%I #27 Sep 08 2022 08:45:19
%S 0,1,100,9801,960400,94109401,9221760900,903638458801,88547347201600,
%T 8676736387298001,850231618608002500,83314021887196947001,
%U 8163923913326692803600,799981229484128697805801,78389996565531285692164900,7681419682192581869134354401,752700738858307491889474566400
%N Member r=100 of the family of Chebyshev sequences S_r(n) defined in A092184.
%C Partial sums of A046173. [_Joerg Arndt_, Jun 10 2013]
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (99,-99,1).
%F a(n) = ((49+20*sqrt(6))^n+(49-20*sqrt(6))^n -2)/96 = 98*a(n-1)-a(n-2)+2 = 99*a(n-1)-99*a(n-2)+a(n-3) = (a(n-1)-1)^2/a(n-2) = A004189(n)^2.
%F G.f.: -x*(x+1)/((x-1)*(x^2-98*x+1)). [_Colin Barker_, Oct 24 2012]
%F From _Wolfdieter Lang_, Feb 01 2016: (Start)
%F a(n) = (T(n, 49) - 1)/48 = (T(2*n, 5) - 1)/48 with Chebyshev's T polynomials A053120. See the name.
%F a(n) = A000217((T(n, 5) - 1)/2)/3. n >= 0.
%F a(n) = S(n-1, 10)^2 = A004189(n)^2, with Chebyshev's S polynomials A049310. This is the triangular number = 3*square number identity. Cf. the famous triangular number = square number identity: A000217((T(n, 3) - 1)/2) = S(n-1, 6)^2. A001109 and A001110. (End)
%t LinearRecurrence[{99, -99, 1}, {0, 1, 100}, 20] (* _Vincenzo Librandi_, Feb 02 2016 *)
%o (Magma) I:=[0,1,100]; [n le 3 select I[n] else 99*Self(n-1)-99*Self(n-2)+Self(n-3): n in [1..20]]; // _Vincenzo Librandi_, Feb 02 2016
%Y Cf. A000217, A004189.
%K nonn,easy
%O 0,3
%A _Henry Bottomley_, Jun 22 2005