by 2, we obtain X Π(q)Ω(q) = (d1,8 (24n + 9) + d3,8 (24n + 9) − d5,8 (24n + 9) − d7,8 (24n + 9) n≥0
− (d1,8 ((8n + 3)/3) + d3,8 ((8n + 3)/3) − d5,8 ((8n + 3)/3) − d7,8 ((8n + 3)/3)))/2 q n , which gives (1.22). (3.8) is equivalent to (Π(q 3 ) + qψ(q 9 ))(φ(q 18 ) + 2q 2 Ω(q 6 )) =
X
(d1,4 (8n + 1) − d3,4 (8n + 1))q n .
n≥0
7
If we extract those terms in which the power of q is congruent to 2 modulo 3, divide by 2q 2 and replace q 3 by q, we obtain X Π(q)Ω(q 2 ) = (d1,4 (24n + 17) − d3,4 (24n + 17))/2 q n , n≥0
which gives (1.23). (3.9) is equivalent to (Π(q 3 ) + qψ(q 9 ))φ(q 6 ) =
X
(d1,3 (8n + 1) − d2,3 (8n + 1))q n .
n≥0
If we extract those terms in which the power of q is congruent to 0 modulo 3 and replace q 3 by q, we obtain X φ(q 2 )Π(q) = (d1,3 (24n + 1) − d2,3 (24n + 1))q n , n≥0
which gives (1.12). (3.10) is equivalent to (Π(q 6 ) + q 2 ψ(q 18 ))(φ(q 9 ) + 2qΩ(q 3 )) =
X
(d1,4 (4n + 1) − d3,4 (4n + 1))q n .
n≥0
If we extract those terms in which the power of q is congruent to 1 modulo 3, divide by 2q and replace q 3 by q, we obtain X Π(q 2 )Ω(q) = (d1,4 (12n + 5) − d3,4 (12n + 5))q n , n≥0
which gives (1.24). (3.11) is equivalent to (Π(q 6 ) + q 2 ψ(q 18 ))φ(q 3 ) =
X
(d1,3 (4n + 1) − d2,3 (4n + 1))q n .
n≥0
If we extract those terms in which the power of q is congruent to 0 modulo 3 and replace q 3 by q, we obtain X φ(q)Π(q 2 ) = (d1,3 (12n + 1) − d2,3 (12n + 1))q n , n≥0
which gives (1.11). (3.12) is equivalent to ψ(q 3 )(φ(q 18 ) + 2q 2 Ω(q 6 )) =
X
(d1,3 (8n + 3) − d2,3 (8n + 3))q n .
n≥0
8
If we extract those terms in which the power of q is congruent to 2 modulo 3, divide by 2q 2 and replace q 3 by q, we obtain X ψ(q)Ω(q 2 ) = (d1,3 (24n + 19) − d2,3 (24n + 19))/2 q n , n≥0
which gives (1.7). (3.13) is equivalent to (Π(q 12 ) + q 4 ψ(q 36 ))(φ(q 9 ) + 2qΩ(q 3 )) X = (d1,8 (2n + 1) + d3,8 (2n + 1) − d5,8 (2n + 1) − d7,8 (2n + 1))q n . n≥0
If we extract those terms in which the power of q is congruent to 0 modulo 3 and replace q 3 by q, we obtain X φ(q 3 )Π(q 4 ) = (d1,8 (6n + 1) + d3,8 (6n + 1) − d5,8 (6n + 1) − d7,8 (6n + 1))q n , n≥0
which gives (1.14), while if we extract those terms in which the power of q is congruent to 2 modulo 3, divide by 2q 5 and replace q 3 by q, we obtain X ψ(q 12 )Ω(q) = (d1,8 (6n+11)+d3,8(6n+11)−d5,8(6n+11)−d7,8(6n+11))/2 q n n≥0
which gives (1.10). If we extract those terms in which the power of q is 1 modulo 3, divide by q and replace q 3 by q, we obtain 2Π(q 4 )Ω(q) + qψ(q 12 )φ(q 3 ) X = (d1,8 (6n + 3) + d3,8 (6n + 3) − d5,8 (6n + 3) − d7,8 (6n + 3))q n , n≥0
while qψ(q 12 )φ(q 3 ) X = (d1,8 (2n + 1) + d3,8 (2n + 1) − d5,8 (2n + 1) − d7,8 (2n + 1))q 3n+1 n≥0
=
X
(d1,8 ((2n + 1)/3) + d3,8 ((2n + 1)/3) − d5,8 ((2n + 1)/3)
n≥0
− d7,8 ((2n + 1)/3))q n . If we subtract the second from the first and divide by 2, we obtain Π(q 4 )Ω(q) 9
=
X
(d1,8 (6n + 3) + d3,8 (6n + 3) − d5,8 (6n + 3) − d7,8 (6n + 3)
n≥0
− (d1,8 ((2n + 1)/3) + d3,8 ((2n + 1)/3) − d5,8 ((2n + 1)/3) − d7,8 ((2n + 1)/3)))/2 q n , which gives (1.25). (3.14) is equivalent to ψ(q 6 )(φ(q 9 ) + 2qΩ(q 3 )) =
X
(d1,3 (4n + 3) − d2,3 (4n + 3))q n .
n≥0
If we extract those terms in which the power of q is congruent to 1 modulo 3, divide by 2q and replace q 3 by q, we obtain X ψ(q 2 )Ω(q) = (d1,3 (12n + 7) − d2,3 (12n + 7))/2 q n , n≥0
which gives (1.8). (3.15) is (essentially) equivalent to (φ(q 9 ) + 2qΩ(q 3 ))2 = 1 + 4
X
(d1,4 (n) − d3,4 (n))q n .
n≥1
If we extract those terms in which the power of q is congruent to 1 modulo 3, divide by 4q and replace q 3 by q, we obtain X φ(q 3 )Ω(q) = (d1,4 (3n + 1) − d3,4 (3n + 1))q n , n≥0
which gives (1.16), while if we extract those terms in which the power of q is congruent to 2 modulo 3, divide by 4q 2 and replace q 3 by q, we obtain X Ω(q)2 = (d1,4 (3n + 2) − d3,4 (3n + 2))q n , n≥0
which gives (1.26). (3.16) is (essentially) equivalent to (φ(q 9 ) + 2qΩ(q 3 ))(φ(q 18 ) + 2q 2 Ω(q 6 )) X =1+2 (d1,8 (n) + d3,8 (n) − d5,8 (n) − d7,8 (n))q n . n≥1
If we extract those terms in which the power of q is congruent to 1 modulo 3, divide by 2q and replace q 3 by q, we obtain X φ(q 6 )Ω(q) = (d1,8 (3n + 1) + d3,8 (3n + 1) − d5,8 (3n + 1) − d7,8 (3n + 1))q n , n≥0
10
which gives (1.18), while if we extract those terms in which the power of q is congruent to 2 modulo 3, divide by 2q 2 and replace q 3 by q, we obtain X φ(q 3 )Ω(q 2 ) = (d1,8 (3n + 2) + d3,8 (3n + 2) − d5,8 (3n + 2) − d7,8 (3n + 2))q n , n≥0
which gives (1.17). If we extract those terms in which the power of q is congruent to 0 modulo 3 and replace q 3 by q, we obtain X φ(q 3 )φ(q 6 )+4qΩ(q)Ω(q 2 ) = 1+2 (d1,8 (3n)+d3,8 (3n)−d5,8 (3n)−d7,8 (3n))q n , n≥1
while φ(q 3 )φ(q 6 ) = 1 + 2
X
(d1,8 (n) + d3,8 (n) − d5,8 (n) − d7,8 (n))q 3n
n≥1
=1+2
X
(d1,8 (n/3) + d3,8 (n/3) − d5,8 (n/3) − d7,8 (n/3))q n .
n≥0
If we subtract the second from the first and divide by 4q, we obtain Ω(q)Ω(q 2 ) X = (d1,8 (3n + 3) + d3,8 (3n + 3) − d5,8 (3n + 3) − d7,8 (3n + 3) n≥0
− (d1,8 ((n + 1)/3) + d3,8 ((n + 1)/3) − d5,8 ((n + 1)/3) − d7,8 ((n + 1)/3)))/2 q n , which gives (1.27). (3.17) is (essentially) equivalent to (φ(q 9 ) + 2qΩ(q 3 ))φ(q 3 ) = 1 + 2
X
(d1,3 (n) − d2,3 (n))q n
n≥1
+4
X
(d1,3 (n/4) − d2,3 (n/4))q n .
n≥1
If we extract those terms in which the power of q is congruent to 1 modulo 3, divide by 2q and replace q 3 by q, we obtain X φ(q)Ω(q) = (d1,3 (3n + 1) − d2,3 (3n + 1))q n n≥0
+2
X
(d1,3 ((3n + 1)/4) − d2,3 ((3n + 1)/4))q n ,
n≥0
which gives (1.15). 11
(3.18) is equivalent to (Π(q 3 ) + qψ(q 9 ))4 =
X
σ(2n + 1)q n .
n≥0
If we extract those terms in which the power of q is congruent to 2 modulo 3, divide by 6q 2 and replace q 3 by q, we obtain X ψ(q 3 )2 Π(q)2 = σ(6n + 5)/6 q n , n≥0
which gives (1.28). (3.20) is (essentially) equivalent to (φ(q 9 ) + 2qΩ(q 3 ))4 = 1 + 8
X
(σ(n) − 4σ(n/4))q n .
n≥1
If we extract those terms in which the power of q is congruent to 2 modulo 3, divide by 24q 2 and replace q 3 by q, we obtain X φ(q 3 )2 Ω(q)2 = (σ(3n + 2) − 4σ((3n + 2)/4))/3 q n , n≥0
which gives (1.29). References [1] M. D. Hirschhorn, The number of representations of a number by various forms, Discrete Math. 298 (2005) 205–211. [2] H. Y. Lam, The number of representations by sums of squares and triangular numbers, Integers 7(1) (2007) #A28. [3] R. S. Melham, Analogues of two classical theorems on the representations of a number, Integers 8(1) (2008) #A51. [4] R. S. Melham, Analogues of Jacobi’s two–square theorem, http://maths.science.uts.edu.au/maths/wiki/RayMelham
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