Ramanujan's $\tau(n)$ and continued fractions

In D.H. Lehmer's paper "Ramanujan's function $\tau(n)$, (Duke J. Math v. 10 1943, pp. 483-492), Lehmer states the Ramanujan conjecture $|\tau( p )|< 2p^{11/2}$, so that $p^{-11/2}\tau( p )=2\cos(\theta_p)$ where $\theta_p$ is real. He observes "It is interesting to note that $2\cos(\theta_{11})=1.000872909\ldots$" ie that $\theta_{11}$ is very nearly $\pi/3$.

In fact, Mathematica computes that $\theta_{11}/\pi=0.333172889904775\ldots$ There's nothing special about the expansion as a decimal, so I looked instead at the continued fraction expansion, which Mathematica can also do. One finds it is $\{0, 3, 692, 5, 4, 1, 2, 3, 1, 2,\ldots\}$ The $3$ is expected, as is the fact that the next term is large. But that it's 692 is surprising to me, recalling that $\tau(n)\equiv \sigma_{11}(n) \bmod 691$ (which is due to Ramanujan himself).

I've looked at other weights and levels with no insight. I've also looked nontraditional continued fraction expansions with alternating $\pm$ signs, and numerators other than 1, to see if I could make 691 (v. 692) appear instead.

The 'eigen-angles' $\theta_p$ clearly carry deep arithmetic information coming from Galois representations, as others on this site can explain much better than I. But can something be proven about these continued fraction expansions? Anything that's true is likely to be quite deep, so a more realistic (but vague) question is:

Are there other examples of this kind of 'numerology'?

• It is probably not as deep as the sort of thing you are getting at but a slightly related thing is the continued fraction for the real root of $x^3-8x-10$. This number has some huge partial quotients very early on (which is uncharacteristic for an algebraic number, also note that the discriminant has 163 as a factor). – muad Sep 14 '10 at 16:40
• @muad: See oeis.org/A002937 and the paper by Stark referenced there. – Douglas Zare Nov 15 '16 at 9:16
• @muad The interesting thing is that Brillhart's continued fraction arises from the value of a q expansion (at a quadratic irrational); the example in my question is coming from the coefficients of a q expansion. – Stopple Nov 15 '16 at 17:53

Since the OP asked for other examples of this kind of numerology,I will give another one to support his observation

The function $\cos(\theta_{11})$ has the following closed form

$\cos(\theta_{11})=\frac{\sigma_{1}(11)}{22\sqrt{11}}-\frac{12\sum_{k=1}^{10} (2178k^2-572k^3+35k^4)\sigma_{1}(11-k)\sigma_{1}(k)}{161051\sqrt{11}}\tag1$

and continued fraction

$\cos(\theta_{11})=\{0;1,1,572,3,2,1,2,1,2,2,4,3,1,6,\dots\}\tag2$

where we clearly see $572$ appearing both as a coefficient in the sum $(1)$ and largest partial quotient in the first few partial quotients of the continued fraction $(2)$

Is this a coincidence?

Edited:03 Sep 2017

And also

$2\cos(\theta_{11})=\frac{\sigma_{1}(11)}{11\sqrt{11}}-\frac{12\sum_{k=1}^{10} (4356k^2-1144k^3+70k^4)\sigma_{1}(11-k)\sigma_{1}(k)}{161051\sqrt{11}}\tag3$

with the following continued fraction

$2\cos(\theta_{11})=\{1;1145,1,1,2,6,2,2,1,1,1,1,1,2,3,\dots\}\tag4$

where $1145$ is the 0th partial qotient in the continued fraction $(4)$ and $1144$ appears as a coefficient in the formula $(3)$

Remark: Identity $(1)$ is a special case of the identity found in A000594 OEIS when $n=11$

$\tau(n)=n^4\sigma_{1}(n)-24\sum_{k=1}^{n-1} (18n^2k^2-52nk^3+35k^4)\sigma_{1}(n-k)\sigma_{1}(k)$

• How on Earth did you get the first closed form ? – Sylvain JULIEN Sep 3 '17 at 18:15
• @ Sylvain JULIEN :I found it using mathematica – Nicco Sep 3 '17 at 19:38
• @Sylvain JULIEN:the identity is also found in the oeis – Nicco Sep 4 '17 at 10:14