ISSN 2148-838Xhttp://dx.doi.org/10.13069/jacodesmath.451218

J. Algebra Comb. Discrete Appl.
5(3) • 137–142

Received: 11 April 2018
Accepted: 16 July 2018

Journal of Algebra Combinatorics Discrete Structures and Applications

Finite Rogers–Ramanujan type continued fractions∗

Research Article

Helmut Prodinger

Abstract: New finite continued fractions related to Bressoud and Santos polynomials are established.

2010 MSC: 05A30, 11A55

Keywords: Bressoud polynomials, Santos polynomials, Rogers–Ramanujan identities

1. Introduction

Define, as it is common today, (x; q)n := (1 − x)(1 − xq) . . . (1 − xqn−1), where we assume that
|q| < 1, and we allow n also to be 0 and infinity. We also need the coefficients

[
n
k

]
:=

(q;q)n
(q;q)k(q;q)n−k

. These
standard notations can be found e. g. in the classic book [1].

The two Rogers-Ramanujan identities [1, 6]

∑
n≥0

qn
2

(q; q)n
=

1

(q; q5)∞(q4; q5)∞
,

∑
n≥0

qn
2+n

(q; q)n
=

1

(q2; q5)∞(q3; q5)∞

are very popular, influential, useful and historically interesting. Let

F(z) =
∑
n≥0

qn
2

zn

(q; q)n
and G(z) =

∑
n≥0

qn
2+nzn

(q; q)n
;

∗ Dedicated to Peter Paule on the occasion of his 60th birthday.
Helmut Prodinger; Department of Mathematics, University of Stellenbosch 7602, Stellenbosch, South Africa

(email: hproding@sun.ac.za).

137

https://orcid.org/0000-0002-0009-8015


H. Prodinger / J. Algebra Comb. Discrete Appl. 5(3) (2018) 137–142

the continued fraction (due to Ramanujan)

zG(z)

F(z)
=

z

1 +
zq

1 +
zq2

1 +
...

is also very well known, see [5, Entry 5] and [6, (6.1)].

There are several families of polynomials that approximate F(z) and G(z). Probably the the most
well known are

fn(z) =
∑
j≥0

qj
2

[
n + 1− j

j

]
zj → F(z) and gn(z) =

∑
j≥0

qj
2+j

[
n− j
j

]
zj → G(z) for n →∞,

because of their link to the Schur polynomials, see [1].

The finite continued fraction

zgn(z)

fn(z)
=

z

1 +
zq

1 +
zq2

... 1 +
zqn

1

is also known [5, Entry 16].

The polynomials

sn(z) =
∑
j≥0

qj
2

[
n

j

]
zj → F(z) and tn(z) =

∑
j≥0

qj
2+j

[
n

j

]
zj → G(z) for n →∞,

due to Bressoud [7], are less well known; see also [8].

2. Bressoud polynomials and continued fractions

In this section, we will establish the following attractive finite continued fraction:

Theorem 2.1.

ztn(z)

sn(z)
=

z

1 +
zq(1−qn)

1 +
zq2

1 +
zq3(1−qn−1)

1 +
zq4

1 +
...

Proof. To prove this statement by induction, define the righthand side by Tn(z). It is plain to see that
T0(z) = z, and

Tn(z) =
z

1 +
zq(1−qn)

1 + Tn−1(zq2)

.

138



H. Prodinger / J. Algebra Comb. Discrete Appl. 5(3) (2018) 137–142

We are left to prove that

ztn(z)

sn(z)
=

z

1 +
zq(1−qn)

1 +
ztn−1(zq

2)

sn−1(zq2)

=
z

1 +
zq(1−qn)sn−1(zq2)

sn−1(zq2) + ztn−1(zq2)

=
z(sn−1(zq

2) + ztn−1(zq
2))

sn−1(zq2) + ztn−1(zq2) + zq(1−qn)sn−1(zq2)
,

which amounts to prove that

tn(z) = sn−1(zq
2) + ztn−1(zq

2),

sn(z) = sn−1(zq
2) + ztn−1(zq

2) + zq(1−qn)sn−1(zq2).

We will show that the coefficients of zj coincide, which is trivial for j = 0, so we assume j ≥ 1:

qj
2+j

[
n

j

]
= qj

2+2j

[
n−1
j

]
+ q(j−1)

2+3(j−1)+2
[
n−1
j −1

]
,

which is equivalent to [
n

j

]
= qj

[
n−1
j

]
+

[
n−1
j −1

]
and therefore true. The second one goes like this:

qj
2

[
n

j

]
= qj

2+2j

[
n−1
j

]
+ q(j−1)

2+3(j−1)+2
[
n−1
j −1

]
+ q(1−qn)q(j−1)

2+2(j−1)
[
n−1
j −1

]
,

which is equivalent to [
n

j

]
= q2j

[
n−1
j

]
+ qj

[
n−1
j −1

]
+ (1−qj)

[
n

j

]
,

and further to [
n

j

]
= qj

[
n−1
j

]
+

[
n−1
j −1

]
,

which finishes the proof.

3. Identities 39 and 38 from Slater’s list

Slater [11] produced a list of Rogers-Ramanujan type series/product identities; Sills [10] in an amaz-
ing effort reworked and annotated this list, providing, in particular, finite versions of all of them.

Arguably the second most popular identities in the Rogers-Ramanujan world are Slater’s [11] iden-
tities (39) and (38)

∑
n≥0

q2n
2

(q; q)2n
=

∏
k≥1, k≡±2,±3,±4,±5 (mod 16)

1

1−qk
,

∑
n≥0

q2n
2+2n

(q; q)2n+1
=

∏
k≥1, k≡±1,±4,±6,±7 (mod 16)

1

1−qk
.

139



H. Prodinger / J. Algebra Comb. Discrete Appl. 5(3) (2018) 137–142

Let

sn(z) =
∑

0≤2h≤n

q2h
2

[
n

2h

]
zh and tn(z) =

∑
0≤2h≤n

q2h
2+2h

[
n

2h + 1

]
zh;

these polynomials are called Santos polynomials [2–4].

In order to describe the finite continued fraction expansion of ztn(z)/sn(z), we define the following
numbers and polynomials (power series) which were originally found by guessing:

a2k :=
(1−q4k+1)(qn+1−2k; q2)2k

q2k(qn−2k; q2)2k+1
,

a2k+1 :=
(1−q4k+3)(qn−2k; q2)2k+1
q2k+2(qn−2k−1; q2)2k+2

;

S2i :=
∑
j≥0

q2(i+j)(i+j+1)(qn−2i−2j; q)2j(q
n−2i; q2)2i+1

(q; q)2j+1(q2j+3; q2)2i
zj,

S2i+1 :=
∑
j≥0

q2(i+j+1)
2

(qn−1−2i−2j; q)2j(q
n−1−2i; q2)2i+2

(q; q)2j+1(q2j+3; q2)2i+1
zj.

Theorem 3.1. The polynomials Si satisfy the second order recurrence zSi+1 = Si−1−aiSi, S−1 = sn(z),
S0 = tn(z). Consequently, we get the finite continued fraction expansion

zt(n)

s(n)
=

z

a0 +
z

a1 +
z

a2 +
z

...

,

or, more elegantly:

zt(n)

s(n)
=

zb0

1 +
zb1

1 +
zb2

1 + . . .

with

b0 =
1

a0
=

1−qn

1−q
and bi =

1

ai−1ai
=

q2i(1−qn−i)(1−qn+i)
(1−q2i−1)(1−q2i+1)

for i ≥ 1.

140



H. Prodinger / J. Algebra Comb. Discrete Appl. 5(3) (2018) 137–142

Proof. The recursion will be shown for even i, the other instance being very similar:

S2i−1 −a2iS2i =
∑
j≥0

q2(i+j)
2

(qn+1−2i−2j; q)2j(q
n+1−2i; q2)2i

(q; q)2j+1(q2j+3; q2)2i−1
zj

−
(1−q4i+1)(qn+1−2i; q2)2i

q2i(qn−2i; q2)2i+1

∑
j≥0

q2(i+j)(i+j+1)(qn−2i−2j; q)2j(q
n−2i; q2)2i+1

(q; q)2j+1(q2j+3; q2)2i
zj

= (qn+1−2i; q2)2i
∑
j≥0

q2(i+j)
2

(qn+1−2i−2j; q)2j
(q; q)2j+1(q2j+3; q2)2i−1

zj

− (1−q4i+1)(qn+1−2i; q2)2i
∑
j≥0

q2(i+j)
2+2j(qn−2i−2j; q)2j

(q; q)2j+1(q2j+3; q2)2i
zj

= (qn+1−2i; q2)2i
∑
j≥0

q2(i+j)
2

(qn+1−2i−2j; q)2j−1(1−qn−2i)(1−q2j+4i+1)
(q; q)2j+1(q2j+3; q2)2i

zj

− (1−q4i+1)(qn+1−2i; q2)2i
∑
j≥0

q2(i+j)
2+2j(qn+1−2i−2j; q)2j−1(1−qn−2i−2j)

(q; q)2j+1(q2j+3; q2)2i
zj

= (qn+1−2i; q2)2i
∑
j≥0

q2(i+j)
2

(qn+1−2i−2j; q)2j−1
(q; q)2j+1(q2j+3; q2)2i

zj

×
[
(1−qn−2i)(1−q2j+4i+1)−q2j(1−q4i+1)(1−qn−2i−2j)

]
= (qn+1−2i; q2)2i

∑
j≥0

q2(i+j)
2

(qn+1−2i−2j; q)2j−1
(q; q)2j+1(q2j+3; q2)2i

zj(1−q2j)(1−qn+1+2i)

= (qn+1−2i; q2)2i+1
∑
j≥1

q2(i+j)
2

(qn+1−2i−2j; q)2j−1
(q; q)2j+1(q2j+3; q2)2i

zj(1−q2j)

= z(qn−1−2i; q2)2i+2
∑
j≥0

q2(i+j+1)
2

(qn−1−2i−2j; q)2j
(q; q)2j+1(q2j+3; q2)2i+1

zj = zS2i+1.

Further,

S−1 =
∑
j≥0

q2j
2

(qn+1−2j; q)2j
(q; q)2j+1(q2j+3; q2)−1

zj =
∑
j≥0

q2j
2

(qn+1−2j; q)2j
(q; q)2j

zj = s0(z)

and

S0 =
∑
j≥0

q2j(j+1)(qn−2j; q)2j(1−qn)
(q; q)2j+1

zj =
∑
j≥0

q2j(j+1)(qn−2j; q)2j+1
(q; q)2j+1

zj = t0(z).

Now we can iterate this relation in the following form:

zt(n)

s(n)
=

zS0
S−1

=
z

a0 +
zS1

S0

=
z

a0 +
z

a1 +
zS2

S1

= . . .

This is the desired (finite) continued fraction expansion.

We remark that for n →∞, the quantities ai and Si appear already in [9].

141



H. Prodinger / J. Algebra Comb. Discrete Appl. 5(3) (2018) 137–142

4. Conclusion

We would like to mention that it is more challenging to find the continued fractions and the relevant
quantities, as the present proofs (and possibly other ones) consist of routine manipulations.

Since there are many Rogers-Ramanujan type identities and polynomials approximating them are
not even unique, there might be additional additional results; compare our previous effort [9] for infinite
versions. Most polynomials from Sill’s list [10] are, however, not expressable in terms of one summation
and thus not candidates for the present approach.

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[3] G. E. Andrews, A. Knopfmacher, P. Paule, An infinite family of Engel expansions of Rogers–
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[4] G. E. Andrews, J. P. O. Santos, Rogers–Ramanujan type identities for partitions with attached odd
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	Introduction
	Bressoud polynomials and continued fractions
	Identities 39 and 38 from Slater's list
	Conclusion
	References