The tri-pentagonal number theorem and related identities

0.19 MB·English

by  Alexander Berkovich

THE TRI-PENTAGONAL NUMBER THEOREM AND RELATED
IDENTITIES
8
0 ALEXANDERBERKOVICH
0
2
Abstract. IrevisitanautomatedproofofAndrews’pentagonalnumberthe-

n orem found by Riese. I uncover a simple polynomial identity hidden behind
a his proof. I explain how to use this identity to prove Andrews’ result along
J withavarietyofnewformulasofsimilartype. Irevealaninterestingrelation

0 betweenthetri-pentagonaltheoremanditems(19),(20),(94),(98)onthecel-
2 ebrated Slater list. Finally,Iestablishanew infinitefamilyofmultipleseries
identities.

]
T
N
. 1. Introduction
h
t
a The Gaussian or q-binomial coefficients are polynomials in q defined by
m
[ n+m := ((qq))nn(+q)mm, if n,m∈N,
1 (cid:20) n (cid:21)q (0, otherwise.
v Here (q) = n (1 qj). We shall require the more general q-shifted factorials
8 n j=1 −
defined by
0
Q
0
1, if n=0,
3
1. (a)n =(a;q)n := jn=−01(1−aqj) if n>0,
0 Q−j=n1 1−a1q−j if n<0.
8
0 We note that Q
1
: =0, if n<0,
v (q)
n
i
X and that
L 1
r lim = .
a L→∞(cid:20)j(cid:21)q (q)j
Here and hereafter q <1. We shall also use the following notations
| |
k
(a ,a ,...,a ) =(a ,a ,...,a ;q) := (a ) ,
1 2 k n 1 2 k n i n
i=1
Y
(a) =(a;q) := lim (a) ,
∞ ∞ n
n→∞
q
[z;q] :=(z, ) .
∞ ∞
z
2000 Mathematics Subject Classification. 33D15,11B65.
Key words and phrases. multidimensional q-binomial identities, quintuple product identity,
Baileylemmas,rationalfunctionidentities,recursionrelations.
ResearchwassupportedinpartbyNSAgrantH98230-07-01-0011.
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2 ALEXANDERBERKOVICH
The literature on q-series abounds with numerous identities of the type
∞
2L
A(i,q) =B(L,q). (1.1)
L i
i=−∞ (cid:20) − (cid:21)q
X
For example,
∞
(−1)jq(2j) L2Lj =δL,0, (1.2)
j=−∞ (cid:20) − (cid:21)q
X
∞
( 1)jq(j2) 2L =( 1)LqL2(q;q2)L, (1.3)
− L j −
j=−∞ (cid:20) − (cid:21)q2
X
∞
( 1)jq2(2j) 2L =qL(q;q2)L. (1.4)
− L j
j=−∞ (cid:20) − (cid:21)q
X
Here δ is the Kronecker delta and i := i(i−1).
i,j 2 2
The observant reader might have recognized (1.2) as a special case of
(cid:0) (cid:1)
∞
(−1)jzjq(j+21) 2LL+ja =(1z)L+a(qz)L, (1.5)
j=−∞ (cid:20) − (cid:21)q
X
where a N. The above formula is due to Cauchy. It is a finite form of the
∈
celebrated Jacobi triple product identity
∞
( 1)jzjq(j+21) =(q)∞[qz;q]∞. (1.6)
−
j=−∞
X
As for the formulas (1.3) and (1.4), they are, essentially, items G(4) and E(3),
respectively, in Slater’s table [9].
Severalyearsago,Andrews[1]revisitedtheumbralmethodsusedbyL.J.Rogers.
In [1], he discussed multidimensional identities of the form
2L
A(L,q) =F(L,q),
L+i
i (cid:20) (cid:21)q
X
where L=(L ,L ,...,L ), i=(i ,i ,...,i ) and
1 2 d 1 2 d
d
2L 2L
k
= .
L+i L +i
(cid:20) (cid:21)q k=1(cid:20) k k(cid:21)q
Y
In particular, he proved that
( 1)i+j+kq(i+j2+k) 2L 2M 2N
− L i M j N k
iX,j,k (cid:20) − (cid:21)q(cid:20) − (cid:21)q(cid:20) − (cid:21)q (1.7)
(q) (q) (q)
= 2L 2M 2N ,
(q) (q) (q)
L+M−N L+N−M M+N−L
and that
( 1)iq(i+2j) 2L 2M =( 1)L(q;q2)L(q2;q2)M(−1)M−L. (1.8)
− L i M j − (q)
i,j (cid:20) − (cid:21)q(cid:20) − (cid:21)q M−L
X
THE TRI-PENTAGONAL NUMBER THEOREM AND RELATED IDENTITIES 3
We note that (1.7) is a three-dimensional generalization of (1.2). Indeed, if we let
N =0, we obtain that
(−1)i+jq(i+2j) L2Li M2Mj =(q)2LδL,M. (1.9)
i,j (cid:20) − (cid:21)q(cid:20) − (cid:21)q
X
If we now set M = 0 we end up with (1.2), as claimed. Also, (1.8) with M = 0
reduces to (1.2). On the other hand, (1.8) with L = 0 becomes (1.5) with z =
1, a=0.
−
In what follows, we will use a small variant of (1.9)
(−1)i+jq(i+j2+1) 2LL+i1 2MM +j1 =−(q)2L+1δL,M. (1.10)
i,j (cid:20) − (cid:21)q(cid:20) − (cid:21)q
X
To verify that (1.10) holds for M =0 (or L=0) we use (1.5) twice as follows
( 1)i+jq(i+j2+1) 2L+1 1 =
− L i j
i,j (cid:20) − (cid:21)q(cid:20)− (cid:21)q
X
( 1)iq(i+21) 2L+1 ( 1)iq(2i) 2L+1 =
− L i − − L i
i (cid:20) − (cid:21)q i (cid:20) − (cid:21)q
X X
(1) (q) (q) (1) =0 (q) δ = (q) δ .
L+1 L L+1 L 1 L,0 2L+1 L,0
− − −
In [8], Riese used his qMultiSum package to provide a simple recurrence proof of
(1.7). In the next section, I will rederive and generalize Riese’s recurrences. As a
bonus, I will get a uniform proof of (1.7) – (1.10). Moreover, I will show that the
same proof can be employed to establish four new identities:
( 1)iq(i+j)2 2L 2M
− L i M j
Xi,j (cid:20) − (cid:21)q2(cid:20) − (cid:21)q2 (1.11)
(q;q2)
=( 1)M L−M (q2;q4) (q2;q4) ,
− ( q;q2) L M
L−M
−
( 1)i+jq(i+2j) 2L 2M =( 1)L+Mq(L−M)2 (q)2M , (1.12)
− L i M j − (q)
i,j (cid:20) − (cid:21)q(cid:20) − (cid:21)q2 M−L
X
( 1)i+jq2(i+2j) 2L 2M
− L i M j
i,j (cid:20) − (cid:21)q(cid:20) − (cid:21)q2 (1.13)
X
=qL−M(q;q2) ( q;q2) (q2(L+1−M);q2) ,
L 2M M
−
and
( 1)i+j+kq(i+j2+k) 2L 2M 2N
− L i M j N k
i,j,k (cid:20) − (cid:21)q2(cid:20) − (cid:21)q2(cid:20) − (cid:21)q2 (1.14)
X
=(q;q2) (q;q2) (q;q2) .
L+M−N L+N−M M+N−L
We remark that (1.14) is a perfect quadratic analogue of (1.7). Setting L = 0 in
(1.11) yields (1.5) with q q2 and a = 0, z = 1. Setting M = 0 there yields
→ −q
(1.5) with q q2, a=0, z = 1. Analogously, we can verify that (1.12) reduces to
→ q
(1.2)and(1.3)andthat(1.13)reducesto(1.4)and(1.5)withq q2, a=0, z =1.
→
4 ALEXANDERBERKOVICH
Finally,(1.14)isathree-dimensionalgeneralizationof(1.3). Indeed,ifweletN =0
in (1.14) we get
(−1)i+jq(i+2j) L2Li M2Mj =(q;q2)L+M(−1)M+Lq(L−M)2. (1.15)
i,j (cid:20) − (cid:21)q2(cid:20) − (cid:21)q2
X
Letting M =0 in (1.15) yields (1.3).
The remainder of this manuscript is organized as follows. In the next section, I
willshowhowtouseasimplepolynomialidentitytoprove(1.7)–(1.15). InSection
3,Iwillemploysomewell-knownq-binomialtransformationstoderivenewtwoand
three-dimensional identities. In Section 4, I will prove, among other results, that
for k 0
≥
qN12+N22+···+Nk2+1
(q) (q) (q) (q2;q2) (q;q2)
n1,...X,nk+1≥0 n1 n2··· nk nk+1 nk+nk+1 (1.16)
(q4+6k;q4+6k) [q2+2k;q4+6k]
∞ ∞
= ,
(q) [q1+k;q4+6k]
∞ ∞
where N :=n +n + +n , 1 i k+1.
i i i+1 k+1
··· ≤ ≤
Finally, in Section 5, I will provide fresh insights into the Tri-Pentagonal
Theorem 1. (Andrews)
1 ( 1)i+j+kq(i+j2+k)+i2+j2+k2 = qi2+j2+k2 . (1.17)
(q)3 − (q) (q) (q)
∞ i,j,k i,j,k≥0 i+j−k i+k−j j+k−i
X X
Recently, this theorem was given a very interesting partition theoretical inter-
pretation in [3].
2. Simple polynomial identity and its implications
It is a fair statement that one does not need a computer to check that
(1 x )(1 x q)+(1 x )(1 x q)
1 1 2 2
− − − −
(1 x x q)(1 x x q2)
1 2 1 2
− − − (2.1)
(1 x )(1 x q)(1 x )(1 x q)
1 1 2 2
− − − − −
+x x (1+q)(1 x q)(1 x q)=0
1 2 1 2
− −
holds true for any x ,x ,q R.
1 2
∈
Next, we divide (2.1) by (1 x x q)(1 x x q2) and let x = qL+a+i, x = qL−i
1 2 1 2 1 2
− −
to get
(1 qL+a+i)(1 qL+a+i+1)
− −
(1 q2L+a+1)(1 q2L+a+2)
− −
(1 qL−i)(1 qL−i+1)
+ − −
(1 q2L+a+1)(1 q2L+a+2)
− − (2.2)
(1 qL+a+i)(1 qL+a+i+1)(1 qL−i)(1 qL−i+1)
− − − −
− (1 q2L+a+1)(1 q2L+a+2)
− −
(1 qL+a+i+1)(1 qL−i+1)
+q2L+a(1+q) − − =1.
(1 q2L+a+1)(1 q2L+a+2)
− −
THE TRI-PENTAGONAL NUMBER THEOREM AND RELATED IDENTITIES 5
2(L+1)+a
If we multiply (2.2) by we deduce that
(L+1) i
(cid:20) − (cid:21)q
2L+a 2L+a 2L+a
+ +q2L+a(1+q)
L (i 1) L (i+1) L i
(cid:20) − − (cid:21)q (cid:20) − (cid:21)q (cid:20) − (cid:21)q (2.3)
2(L 1)+a 2(L+1)+a
(1 q2L+a)(1 q2L+a−1) − = .
− − − (L 1) i (L+1) i
(cid:20) − − (cid:21)q (cid:20) − (cid:21)q
We now define
L,M L,M,q ,q 2L+a 2M +a
f =f 1 2 := .
a a
i,j i,j L i M j
(cid:18) (cid:19) (cid:18) (cid:19) (cid:20) − (cid:21)q1(cid:20) − (cid:21)q2
Obviously, (2.3) implies that
L,M L,M L,M
f +f +q2L+a(1+q )f
a i 2,j a i,j 1 1 a i 1,j
(cid:18) − (cid:19) (cid:18) (cid:19) (cid:18) − (cid:19) (2.4)
L 1,M L+1,M
(1 q2L+a)(1 q2L+a−1)f − =f
− − 1 − 1 a i 1,j a i 1,j
(cid:18) − (cid:19) (cid:18) − (cid:19)
and that
L,M L,M L,M
f +f +q2M+a(1+q )f
a i,j 2 a i,j 2 2 a i,j 1
(cid:18) − (cid:19) (cid:18) (cid:19) (cid:18) − (cid:19) (2.5)
L,M 1 L,M +1
(1 q2M+a)(1 q2M+a−1)f − =f .
− − 2 − 2 a i,j 1 a i,j 1
(cid:18) − (cid:19) (cid:18) − (cid:19)
It is easy to combine (2.4) and (2.5) as follows
L,M L,M
f +q2L+a(1+q )f
a i 2,j 1 1 a i 1,j
(cid:18) − (cid:19) (cid:18) − (cid:19)
L 1,M L+1,M
(1 q2L+a)(1 q2L+a−1)f − f
− − 1 − 1 a i 1,j − a i 1,j
(cid:18) − (cid:19) (cid:18) − (cid:19) (2.6)
L,M L,M
=f +q2M+a(1+q )f
a i,j 2 2 2 a i,j 1
(cid:18) − (cid:19) (cid:18) − (cid:19)
L,M 1 L,M +1
(1 q2M+a)(1 q2M+a−1)f − f .
− − 2 − 2 a i,j 1 − a i,j 1
(cid:18) − (cid:19) (cid:18) − (cid:19)
To proceed, we require one more definition
L,M,q ,q
F (L,M)=F (L,M,x,y,q ,q ,q ):= xiyjqP(i+j)f 1 2 , (2.7)
a a 0 1 2 0 a i,j
i,j (cid:18) (cid:19)
X
where P(z) is some polynomial in z.
Clearly,
L,M L,M
y2 xiyjqP(i+j−1)f =x2 xiyjqP(i+j−1)f . (2.8)
0 a i 2,j 0 a i,j 2
i,j (cid:18) − (cid:19) i,j (cid:18) − (cid:19)
X X
6 ALEXANDERBERKOVICH
Next, we multiply (2.6) by xiyjqP(i+j−1) and sum over i,j. Taking advantage of
0
(2.8) we derive that
x q2L+a(1+q )F (L,M) (1 q2L+a)(1 q2L+a−1)F (L 1,M)
1 1 a − − 1 − 1 a −
(cid:8) Fa(L+1,M)
− (2.9)
=y q22M+a(1+q2)Fa(L,M)−(1−q22M+(cid:9)a)(1−q22M+a−1)Fa(L,M −1)
(cid:8) Fa(L,M +1) ,
−
provided x = y . (cid:9)
| | | |
Observethatfora=0,1the recurrence(2.9)togetherwiththe boundaryvalues
F (L,0), L 0 and F (0,M), M 0 specifies F (L,M) completely for L
a a a
≥ ≥ ≥
0, M 0.
≥
Itisplainthatthelefthandsidesin(1.8)–(1.13)and(1.15)areoftheform(2.7)
with x = y = 1. It is also straightforward to check that the right hand sides
| | | |
there satisfy (2.9). This implies that (1.8)–(1.13) and (1.15) hold true if they hold
when L 0, M = 0 and L = 0, M 0. But this is indeed the case as we saw in
≥ ≥
the Introduction.
Fortunately, more is true. Define
Ym(L,M,N):= ( 1)i+j+kq(i+j2+k) 2L 2M 2N . (2.10)
− L+i M +j N +k
i,j,k (cid:20) (cid:21)qm(cid:20) (cid:21)qm(cid:20) (cid:21)qm
X
Clearly,Y andY arethe lefthandsidesof(1.7)and(1.14),respectively. Multiply
1 2
(2.6) with q1 = q2 = qm by (−1)i+j+kq(i+j+2k−1) N2Nk and sum over i,j,k to
(cid:20) − (cid:21)qm
derive that
q2Lm(1+qm)Y (L,M,N) (1 q2Lm)(1 q2Lm−m)Y (L 1,M,N)
m m
− − − −
Y (L+1,M,N)
m
−
=q2Mm(1+qm)Y (L,M,N) (1 q2Mm)(1 q2Mm−m)Y (L,M 1,N)
m m
− − − −
Y (L,M +1,N).
m
−
(2.11)
We remark that (2.11) with m = 1 is, essentially, the recurrence derived by Riese
in [8]. Once again, (2.11) together with the boundary values Y (0,M,N), M
m
≥
0, N 0 and Y (L,0,N), L 0, N 0 specify Y (L,M,N) completely for
m m
≥ ≥ ≥
L,M,N 0. Next, we check that the right hand sides of (1.7) and (1.14) satisfy
≥
(2.11) with m = 1 and m = 2, respectively. Moreover, on the boundary these
identities reduce to the two proven identities: (1.9) and (1.15). And so, (1.7) and
(1.14) hold true, as claimed.
The reader may wonder if the polynomial identity (2.1) can be extended to n
variables : x ,x ,x ,...,x . This is indeed possible. The following generalization
1 2 3 n
was suggested to me by Alain Lascoux:
n n n n−1
(x ) + ( 1)t(qt−1x x x ) = (qx ) (qt ( 1)t)e ,
i 2 − i1 i2··· it 2 i 1 − − t+1
iY=1 Xt=1i1<iX2<···<it Yi=1 Xt=1
(2.12)
where e ’s are the elementary symmetric functions in x ,x ,...,x .
i 1 2 n
THE TRI-PENTAGONAL NUMBER THEOREM AND RELATED IDENTITIES 7
3. q-binomial transformations
We begin by recalling some well-known formulas
qr2 (q)2L 2r =qj2 2L , (3.1)
(q) (q) r j L j
r≥0 L−r 2r(cid:20) − (cid:21)q (cid:20) − (cid:21)q
X
1 L 2r 2L
qL−2r( ;q2) = , (3.2)
−q L−2r 2r r j L 2j
r≥0 (cid:20) (cid:21)q2(cid:20) − (cid:21)q4 (cid:20) − (cid:21)q
X
L 2r+a 2L
q2r(r+a)( q) =q2j(j+a) , (3.3)
− L−2r−a 2r+a r j L 2j a
r≥0 (cid:20) (cid:21)q2(cid:20) − (cid:21)q2 (cid:20) − − (cid:21)q
X
where a=0,1.
Weremarkthat(3.1)wasusedbyBressoud[5]togiveasimpleproofoftheRogers–
Ramanujan identities. It can be recognized as a special case of the Bailey Lemma
initsversionduetoAndrews[2]andPaule[7]. Thetransformations(3.2)and(3.3)
were introduced by Berkovichand Warnaar in [4].
In [1], Andrews applied (3.1) to (1.7) three times to obtain the tri-pentagonal
theorem (1.17). It is interesting that a single application of (3.1) to (1.7) yields a
new three-dimensional identity
( 1)i+j+kq(i+j2+k)+i2 2L 2M 2N
− L i M j N k
i,j,k (cid:20) − (cid:21)q(cid:20) − (cid:21)q(cid:20) − (cid:21)q
X (3.4)
=q(N−M)2(q) 2L .
L+M+N L+N M
(cid:20) − (cid:21)q
Indeed, we have that
( 1)i+j+kq(i+j2+k)+i2 2L 2M 2N
− L i M j N k
i,j,k (cid:20) − (cid:21)q(cid:20) − (cid:21)q(cid:20) − (cid:21)q
X (3.5)
qr2 (q) (q) (q)
= 2L 2M 2N .
(q) (q) (q) (q)
L−r r+M−N r+N−M M+N−r
r≥0
X
The right hand side of (3.5) can be written in the form
qn2(q) (q) (q) (qn−L,qn−M−N)
RHS(3.5)= 2L 2M 2N θ(L n) rqr(L+M+N+1),
(q) (q) (q) ≥ (q,q2n+1)
2n L−n M+N−n r
r≥0
X
(3.6)
where n:= N M and
| − |
1, if L n,
θ(L n)= ≥ (3.7)
≥ (0, otherwise.
The sum in (3.6) can be evaluated by the q-Chu-Vandermonde formula [[6], (II.7)]
as
(q)
M+N+L .
(q2n+1) (q)
M+N−n n+L
8 ALEXANDERBERKOVICH
And so,
qn2(q) (q) (q) (q)
RHS(3.5)= 2L 2M 2N M+N+L
(q) (q) (q) (q)
M+N+n M+N−n L−n L+n
=q(N−M)2 2L (q) ,
L+N M L+M+N
(cid:20) − (cid:21)q
as claimed.
Obviously,(3.2)and(3.3)canalsobeemployedtoproducenewtwo-dimensional
identities. For example, we can replace q by q4 in (1.9) and apply (3.2) to obtain
(−1)i+jq4(i+2j) L2Li M2M2j =qM−2L(−1q;q2)M−2L 2ML (q4;q4)2L.
i,j (cid:20) − (cid:21)q4(cid:20) − (cid:21)q (cid:20) (cid:21)q2
X
(3.8)
Or we can replace q by q2 in (1.9) and apply (3.3) with a=0 twice to get
( 1)i+jq(i+j)(i+j+1)+2i2+2j2 2L 2M
− L 2i M 2j
i,j (cid:20) − (cid:21)q(cid:20) − (cid:21)q
X
= qr2 (q2;q2)L(q2;q2)M . (3.9)
(q2;q2) (q) (q)
r L−r M−r
r≡0 mod2,
X
r≥0
Analogously, replacing q by q2 in (1.10) and using (3.3) with a=1 we obtain that
2L 2M
q ( 1)i+jq(i+j)(i+j+1)+2i(i+1)+2j(j+1)
− − L (2i+1) M (2j+1)
i,j (cid:20) − (cid:21)q(cid:20) − (cid:21)q
X
= qr2 (q2;q2)L(q2;q2)M .
(q2;q2) (q) (q)
r L−r M−r
r≡1 mod2,
X
r>0
(3.10)
4. Two infinite families of multiple series identities
If we let L in (3.9) and (3.10) we end up with the following result
→∞
qr2 (q2;q2)M = (−q)a
(q2;q2) (q) (q2;q2)
r M−r ∞
r≡a mod2,
X
r≥0 (4.1)
2M
( 1)i+jq(i+j)(i+j+1)+2i(i+a)+2j(j+a) ,
− M 2j a
i,j (cid:20) − − (cid:21)q
X
where a=0,1. Remarkably, the double sum on the righthand side of (4.1) can be
reducedtoasinglesum. Tothis end,weperformacleverchangeofthe summation
variables j 3j+r a with r =0, 1 and i i j. This yields
→ − ± → −
1 ∞
RHS(4.1)= 1 ( 1)r ( 1)iq3i2+i(1+2r)
(q2;q2) − −
∞
r=−1 i=−∞
X X (4.2)
∞
q24j2+2j(8r+1−6δa,1)+3r2+r+δa,1(1−4r) 2M .
M 6j 2r+a
j=−∞ (cid:20) − − (cid:21)q
X
THE TRI-PENTAGONAL NUMBER THEOREM AND RELATED IDENTITIES 9
We now make use of a special case of (1.6)
∞
( 1)iq3i2+i(1+2r) =(q2;q2) (1 δ ), r =0, 1 (4.3)
∞ r,1
− − ±
i=−∞
X
to simplify (4.2) further. This way we obtain
qr2 (q2;q2)M =
(q2;q2) (q)
r M−r
r≡a mod2,
X
r≥0
∞
q24j2+2j(1−6δa,1)+δa,1 2M (4.4)
M 6j+a −
j=−∞ (cid:20) − (cid:21)q
X
∞
q24j2−2j(7+6δa,1)+5δa,1 2M .
M 6j+a+2
j=−∞ (cid:20) − (cid:21)q
X
Clearly, one could have arrived at (4.4) by taking a more direct route by applying
(3.3) to the polynomial identity
∞
2M +a 2M +a
q(3j+1+a)2j =q2M(M+a), a=0,1,
j=−∞ (cid:20)M −3j(cid:21)q2 −(cid:20)M −3j−1(cid:21)q2!
X
which is, essentially, A(5) and A(8) in [9].
However, I feel that the passage from (4.1) to (4.4) is a good warm-up exercise to
prepare the reader for the development in the next section.
We cannow follow a well trodden path [2] anditerate (3.1) to get for k 1 and
≥
a=0,1
qN12+···+Nk2+1(q)2M
=
(q) (q) (q) (q2;q2) (q;q2)
n1,...,Xnk+1≥0, M−N1 n1··· nk nk+1 nk+nk+1
nk+1≡a mod2
∞
q12(2+3k)j2+2j(1−δa,16(1+k))+(1+k)δa,1 2M (4.5)
M 6j+a −
j=−∞ (cid:20) − (cid:21)q
X
∞
q12(2+3k)j2+2j(−7−12k−δa,16(1+k))+2+k(a+2)2+5δa,1 2M ,
M 6j+a+2
j=−∞ (cid:20) − (cid:21)q
X
whereN =n + +n , 1 i k+1. If wenowletM in(4.5)wearrive
i i k+1
··· ≤ ≤ →∞
at
qN12+···+Nk2+1
=
(q) (q) (q2;q2) (q;q2)
n1,...,Xnk+1≥0, n1··· nk nk+1 nk+nk+1
nk+1≡a mod2 (4.6)
q(1+k)δa,1 ∞
q4(2+3k)(3j−1)jz3j(1 z q8(2+3k)j),
(q) a − a
∞
j=−∞
X
10 ALEXANDERBERKOVICH
where a = 0,1 and za := q2+4a+4k(1+δa,1). At this stage we recall the quintuple
product identity [[6], Ex. 5.6]
∞
(−1)nq3n2−1nz3n(1+zqn)=(q,−z,−zq)∞(qz2,zq2;q2)∞
n=X−∞ (4.7)
[z2;q]
∞
=(q) .
∞
[z;q]
∞
This identity enables us to rewrite (4.5) as
qN12+···+Nk2+1
=
(q) (q) (q2;q2) (q;q2)
nnk1+,1..≡.,Xank+m1≥od0,2 n1··· nk nk+1 nk+nk+1 (4.8)
(q16+24k;q16+24k)
q(1+k)δa,1 (q) ∞[za;q16+24k]∞[q16+24kza2;q32+48k]∞,
∞
where a=0,1.
It remains to establish (1.16). To this end we add together (4.5) with a=0 and
(4.5) with a = 1. This way we immediately obtain the correct left hand side of
(1.16). Making use of (4.7) on the right we derive that
∞
1
q4(2+3k)(3j−1)jz3j(1 z q8(2+3k)j)+
(q) 1 − 1
∞
j=−∞
X
q1+k ∞
q4(2+3k)(3j−1)jz3j(1 z q8(2+3k)j)=
(q) 2 − 2
∞
j=X−∞ (4.9)
∞
1
( 1)jq(2+3k)(3j−1)jq3(1+k)j(1+q1+kq(2+3k)2j)=
(q) −
∞
j=−∞
X
(q4+6k;q4+6k) [q2+2k;q4+6k]
∞ ∞
,
(q) [q1+k;q4+6k]
∞ ∞
asdesired. Itisinstructivetocompare(1.16)withasomewhatsimilarformula[[2],
(1.8)]
qN˜12+···+N˜k2
=
(q) (q) (q) (q)
n1,.X..,nk≥0 n1 n2··· nk−1 2nk (4.10)
(q4+6k;q4+6k) [q2+2k;q4+6k]
∞ ∞
,
(q) [ q1+k;q4+6k]
∞ ∞
−
where N˜ :=n + +n , 1 i k.
i i k
··· ≤ ≤
5. One-dimensional version of the tri-pentagonal theorem
This paper arose from my attempt to ascertain if Andrews’ formula (1.17) was
“genuinely” three-dimensional. In the course of this investigation I found that