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 A098301 #59 Feb 24 2021 02:43:39
%S A098301 0,1,16,225,3136,43681,608400,8473921,118026496,1643897025,
%T A098301 22896531856,318907548961,4441809153600,61866420601441,
%U A098301 861688079266576,12001766689130625,167163045568562176,2328280871270739841,32428769152221795600,451674487259834398561
%N A098301 Member r=16 of the family of Chebyshev sequences S_r(n) defined in A092184.
%C A098301 Also m such that (3*m^2 + m)/4 = m*(3*m + 1)/4 is a perfect square. - _Ctibor O. Zizka_, Oct 15 2010
%C A098301 Consequently A049451(k) is a square if and only if k = a(n). - _Bruno Berselli_, Oct 14 2011
%H A098301 Colin Barker, <a href="/A098301/b098301.txt">Table of n, a(n) for n = 0..874</a>
%H A098301 S. Barbero, U. Cerruti, and N. Murru, <a href="http://www.seminariomatematico.polito.it/rendiconti/78-1/BarberoCerrutiMurru.pdf">On polynomial solutions of the Diophantine equation (x + y - 1)^2 = wxy</a>, Rendiconti Sem. Mat. Univ. Pol. Torino (2020) Vol. 78, No. 1, 5-12.
%H A098301 <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%H A098301 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (15,-15,1).
%F A098301 a(n) = (T(n, 7)-1)/6 with Chebyshev's polynomials of the first kind evaluated at x=7: T(n, 7) = A011943(n) = ((7 + 4*sqrt(3))^n + (7 - 4*sqrt(3))^n)/2; therefore: a(n) = ((7 + 4*sqrt(3))^n + (7 - 4*sqrt(3))^n - 2)/12.
%F A098301 a(n) = A001353(n)^2 = S(n-1, 4)^2 with Chebyshev's polynomials of the second kind evaluated at x=4, S(n, 4):=U(n, 2).
%F A098301 a(n) = 14*a(n-1) - a(n-2) + 2, n >= 2, a(0)=0, a(1)=1.
%F A098301 a(n) = 15*a(n-1) - 15*a(n-2) + a(n-3), n >= 3.
%F A098301 G.f.: x*(1+x)/((1-x)*(1 - 14*x + x^2)) = x*(1+x)/(1 - 15*x + 15*x^2 - x^3) (from the Stephan link, see A092184).
%F A098301 Conjecture: 4*A007655(n+1) + A046184(n) = A055793(n+2) + a(n+1). - _Creighton Dement_, Nov 01 2004
%F A098301 a(n) = (A001075(n)^2-1)/3. - _Parker Grootenhuis_, Nov 28 2017
%t A098301 LinearRecurrence[{# - 1, -# + 1, 1}, {0, 1, #}, 20] &[16] (* _Michael De Vlieger_, Feb 23 2021 *)
%o A098301 (PARI) concat(0, Vec(x*(1+x)/((1-x)*(1-14*x+x^2)) + O(x^50))) \\ _Colin Barker_, Jun 15 2015
%Y A098301 Cf. A001075, A001353, A049451, A092184.
%K A098301 nonn,easy
%O A098301 0,3
%A A098301 _Wolfdieter Lang_, Oct 18 2004