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%I #28 Jun 19 2015 11:19:41
%S 1,12,119,1178,11661,115432,1142659,11311158,111968921,1108378052,
%T 10971811599,108609737938,1075125567781,10642645939872,
%U 105351333830939,1042870692369518,10323355589864241,102190685206272892,1011583496472864679,10013644279522373898
%N Bisection (even part) of Chebyshev sequence with Diophantine property.
%C b(n)^2 - 24*a(n)^2 = 25, with the companion sequence b(n) = A077409(n).
%C The odd part is A077249(n) with Diophantine companion A077250(n).
%H Colin Barker, <a href="/A077251/b077251.txt">Table of n, a(n) for n = 0..1000</a>
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H <a href="/index/Ch#Cheby">Index entries for sequences related to Chebyshev polynomials.</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (10,-1).
%F a(n) = 10*a(n-1)- a(n-2), a(-1)=-2, a(0)=1.
%F a(n) = S(n, 10)+2*S(n-1, 10), with S(n, x) = U(n, x/2), Chebyshev's polynomials of the 2nd kind, A049310. S(n, 10)= A004189(n+1).
%F a(n) = sqrt((A077409(n)^2 - 25)/24).
%F G.f.: (1+2*x)/(1-10*x+x^2).
%e 24*a(1)^2 + 25 = 24*12^2 + 25 = 3481 = 59^2 = A077409(1)^2.
%t CoefficientList[Series[(2 z + 1)/(z^2 - 10 z + 1), {z, 0, 200}], z] (* _Vladimir Joseph Stephan Orlovsky_, Jun 11 2011 *)
%o (PARI) Vec((1+2*x)/(1-10*x+x^2)+O(x^99)) \\ _Charles R Greathouse IV_, Jun 11 2011
%o (PARI) a(n)=([0,1;-1,10]^n*[1;12])[1,1] \\ _Charles R Greathouse IV_, Jun 15 2015
%K nonn,easy
%O 0,2
%A _Wolfdieter Lang_, Nov 08 2002