First, an irrelevant puzzle: Where is the "full-size, 170k" picture of "Warren's" Libyan meteorite? http://www.sciencedirect.com/science/article/pii/S0012821X13004998 Spoiler at bottom. On Wed, Oct 9, 2013 at 2:44 AM, Bill Gosper <billgosper@gmail.com> wrote:
I mentioned f(nth convergent(√3)) = 1/56 (-40 + 2^(3 n/2) (13 + 12 Sqrt[2] + (-1)^n (13 - 12 Sqrt[2]))) where f is Newman's rational counter. Why didn't I say f(1 + (-2 (1 - Sqrt[3])^(-1 + n) + 2 (1 + Sqrt[3])^(-1 + n))/(-(1 - Sqrt[3])^n + (1 + Sqrt[3])^n)) ? Because instead of this formula, I got the puzzling expression In[83]:= FullSimplify[ FindSequenceFunction[Convergents[Sqrt[3], 22]]@n, n \[Element] Integers]
Out[83]= (2 Sqrt[3] (-2 - Sqrt[3])^n QPochhammer[2 - Sqrt[3], -2 + Sqrt[3], n])/ ((-1 + (-2 - Sqrt[3])^n) QPochhammer[-1, -2 + Sqrt[3], n])
(Strangely, the supposedly equivalent FullSimplify[FindSequenceFunction[Convergents[Sqrt[3], 22], n], n \[Element] Integers] simply gives up.)
[...]
Simplifying the QPochhammer expression down to the powers of surds requires (((-5 - 3 Sqrt[3])^(1 + n) - (1 - Sqrt[3])^(1 + n)) ((-5 - 3 Sqrt[3])^ n - (1 + Sqrt[3])^n))/ (((-5 - 3 Sqrt[3])^n - (1 - Sqrt[3])^n) ((-5 - 3 Sqrt[3])^(1 + n) - (1 + Sqrt[3])^(1 + n))) -> (1 + (-2 + Sqrt[3])^(1 + n))/(1 + (-2 + Sqrt[3])^n) (integer n). This seems pretty hard. NeilB did it with undetermined "coefficients": Suppose the answer is (a+b s^n)/(c+s^n). Cross multiply. Eliminate a, b, and c from the special cases n=1,2,3,4. Solve for s. Now solve for a,b,c. Cross multiply and verify for general integer n. How long before Mathematica, e.g., can do these? --rwg http://ars.els-cdn.com/content/image/1-s2.0-S0012821X13004998-gr001.jpg