The map P-->3P should give an explicit recursion, but of course misses lots of points. On Mon, Jun 30, 2014 at 6:06 AM, Bill Gosper <billgosper@gmail.com> wrote:
In what was probably Europe's first public math contest, the great Fibonacci was asked for three rational squares with common difference 5. (This is equivalent to finding a rational Pythagorean triangle with area 5.) Somehow, Fibonacci found (31/12)^2, (41/12)^2, (49/12)^2. This is the first of an infinite sequence of such solutions, found these days by repeated "addition" of the point {5*5/4, 6*5^2/4^2} along the "elliptic" curve y^2 = (x-5) x (x+5), producing the sequence 5/2, 41/12, 11285/1562, 3344161/1494696, 44572169525/7118599318, 654686219104361/178761481355556, ..., whose elements *of even index* are Fibonacci's desire. Elliptic curve addition doesn't quite provide a nice recurrence formula. Empirically, there's a mysterious, period 8 sign pattern: pisano[1] = 5/2; pisano[2] = 41/12; pisano[n_Integer] := (5 \[Sqrt](-100 + 9 pisano[-1 + n]^2 + 4 pisano[-1 + n]^4 - 12 (-1)^(1/8 (-2 + n) (1 + n) (3 + n) (4 + n)) Sqrt[-25 pisano[-1 + n]^2 + pisano[-1 + n]^6]))/ (Abs[-5 + 2 pisano[-1 + n]] (5 + 2 pisano[-1 + n]))
Try a few: Table[pisano[n], {n, 8}]
{5/2, 41/12, 11285/1562, 3344161/1494696, 44572169525/7118599318, 654686219104361/178761481355556, 312098738002194296165/128615821825334210638, 249850594047271558364480641/5354229862821602092291248}
Test the "even" ones: Sqrt[#^2 + {-5, 5}] & /@ %[[Range[2, 8, 2]]]
{{31/12, 49/12}, {113279/1494696, 4728001/1494696}, {518493692732129/178761481355556, 767067390499249/178761481355556}, {249563579992463717493803519/5354229862821602092291248, 250137278774864229623059201/5354229862821602092291248}}
Rich suggested the solutions might obey a Somos recurrence. So far, I've only found one (Somos5) for the numerators: a[n] -> (124558 a[-3 + n] a[-2 + n] - 781 a[-4 + n] a[-1 + n])/a[-5 + n] The 124558 and 781 are very probably minimal. "Fitting" a 𝝑 fcn to these numerators held a surprise: a[0], a[1], a[2], ... , a[n] = 1,5,41,11285,... =
(I (-1)^n 5^(1/2 Mod[n, 2]) E^(- I ArcTan[3/5]/2) (π/2)^(3/4) u^n^2* EllipticTheta[2, n z, (-1)^(-(13/34) + I/34)])/(2 17^(1/4) Gamma[5/4])
(Note the crazy (constructible!) 34th root of unity. Unfortunately, u and z are simply what they need to be to make a[1] and a[2] work. ISC doesn't recognize them, and I can't find a way to symbolically separate them. z ->
4.908105290682877137963508159098016296309287655031188041917052851155165374957836660235707132115102`69. -
0.197404925101320072627868756915313864010305639948302119404764160895568125915396715987289924743617`69. I, u ->
-0.910971694653082823579468669395203737157839231764166792826687755862129892782777001356185727331806`65.8+
1.430029427800740885258427207242669226859970344275381501850156729643487583694210398313066894949425`66. I}
Mathematica's Root notation lets you exactly specify an implicitly defined root, say, the solution to a transcendental equation. Newton's method is getting to be like another button on your calculator. But how do we handle the solution vectors of simultaneous transcendental (say) equations, like for u and z here? --rwg (with much help from NeilB) _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun