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A339325 Numerators of s in rational solutions of s^4 + s^3 + s^2 + s + 1 = y^2 with |s| <= 1.

%I A339325 #20 Nov 15 2021 09:32:53
%S A339325 -1,0,1,-8,-35,627,-20965,-761577,72676071,-3470319335,-4692803610731,
%T A339325 418741237461085,180890439981934931,-2244276655546627749157,
%U A339325 -764583000726654718413105,12199909914654265034887926688,-296226554714255082163286916350895,-1802246724473363548181037369907741088
%N A339325 Numerators of s in rational solutions of s^4 + s^3 + s^2 + s + 1 = y^2 with |s| <= 1.
%C A339325 If (s, y) is a solution then so is (s, -y); if s != 0 then (1/s, y/s^2) is also a solution.
%C A339325 The quartic elliptic curve is birationally equivalent to v^2 = u^3 - 5*u^2 + 5*u with s = (2*v-u) / (4*u-5). All solutions can be generated from multiples of (u,v) = P = (1,1) and the two transformations above.
%C A339325 Let (s, y) be a solution, a = 1 + s, b = 1 + 1/s and c = |y/s|. Then the distance between a*exp(3*i*Pi/5) and 1 + b*exp(2*i*Pi/5) is c, with a, b, c all rational. This allows creating a rigid regular pentagon with idealized Meccano strips - see 't Hooft for the solution corresponding to 3P, and the Mathematics Stack Exchange link for the derivation and solution corresponding to 4P.
%H A339325 Jeremy Tan, <a href="/A339325/b339325.txt">Table of n, a(n) for n = 1..50</a>
%H A339325 Jeremy Tan, <a href="https://math.stackexchange.com/q/3604438">Rigid pentagons and rational solutions of s^4+s^3+s^2+s+1=y^2</a>, Mathematics Stack Exchange, Apr 1 2020.
%H A339325 Gerard 't Hooft, <a href="https://webspace.science.uu.nl/~hooft101/lectures/meccano.pdf">Meccano Math I</a>
%e A339325 The values of s in solutions (s, y) with |s| <= 1 begin -1, 0, 1/3, -8/11, -35/123, 627/808, -20965/43993, ...
%t A339325 a[1] = -1; a[2] = 0; a[n_] := Module[{x = 4, y = 2, s, xr}, Do[s = (y-1) / (x-1); xr = s^2 - x + 4; {x, y} = {xr, s(x-xr) - y}, n-2]; s = (2y-x) / (4x-5); Numerator[MinimalBy[{s, 1/s}, Abs][[1]]]]; Table[a[k], {k, 20}] (* _Jeremy Tan_, Nov 15 2021 *)
%o A339325 (PARI)
%o A339325 a(n) = {
%o A339325     [u,v] = ellmul(ellinit([0,-5,0,5,0]), [1,1], n);
%o A339325     s = (2*v-u) / (4*u-5);
%o A339325     if(abs(s)>1, s=1/s);
%o A339325     numerator(s)
%o A339325 }
%Y A339325 Cf. A339326 (denominators).
%K A339325 sign,frac
%O A339325 1,4
%A A339325 _Jeremy Tan_, Nov 30 2020