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Bisection (odd part) of Chebyshev sequence with Diophantine property.
6

%I #27 Jun 19 2015 11:19:41

%S 2,21,208,2059,20382,201761,1997228,19770519,195707962,1937309101,

%T 19177383048,189836521379,1879187830742,18602041786041,

%U 184141230029668,1822810258510639,18043961355076722,178616803292256581,1768124071567489088,17502623912382634299

%N Bisection (odd part) of Chebyshev sequence with Diophantine property.

%C -24*a(n)^2 + b(n)^2 = 25, with the companion sequence b(n) = A077250(n).

%C The even part is A077251(n) with Diophantine companion A077409(n).

%H Colin Barker, <a href="/A077249/b077249.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) := -1, a(0)=2.

%F a(n) = 2*S(n, 10)+S(n-1, 10), with S(n, x) := U(n, x/2), Chebyshev polynomials of 2nd kind, A049310. S(n, 10)= A004189(n+1).

%F G.f.: (2+x)/(1-10*x+x^2).

%e 24*a(1)^2 + 25 = 24*21^2+25 = 10609 = 103^2 = A077250(1)^2.

%t CoefficientList[Series[(z + 2)/(z^2 - 10 z + 1), {z, 0, 200}], z] (* _Vladimir Joseph Stephan Orlovsky_, Jun 11 2011 *)

%t LinearRecurrence[{10,-1},{2,21},40] (* _Harvey P. Dale_, Apr 08 2012 *)

%o (PARI) a(n)=if(n<0,0,subst(-7*poltchebi(n)+11*poltchebi(n+1),x,5)/24)

%o (PARI) a(n)=2*polchebyshev(n,2,5)+polchebyshev(n-1,2,5) \\ _Charles R Greathouse IV_, Jun 11 2011

%o (PARI) Vec((2+x)/(1-10*x+x^2) + O(x^30)) \\ _Colin Barker_, Jun 15 2015

%K nonn,easy

%O 0,1

%A _Wolfdieter Lang_, Nov 08 2002