[math-fun] The first congruent number
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)
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
GAAAA--Rich privately suggested that the sign pattern in my 1st order, nonlinear recursion was not period 8, not even periodic. And he is right!! Watch the recurrence slam into reverse at n=57! In[17]:= Reap[Nest[{#[[1]] + 1, Sow[N[Numerator[#[[2]]]]]; (5 \[Sqrt](-100 + 9 #[[2]]^2 + 4 #[[2]]^4 - 12 (-1)^( 1/8 (-2 + #[[1]]) (1 + #[[1]]) (3 + #[[1]]) (4 + #[[ 1]])) #[[2]] Sqrt[-25 + #[[2]]^4]))/(Abs[-5 + 2 #[[2]]] (5 + 2 #[[2]]))} &, {3, 41/12}, 57]][[2,1]] // tim During evaluation of In[17]:= 0.042828,57 Out[17]= {41., 11285., 3.34416*10^6, 4.45722*10^10, 6.54686*10^14, 3.12099*10^20, 2.49851*10^26, 5.00963*10^33, 1.60444*10^41, 1.784*10^50, 2.09239*10^59, 1.09929*10^70, 6.53153*10^80, 1.21849*10^93, 3.89694*10^105, 3.13867*10^119, 3.92379*10^133, 1.76084*10^149, 8.19116*10^164, 1.69312*10^182, 4.07077*10^199, 2.97577*10^218, 3.79435*10^237, 1.22952*10^258, 5.99781*10^278, 1.08505*10^301, 2.006209100334819*10^323, 1.628644108339963*10^347, 1.584775641701988*10^371, 4.546385929353489*10^396, 2.306385900107065*10^422, 3.011027064913547*10^449, 5.731163695214867*10^476, 4.174132093433851*10^505, 3.074040656862112*10^534, 9.78526773250994*10^564, 3.853394734453615*10^595, 4.345605709540367*10^627, 8.752339001418554*10^659, 4.609196830221771*10^693, 3.423848362005287*10^727, 1.002440705228261*10^763, 2.946534125540071*10^798, 3.672615511787510*10^835, 5.851420618448239*10^872, 2.598805008061456*10^911, 2.073644300832899*10^950, 4.409690172448374*10^990, 1.278984026197366*10^1031, 1.502854321987985*10^1073, 1.766611469306492*10^1115, 8.611636998530298*10^1158, 5.548598132260577*10^1202, 9.72410563316432*10^1247, 5.548598132260577*10^1202, 8.611636998530298*10^1158, 1.766611469306492*10^1115} The usual penalty for choosing the wrong sign is surds. Before focusing on 57, I was running out to 9999 trying to figure out why the numerators weren't growing faster. How did it never surd? At any rate, the formula can be (grotesquely} rescued by trying both signs and choosing the rational with larger numerator. --rwg On Mon, Jun 30, 2014 at 4:06 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)
Here is the recursion I was talking about. If (a,b) is a rational point on y^2 = (x-5) x (x+5) then so is (A,B) where A = a (a^4+150a^2-1875)^2 / (3a^4-150a^2-625)^2 and B = b (a^4-20a^3-150a^2-500a+625) (a^4+20a^3-150a^2+500a+625) (a^4+150a^2-1875) / (3a^4-150a^2-625)^3 Of course not all rational points occur. On Tue, Jul 1, 2014 at 3:49 AM, Bill Gosper <billgosper@gmail.com> wrote:
GAAAA--Rich privately suggested that the sign pattern in my 1st order, nonlinear recursion was not period 8, not even periodic. And he is right!! Watch the recurrence slam into reverse at n=57! In[17]:= Reap[Nest[{#[[1]] + 1, Sow[N[Numerator[#[[2]]]]]; (5 \[Sqrt](-100 + 9 #[[2]]^2 + 4 #[[2]]^4 - 12 (-1)^( 1/8 (-2 + #[[1]]) (1 + #[[1]]) (3 + #[[1]]) (4 + #[[ 1]])) #[[2]] Sqrt[-25 + #[[2]]^4]))/(Abs[-5 + 2 #[[2]]] (5 + 2 #[[2]]))} &, {3, 41/12}, 57]][[2,1]] // tim
During evaluation of In[17]:= 0.042828,57
Out[17]= {41., 11285., 3.34416*10^6, 4.45722*10^10, 6.54686*10^14, 3.12099*10^20, 2.49851*10^26, 5.00963*10^33, 1.60444*10^41, 1.784*10^50, 2.09239*10^59, 1.09929*10^70, 6.53153*10^80, 1.21849*10^93, 3.89694*10^105, 3.13867*10^119, 3.92379*10^133, 1.76084*10^149, 8.19116*10^164, 1.69312*10^182, 4.07077*10^199, 2.97577*10^218, 3.79435*10^237, 1.22952*10^258, 5.99781*10^278, 1.08505*10^301, 2.006209100334819*10^323, 1.628644108339963*10^347, 1.584775641701988*10^371, 4.546385929353489*10^396, 2.306385900107065*10^422, 3.011027064913547*10^449, 5.731163695214867*10^476, 4.174132093433851*10^505, 3.074040656862112*10^534, 9.78526773250994*10^564, 3.853394734453615*10^595, 4.345605709540367*10^627, 8.752339001418554*10^659, 4.609196830221771*10^693, 3.423848362005287*10^727, 1.002440705228261*10^763, 2.946534125540071*10^798, 3.672615511787510*10^835, 5.851420618448239*10^872, 2.598805008061456*10^911, 2.073644300832899*10^950, 4.409690172448374*10^990, 1.278984026197366*10^1031, 1.502854321987985*10^1073, 1.766611469306492*10^1115, 8.611636998530298*10^1158, 5.548598132260577*10^1202, 9.72410563316432*10^1247, 5.548598132260577*10^1202, 8.611636998530298*10^1158, 1.766611469306492*10^1115}
The usual penalty for choosing the wrong sign is surds. Before focusing on 57, I was running out to 9999 trying to figure out why the numerators weren't growing faster. How did it never surd? At any rate, the formula can be (grotesquely} rescued by trying both signs and choosing the rational with larger numerator. --rwg
On Mon, Jun 30, 2014 at 4:06 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
I'd reserve the exclamation point for when Rich is wrong. --Dan ----- Rich privately suggested that .... ... ... And he is right!! -----
Much as I wish it were true, alas, ... ------- Quoting Dan Asimov <dasimov@earthlink.net>:
I'd reserve the exclamation point for when Rich is wrong.
--Dan
----- Rich privately suggested that .... ... ... And he is right!! -----
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
I am amazed. If our sequence is a[n] := 5/2, 41/12, 11285/1562, 3344161/1494696, 44572169525/7118599318, ... then 5 Abs[(3 a[n] - 2 Sqrt[-25 + a[n]^4])/(-25 + 4 a[n]^2)] is either a[n-1] or a[n+1], and changing the sign of the √ gives the other one! {5 Abs[(3 #1 - 2 Sqrt[-25 + #1^4])/(-25 + 4 #1^2)], 5 Abs[(3 #1 + 2 Sqrt[-25 + #1^4])/(-25 + 4 #1^2)]} & /@ {5/2, 41/12, 11285/1562, 3344161/1494696, 44572169525/7118599318, 654686219104361/178761481355556, 312098738002194296165/128615821825334210638, 249850594047271558364480641/ 5354229862821602092291248, 5009631795998363486645417214683045/1935878334514951131830244285524398} {{Indeterminate, \[Infinity]}, {5/2, 11285/1562}, {3344161/1494696, 41/12}, {44572169525/7118599318, 11285/1562}, {3344161/1494696, 654686219104361/178761481355556}, {312098738002194296165/128615821825334210638, 44572169525/7118599318}, {654686219104361/178761481355556, 249850594047271558364480641/5354229862821602092291248}, {312098738002194296165/128615821825334210638, 5009631795998363486645417214683045/1935878334514951131830244285524398}, {160443526614433014168714029147613242401001/50016678000996026579336936742637753055940, 249850594047271558364480641/5354229862821602092291248}} So now we have *two* mysterious bit sequences--the signs of the Absand and √ that get us a[n+1] from a[n]. On Tue, Jul 1, 2014 at 1:49 AM, Bill Gosper <billgosper@gmail.com> wrote:
GAAAA--Rich privately suggested that the sign pattern in my 1st order, nonlinear recursion was not period 8, not even periodic. And he is right!!
How the freep did he know that? --rwg Watch the recurrence slam into reverse at n=57!
In[17]:= Reap[Nest[{#[[1]] + 1, Sow[N[Numerator[#[[2]]]]]; (5 \[Sqrt](-100 + 9 #[[2]]^2 + 4 #[[2]]^4 - 12 (-1)^( 1/8 (-2 + #[[1]]) (1 + #[[1]]) (3 + #[[1]]) (4 + #[[ 1]])) #[[2]] Sqrt[-25 + #[[2]]^4]))/(Abs[-5 + 2 #[[2]]] (5 + 2 #[[2]]))} &, {3, 41/12}, 57]][[2,1]] // tim
During evaluation of In[17]:= 0.042828,57
Out[17]= {41., 11285., 3.34416*10^6, 4.45722*10^10, 6.54686*10^14, 3.12099*10^20, 2.49851*10^26, 5.00963*10^33, 1.60444*10^41, 1.784*10^50, 2.09239*10^59, 1.09929*10^70, 6.53153*10^80, 1.21849*10^93, 3.89694*10^105, 3.13867*10^119, 3.92379*10^133, 1.76084*10^149, 8.19116*10^164, 1.69312*10^182, 4.07077*10^199, 2.97577*10^218, 3.79435*10^237, 1.22952*10^258, 5.99781*10^278, 1.08505*10^301, 2.006209100334819*10^323, 1.628644108339963*10^347, 1.584775641701988*10^371, 4.546385929353489*10^396, 2.306385900107065*10^422, 3.011027064913547*10^449, 5.731163695214867*10^476, 4.174132093433851*10^505, 3.074040656862112*10^534, 9.78526773250994*10^564, 3.853394734453615*10^595, 4.345605709540367*10^627, 8.752339001418554*10^659, 4.609196830221771*10^693, 3.423848362005287*10^727, 1.002440705228261*10^763, 2.946534125540071*10^798, 3.672615511787510*10^835, 5.851420618448239*10^872, 2.598805008061456*10^911, 2.073644300832899*10^950, 4.409690172448374*10^990, 1.278984026197366*10^1031, 1.502854321987985*10^1073, 1.766611469306492*10^1115, 8.611636998530298*10^1158, 5.548598132260577*10^1202, 9.72410563316432*10^1247, 5.548598132260577*10^1202, 8.611636998530298*10^1158, 1.766611469306492*10^1115}
The usual penalty for choosing the wrong sign is surds. Before focusing on 57, I was running out to 9999 trying to figure out why the numerators weren't growing faster. How did it never surd? At any rate, the formula can be (grotesquely} rescued by trying both signs and choosing the rational with larger numerator. --rwg
On Mon, Jun 30, 2014 at 4:06 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)
On Tue, Jul 1, 2014 at 11:50 AM, Bill Gosper <billgosper@gmail.com> wrote:
I am amazed. If our sequence is a[n] := 5/2, 41/12, 11285/1562, 3344161/1494696, 44572169525/7118599318, ... then 5 Abs[(3 a[n] - 2 Sqrt[-25 + a[n]^4])/(-25 + 4 a[n]^2)] is either a[n-1] or a[n+1], and changing the sign of the √ gives the other one!
{5 Abs[(3 #1 - 2 Sqrt[-25 + #1^4])/(-25 + 4 #1^2)], 5 Abs[(3 #1 + 2 Sqrt[-25 + #1^4])/(-25 + 4 #1^2)]} & /@
{5/2, 41/12, 11285/1562, 3344161/1494696, 44572169525/7118599318, 654686219104361/178761481355556, 312098738002194296165/128615821825334210638, 249850594047271558364480641/ 5354229862821602092291248, 5009631795998363486645417214683045/1935878334514951131830244285524398}
{{Indeterminate, \[Infinity]}, {5/2, 11285/1562}, {3344161/1494696, 41/12}, {44572169525/7118599318, 11285/1562}, {3344161/1494696, 654686219104361/178761481355556}, {312098738002194296165/128615821825334210638, 44572169525/7118599318}, {654686219104361/178761481355556, 249850594047271558364480641/5354229862821602092291248}, {312098738002194296165/128615821825334210638, 5009631795998363486645417214683045/1935878334514951131830244285524398}, {160443526614433014168714029147613242401001/50016678000996026579336936742637753055940,
249850594047271558364480641/5354229862821602092291248}}
So now we have *two* mysterious bit sequences--the signs of the Absand and √ that get us a[n+1] from a[n].
But they only amount to trits: In[61]:= MapAt[N, Reap[Nest[ Block[{L = {(5 (-3 #1 - 2 Sqrt[-25 + #1^4]))/(-25 + 4 #1^2), ( 5 (3 #1 - 2 Sqrt[-25 + #1^4]))/(-25 + 4 #1^2), ( 5 (-3 #1 + 2 Sqrt[-25 + #1^4]))/(-25 + 4 #1^2), ( 5 (3 #1 + 2 Sqrt[-25 + #1^4]))/(-25 + 4 #1^2)}}, (Sow[Position[L, #][[1, 1]]]; #) &@ Sort[L, Numerator[#1] > Numerator[#2] &][[1]]] &, 41/12, 999]], 1] // tim During evaluation of In[61]:= 3322.327528,2 Out[61]= {9.65752, {{4, 3, 3, 4, 3, 1, 4, 3, 4, 3, 3, 4, 3, 1, 4, 3, 4, 3, 3, 4, 3, 1, 4, 3, 4, 3, 3, 4, 3, 1, 4, 3, 4, 3, 3, 4, 3, 1, 4, 3, 4, 3, 3, 4, 3, 1, 4, 3, 4, 3, 3, 4, 3, 3, 4, 3, 1, 4, 3, 4, 3, 3, 4, 3, 1, 4, 3, 4, 3, 3, 4, 3, 1, 4, 3, 4, 3, 3, 4, 3, 1, 4, [...] 3, 4, 3, 3, 4, 3, 1, 4, 3, 4, 3, 3, 4, 3, 1, 4, 3, 4, 3, 3, 4, 3, 3, 4, 3, 1, 4, 3, 4, 3, 3, 4, 3, 1, 4, 3, 4}}} In[62]:= FreeQ[%, 2] Out[62]= True I.e., for some reason, the 2nd sign pattern is never the way forward.
On Tue, Jul 1, 2014 at 1:49 AM, Bill Gosper <billgosper@gmail.com> wrote:
GAAAA--Rich privately suggested that the sign pattern in my 1st order, nonlinear recursion was not period 8, not even periodic. And he is right!!
How the freep did he know that? --rwg
The reason I got so exclamatory was the early (and middle)
lack of supporting evidence or motivation for his claim. --rwg
Watch the recurrence slam into reverse at n=57!
[big chop]
On Tue, Jul 1, 2014 at 5:14 PM, Bill Gosper <billgosper@gmail.com> wrote:
[Chop]
On Tue, Jul 1, 2014 at 1:49 AM, Bill Gosper <billgosper@gmail.com> wrote:
GAAAA--Rich privately suggested that the sign pattern in my 1st order, nonlinear recursion was not period 8, not even periodic. And he is right!!
How the freep did he know that? --rwg
The reason I got so exclamatory was the early (and middle) lack of supporting evidence or motivation for his claim. --rwg
Watch the recurrence slam into reverse at n=57!
[big chop]
NeilB seems to have it completely psyched. You just need to carry along your subscript (i.e., a counter): In[784]:= pisano[{n_, x_}] := {n + 1, Abs[(5 (3 x - psign[n]*2 Sqrt[x^4 - 25]))/(4 x^2 - 25)]}
[Terse, no?] In[787]:= NestList[pisano, {2, 41/12}, 4] Out[787]= {{2, 41/12}, {3, 11285/1562}, {4, 3344161/1494696}, {5, 44572169525/7118599318}, {6, 654686219104361/178761481355556}} This list actually starts {0,∞}, {1,5/2}, {2,41/12},..., if pisano bothered to take limits. In[790]:= Floor[Log[Numerator[#[[2]]]]] & /@ NestList[pisano, {2, 41/12}, 69] Verify forward progress: Out[790]= {3, 9, 15, 24, 34, 47, 60, 77, 94, 115, 136, 161, 186, 214, 243, 275, 307, 343, 379, 419, 459, 503, 547, 594, 641, 693, 744, 799, 854, 913, 972, 1034, 1097, 1164, 1230, 1300, 1371, 1445, 1519, 1597, 1675, 1756, 1838, 1923, 2009, 2098, 2188, 2281, 2374, 2471, 2567, 2668, 2769, 2873, 2978, 3086, 3194, 3306, 3418, 3534, 3650, 3770, 3890, 4013, 4136, 4264, 4391, 4522, 4653, 4788} This wouldn't be monotonic if psign ever lied. But wait, what's psign?? psign[n_] := Sign[Mod[n*-0.37271561316037777`, 1] - 1/2] But what's -0.37271561316037777`?? In[775]:= N[InverseWeierstrassP[5^2/4*1/4, {5^2/4, 0}]/ (2 WeierstrassHalfPeriods[{5^2/4, 0}])][[1]] // Chop Out[775]= -0.37271561316037793 which Neil got from staring at a very spongy Riemann surface. Note that this is "only" good out to about n=10^15. So why not leave it symbolic and let Mathematica gin up the requisite precision for the Mod and Sign? Notice the [[1]] near the end of %775. That quotient "N"s to an ordered pair instead of a number! And N seems to be the only access to it--you can't cleave the pair without committing to imprecision. And screwing around with Hold and ReleaseHold. --rwg
On Fri, Jul 18, 2014 at 12:20 AM, Bill Gosper <billgosper@gmail.com> wrote:
On Tue, Jul 1, 2014 at 5:14 PM, Bill Gosper <billgosper@gmail.com> wrote:
[Chop]
On Tue, Jul 1, 2014 at 1:49 AM, Bill Gosper <billgosper@gmail.com> wrote:
GAAAA--Rich privately suggested that the sign pattern in my 1st order, nonlinear recursion was not period 8, not even periodic. And he is right!!
How the freep did he know that? --rwg
The reason I got so exclamatory was the early (and middle) lack of supporting evidence or motivation for his claim. --rwg
Watch the recurrence slam into reverse at n=57!
[big chop]
NeilB seems to have it completely psyched. You just need to carry along your subscript (i.e., a counter): In[784]:= pisano[{n_, x_}] := {n + 1, Abs[(5 (3 x - psign[n]*2 Sqrt[x^4 - 25]))/(4 x^2 - 25)]}
[Terse, no?] In[787]:= NestList[pisano, {2, 41/12}, 4]
Out[787]= {{2, 41/12}, {3, 11285/1562}, {4, 3344161/1494696}, {5, 44572169525/7118599318}, {6, 654686219104361/178761481355556}}
This list actually starts {0,∞}, {1,5/2}, {2,41/12},..., if pisano bothered to take limits. In[790]:= Floor[Log[Numerator[#[[2]]]]] & /@ NestList[pisano, {2, 41/12}, 69]
Verify forward progress: Out[790]= {3, 9, 15, 24, 34, 47, 60, 77, 94, 115, 136, 161, 186, 214, 243, 275, 307, 343, 379, 419, 459, 503, 547, 594, 641, 693, 744, 799, 854, 913, 972, 1034, 1097, 1164, 1230, 1300, 1371, 1445, 1519, 1597, 1675, 1756, 1838, 1923, 2009, 2098, 2188, 2281, 2374, 2471, 2567, 2668, 2769, 2873, 2978, 3086, 3194, 3306, 3418, 3534, 3650, 3770, 3890, 4013, 4136, 4264, 4391, 4522, 4653, 4788}
This wouldn't be monotonic if psign ever lied. But wait, what's psign??
psign[n_] := Sign[Mod[n*-0.37271561316037777`, 1] - 1/2]
But what's -0.37271561316037777`??
In[775]:= N[InverseWeierstrassP[5^2/4*1/4, {5^2/4, 0}]/ (2 WeierstrassHalfPeriods[{5^2/4, 0}])][[1]] // Chop
Out[775]= -0.37271561316037793
which Neil got from staring at a very spongy Riemann surface. Note that this is "only" good out to about n=10^15. So why not leave it symbolic and let Mathematica gin up the requisite precision for the Mod and Sign? Notice the [[1]] near the end of %775. That quotient "N"s to an ordered pair instead of a number! And N seems to be the only access to it--you can't cleave the pair without committing to imprecision. And screwing around with Hold and ReleaseHold. --rwg
Neil's email (which he quietly sent while sitting right next to me) concludes: Additionally, given k=InverseWeierstrassP[<initial x>/4, {n^2/4,0}], we can compute the m'th elliptic curve x-coordinate with 4*WeierstrassP[m*k,{n^2/4,0}] and the m'th y-coordinate (y^2=x^3-n^2*x) with 4*WeierstrassPPrime[m*k,{n^2/4,0}] --Neil This completely finishes Fibonacci's problem of finding all triples of square rationals with common difference 5, the mth having √(central term) Out[811]= 2 Sqrt[WeierstrassP[2 m InverseWeierstrassP[{25/16, 75/32}, {25/4, 0}], {25/4, 0}]] The first eight are In[819]:= Table[Rationalize[%811, 69^-239], {m, 8}] Out[819]= {41/12, 3344161/1494696, 654686219104361/178761481355556, \ 249850594047271558364480641/5354229862821602092291248, \ 160443526614433014168714029147613242401001/\ 50016678000996026579336936742637753055940, \ 209239116668342644167838867143329714389679018137228536721441/\ 93092380947563478644577555596900542802151091304399908363272, \ 6531529540091581433784275365607334203207979539157560140956253170969369\ 04977985961/\ 1652129432363645568174549601966221192003955894694678397387353366810911\ 41974929836, 389694104145848748532083272246996368636625626448600416977271058482117\ 6712668535259971051251201565099266561/ 16701943947756425480481933369219751647526957097882440351972170228225\ 9070147808848039273138602363503527584} Test: In[820]:= Sqrt[#^2 + {5, -5}] & /@ % Out[820]= {{49/12, 31/12}, {4728001/1494696, 113279/ 1494696}, {767067390499249/178761481355556, 518493692732129/ 178761481355556}, {250137278774864229623059201/ 5354229862821602092291248, 249563579992463717493803519/ 5354229862821602092291248}, \ {195577262542844878506138849501555847171249/ 50016678000996026579336936742637753055940, 115038188620995226180802686473825513089249/ 50016678000996026579336936742637753055940}, \ {295147361324101461665473218814630755582253386512598845803201/ 93092380947563478644577555596900542802151091304399908363272, 21214405287844054428542609853501469645112322962237848990081/ 93092380947563478644577555596900542802151091304399908363272}, \ {750390141456429675692975482159216288564649702558340978265068525529492\ 577197267249/ 16521294323636455681745496019662211920039558946946783973873533668109\ 1141974929836, 53863920973762981989969801391321803641388652858263786597026709272685\ 4572430487071/ 16521294323636455681745496019662211920039558946946783973873533668109\ 1141974929836}, \ {391479590099159062446526431532215958267722907243296693773538849552903\ 3628971288541170127075160953800576001/ 1670194394775642548048193336921975164752695709788244035197217022822\ 59070147808848039273138602363503527584, 38790039977791415434943347292124944095416434812036821900993615287736\ 30968811107453857812241981709854955521/ 1670194394775642548048193336921975164752695709788244035197217022822\ 59070147808848039273138602363503527584}} I guess he emailed because he realized it would take a couple of days to sink in. --rwg All along, I was looking for a Somos-like solution with 𝝑 functions. So now my question is: Can ℘ express Somos?
participants (4)
-
Bill Gosper -
Dan Asimov -
James Buddenhagen -
rcs＠xmission.com