OFFSET
1,3
COMMENTS
If abs(A339325(n)) = 1 or a(n) = 1 then n <= 3, i.e., the only integer solutions to s^4 + s^3 + s^2 + s + 1 = y^2 are (s, y) = (-1, +-1), (0, +-1), (3, +-11). This may easily be shown by bounding the LHS between two consecutive perfect squares.
LINKS
Jeremy Tan, Table of n, a(n) for n = 1..50
Jeremy Tan, Rigid pentagons and rational solutions of s^4+s^3+s^2+s+1=y^2, Mathematics Stack Exchange, Apr 1 2020.
Gerard 't Hooft, Meccano Math I
EXAMPLE
The values of s in solutions (s, y) with |s| <= 1 begin -1, 0, 1/3, -8/11, -35/123, 627/808, -20965/43993, ...
MATHEMATICA
a[1] = 1; a[2] = 1; 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); Denominator[MinimalBy[{s, 1/s}, Abs][[1]]]]; Table[a[k], {k, 20}] (* Jeremy Tan, Nov 15 2021 *)
PROG
(PARI)
a(n) = {
[u, v] = ellmul(ellinit([0, -5, 0, 5, 0]), [1, 1], n);
s = (2*v-u) / (4*u-5);
if(abs(s)>1, s=1/s);
denominator(s)
}
CROSSREFS
KEYWORD
nonn,frac
AUTHOR
Jeremy Tan, Nov 30 2020
STATUS
approved