This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.
%I A157880 #13 Dec 04 2019 19:06:56 %S A157880 0,136,157080,181270320,209185792336,241400223085560, %T A157880 278575648254944040,321476056685982336736,370983090839975361649440, %U A157880 428114165353274881361117160,494043375834588373115367553336,570125627598949629300252795432720,657924480205812037624118610561805680 %N A157880 Expansion of 136*x^2 / (-x^3+1155*x^2-1155*x+1). %C A157880 This sequence is part of a solution of a more general problem involving two equations, three sequences a(n), b(n), c(n) and a constant A: %C A157880 A * c(n)+1 = a(n)^2, %C A157880 (A+1) * c(n)+1 = b(n)^2, for details see comment in A157014. %C A157880 A157880 is the c(n) sequence for A=8. %H A157880 Colin Barker, <a href="/A157880/b157880.txt">Table of n, a(n) for n = 1..325</a> %H A157880 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (1155,-1155,1). %F A157880 G.f.: 136*x^2/(-x^3+1155*x^2-1155*x+1). %F A157880 c(1) = 0, c(2) = 136, c(3) = 1155*c(2), c(n) = 1155 * (c(n-1)-c(n-2)) + c(n-3) for n>3. %F A157880 a(n) = -((577+408*sqrt(2))^(-n)*(-1+(577+408*sqrt(2))^n)*(17+12*sqrt(2)+(-17+12*sqrt(2))*(577+408*sqrt(2))^n))/288. - _Colin Barker_, Jul 25 2016 %t A157880 LinearRecurrence[{1155,-1155,1},{0,136,157080},20] (* _Harvey P. Dale_, Dec 04 2019 *) %o A157880 (PARI) concat(0, Vec(136*x^2/(-x^3+1155*x^2-1155*x+1) + O(x^20))) \\ _Charles R Greathouse IV_, Sep 26 2012 %o A157880 (PARI) a(n) = round(-((577+408*sqrt(2))^(-n)*(-1+(577+408*sqrt(2))^n)*(17+12*sqrt(2)+(-17+12*sqrt(2))*(577+408*sqrt(2))^n))/288) \\ _Colin Barker_, Jul 25 2016 %Y A157880 8*A157880(n)+1 = A077420(n-1)^2. %Y A157880 9*A157880(n)+1 = A046176(n)^2. %K A157880 nonn,easy %O A157880 1,2 %A A157880 _Paul Weisenhorn_, Mar 08 2009 %E A157880 Edited by _Alois P. Heinz_, Sep 09 2011