Suppose you are asked to describe an infinite sequence of signs + or -. How likely is it that you will end up with the same sequence as the one I am now thinking about? Well, according to all natural probabilistic models, the probability should be zero. Read on, then…
Quite a while ago, I wrote a short note to try to understand the distribution of the root number (also known as the sign of the functional equation) for the Jacobian of the modular curve X0(p), where p is prime. My guess was that this should be +1 or -1 more or less equally often, and it should have been easy to confirm it by computing explicitly this sign using the Eichler-Shimura relation for the L-function of this jacobian and the trace formula for the Fricke involution, except that the latter turns out to involve class numbers, and after watching the dust settle, one sees that the approximate equidistribution is equivalent to asking some questions about the equidistribution modulo 4 or 8 of some class numbers of imaginary quadratic fields for which one knows the residue modulo 2 or 4, respectively.
And in particular, one needs to know the distribution modulo 4 of h(-p) for p congruent to 3 modulo 4, in which case genus theory says that the class number is odd.
And now for the coincidence: having reached this question, it seemed fairly natural to drop by the office (two doors up on the other side of the corridor) of one of the foremost expert on the distribution of class numbers, Henri Cohen, and ask him if the answer was already known? As it turned out, Henri was just in the midst of using Pari/GP to compute the values
of the p-adic Gamma function at 1/2. These, although they should (in theory) be elements of p-adic fields, satisfy
or, in other words, these values form a sequence of signs. And the two sequences are the same!: we have
After this, I can not be impressed when hearing of people who just happen to think of their great-aunt who has been lost in the jungle of New Guinea for twenty years just a few minutes before receiving a phone call from her.