I said

Now I can apply the exact same process to Heine's series instead of Gauss's and get a q-extension [of Ramanujan's dyadic 1/pi = pFq[1/64]]. Here it is cleaned up.

2 2k+1 3 6k+1 6k+1 6k+3 3 6 k (q + 1) (1 - q ) q (1 - q ) [ 2 k ] inf q (------------------------ - -----------------) ([ ]) ==== 2 2 [ k ] \ 1 - q 1 - q q

-------------------------------------------------------------- / 6 3 ==== (- q; q) (- q; q) k = 0 k 2 k + 1 2 2 (q; q ) 1/4 inf q = -------------------- = ---- 2 pi 2 2 2 q (q ; q ) (1 - q ) inf

where the (q; q) [ 2 k ] 2 k ([ ]) :=--------- [ k ] q 2 (q; q) k are the central q-binomial coefficients, and are polynomials in q: k qbin(2k,k) 0 1 1 q + 1 2 2 2 (q + 1) (q + q + 1) 2 2 4 3 2 3 (q + 1) (q + 1) (q - q + 1) (q + q + q + q + 1) 2 4 4 3 2 4 (q - q + 1) (q + 1) (q + q + q + q + 1) 6 5 4 3 2 (q + q + q + q + q + q + 1) 2 2 4 4 3 2 5 (q + 1) (q - q + 1) (q + q + 1) (q + 1) (q - q + q - q + 1) 6 3 6 5 4 3 2 (q + q + 1) (q + q + q + q + q + q + 1) 2 4 4 2 4 3 2 6 3 6 (q + 1) (q + 1) (q - q + 1) (q - q + q - q + 1) (q + q + 1) 6 5 4 3 2 (q + q + q + q + q + q + 1) 10 9 8 7 6 5 4 3 2 (q + q + q + q + q + q + q + q + q + q + 1) It's interesting where the 12k+4 powers of 2 come from. One from the 1-q^2 in the "numerator", 6k and 6k+3 from the q-pochhammers in the denominator. I said

[...]which will obey nonlinear, three term relations among pi(q), pi(q^a), and pi(q^b), for rational a and b. (Actually, we want Pi(q):=pi(q)/(1-q^2), which changes q^(6k^2) to q^(6(k^2-1/24)) in our series.) E.g., 2 2 2 Pi (q) Pi (q ) 4 = ------------- - -------, 2 4 2 4 Pi(q ) Pi(q ) Pi (q )

9 2 3 9 9 3 Pi(q) Pi(q ) + Pi (q ) = sqrt(Pi(q) Pi(q )) (3 Pi(q ) + Pi(q)), . . . which Gene pointed out (in the 70s) give quadratic and cubic methods for pi computation. --rwg --------------------------------- Looking for a deal? Find great prices on flights and hotels with Yahoo! FareChase.