Plus six more hilbs <http://gosper.org/4hilbthets.png>. (And no dragun thetas)—: It looks like we can always do hilbert[EllipticTheta[3,0,2^-n]/2^k+b/3/2^j. A "general formula" looks guessable in terms of EllipticTheta[3 or 4, 0, 2^-n or 2/2^-n] and hilbert[certain rationals depending on b, k, and n]. Likewise for hilbert[EllipticTheta[4,...]. Note the last identity: hilbert[EllipticTheta[4, 1/4]] == 1/2 (I EllipticTheta[3, 1/2] + EllipticTheta[4, 1/2] + (1 - I)) Valuations of dragun[EllipticTheta[...]] seem much rarer. terDragon[EllipticTheta[3,0,1/9] failed completely. Also, no luck with hilbert@QPochhammer[1/4,1/4]. —rwg The following remarks are pink due to embarrassment. —rwg On Sun, Aug 9, 2020 at 10:15 PM Bill Gosper <billgosper@gmail.com> wrote:
from three hilberts <http://gosper.org/3hilbthets.png> and three draguns <http://gosper.org/3dragthets.png> at theta constants. Such identities seem rather abundant, but probably not reliably enough to find a formula with a parameter.
There's a pretty good chance I can find a closed form for a terDragon[𝜗[]] or two. —rwg
I think these formulas can be proved with finite-state machines, but I'll just settle for a hundred digits.