18 Dec
2013
18 Dec
'13
1:08 a.m.
(I never received my copy of this, but I assume most of you did, since I got an offlist response.) On Tue, Dec 17, 2013 at 6:20 PM, Bill Gosper <billgosper@gmail.com> wrote:
Neil[B] and I recently noticed
(2^(2*k + 1) - 1)^2 + 8 == (2^(2*k + 1) + 2^(k + 2) + 3) * (2^(2*k + 1) - 2^(k + 2) + 3) Old? --rwg
This is pretty dull, since Factor[...] can simply do it. Slightly more interesting is In[182]:= Factor[(2^(k + 1) - 3)^2 - 8]
Out[182]= 1 - 3 2^(2 + k) + 2^(2 + 2 k) (Fails) but both the "odd" and "even" subcases work: In[185]:= Factor[(2^(2*k + 1) - 3)^2 - 8] Out[185]= (1 - 2^(2 + k) + 2^(1 + 2 k)) (1 + 2^(2 + k) + 2^(1 + 2 k)) In[186]:= Factor[(2^(2*k) - 3)^2 - 8] Out[186]= (-1 + 2^(2 k) - 2^(1 + k)) (-1 + 2^(2 k) + 2^(1 + k)) --rwg