# [seqfan] Peculiar Continued Fractions (for theta etc.)

Joerg Arndt arndt at jjj.de
Mon Apr 5 15:15:27 CEST 2010

```* Paul D Hanna <pauldhanna at juno.com> [Mar 25. 2010 18:01]:
> SeqFans,
>      Should have added that, if we set y=x=q, then the prior conjecture gives a nice continued fraction expansion of the elliptic function:
> Sum_{n>=0} q^(n(n+1)/2) = 1/(1 - [q/(q+1)]/(1 - [q^2/((q+1)*(q^2+1))]/(1 - [q^3/((q^2+1)*(q^3+1))]/(1 - [q^4/((q^3+1)*(q^4+1))]/(1 -...))))).
> and equals  theta2(q^(1/2))/(2*q^(1/8)).
>
> Is this continued fraction expression known, and is it useful in evaluating special values of theta2?
>     Paul

[1, 0]
[2, 1]
[3, -q/(q + 1)]
[4, -q^2/(q^3 + q^2 + q + 1)]
[5, -q^3/(q^5 + q^3 + q^2 + 1)]
[6, -q^4/(q^7 + q^4 + q^3 + 1)]
[7, -q^5/(q^9 + q^5 + q^4 + 1)]
[8, -q^6/(q^11 + q^6 + q^5 + 1)]
[9, -q^7/(q^13 + q^7 + q^6 + 1)]

The simple CF is:
[1, 0]
[2, 1]
[3, (-q - 1)/q]
[4, (q^2 + 1)/q]
[5, (-q^3 - 1)/q^2]
[6, (q^4 + 1)/q^2]
[7, (-q^5 - 1)/q^3]
[8, (q^6 + 1)/q^3]
[9, (-q^7 - 1)/q^4]

For theta3(), you form:
[1, 0]
[2, 1]
[3, -2*q/(2*q + 1)]
[4, -q^3/(2*q^4 + q^3 + 2*q + 1)]
[5, -q^5/(q^8 + q^5 + q^3 + 1)]
[6, -q^7/(q^12 + q^7 + q^5 + 1)]
[7, -q^9/(q^16 + q^9 + q^7 + 1)]
[8, -q^11/(q^20 + q^11 + q^9 + 1)]
[9, -q^13/(q^24 + q^13 + q^11 + 1)]

Simple CF:
[1, 0]
[2, 1]
[3, (-2*q - 1)/(2*q)]
[4, (2*q^3 + 2)/q^2]
[5, (-q^5 - 1)/(2*q^3)]
[6, (2*q^7 + 2)/q^4]
[7, (-q^9 - 1)/(2*q^5)]
[8, (2*q^11 + 2)/q^6]
[9, (-q^13 - 1)/(2*q^7)]

[1, 0]
[2, 1]
[3, -q]
[4, q^3 - q^2]
[5, q^5 - q^3]
[6, q^7 - q^4]
[7, q^9 - q^5]
[8, q^11 - q^6]
[9, q^13 - q^7]

Simple CF isn't that simple:
[1, 0]
[2, 1]
[3, 1/-q]
[4, -1/(q^2 - q)]
[5, 1/(-q^3 - q^2)]
[6, (q + 1)/(-q^5 + q^2)]
[7, (q^2 + q + 1)/(-q^7 - 2*q^6 - 2*q^5 - 2*q^4 - q^3)]
[8, (q^4 + 2*q^3 + 2*q^2 + 2*q + 1)/(-q^10 - q^9 - q^8 + q^5 + q^4 + q^3)]
[9, (q^4 + q^3 + q^2 + q + 1)/(-q^11 - 2*q^10 - 2*q^9 - 3*q^8 - 3*q^7 - 2*q^6 - 2*q^5 - q^4)]

Does what I referred to as "your form" have a generally accepted name?

```