Here's something not directly related, but that might go somewhere (I'm certainly going to research it some more): For the sake of brevity below, F means GF(2^n). You can talk about functions that act on elements of F, but the elements of F can be treated as functions themselves, in a limited sense. Usually we think of them as polynomials in x with coefficients in {0,1}. That suggested to me that I could take the q-derivative of a single element of F rather than the q-derivative of a function acting on elements of F. Then you can identify an element e^x with the property that D((e^x)^k) = D(e^kx) = [k]e^kx, where [k] is the q-number (q^k - 1)/(q-1). So in addition to treating elements of F as polynomials, we've can treat them as trig functions, which all have nice continued fraction representations. Continued fractions lose uniqueness of the representation when there's no concept of magnitude (and therefore no "floor" function) but a lot of the theory still works. Of course, AES is just an overblown continued fraction, so it would be fun to see if there's some way to write it as, say, a sum of Bessel functions or something.
Gabor Hetyei <ghetyei@uncc.edu> writes
While browsing the internet, I came accross your remarkable formula http://www.inwap.com/pdp10/hbaker/hakmem/cf.html#item99 expressing the value of a continued fraction with partial quotients increasing in arithmetic progression. I would be interested in the known proof of this formula. Is there any
published paper or preprint on the issue that you could share with me or
point me to? The reason of my interest is that I think it is possible to prove this
formula using combinatorial enumeration, and I wonder if this has already
been done or would be interesting to do.
I'd been interested in the value of the continued fraction 1+1/2+1/3+1/4+...
since high school. In college, I came across the Dover reprint of the NIST
tables by Abramowicz & Stegun, and spent a fair amount of time studying the
special functions. I came across the Bessel-CF formula 9.1.73 for J, and recognized its relevance to my puzzle. Then I found the recurrence for I (first formula in section 9.6.26) and a litle manipulation does the rest.
I was later able to find a direct proof using the power series for I, the one with a product of factorials in the denominator. I don't recall details,
but it was pretty straightforward, just lining up terms to check the recurrence. You need to add something about convergence, I don't recall if I finished this step.
I suspect "my" formula is actually well-known among continued fraction experts.
I'm not a professional combinatorist, so I don't know if your work is new or not. I'd think that anything relating combinatorics to Bessel functions
would count as interesting, since there's no obvious connection. (Although
I think they turn up in the formula for partitions into unequal parts.)
There are some people I can ask. I'm forwarding this reply to them.
Rich Schroeppel rcs@cs.arizona.edu
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