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%I #47 Sep 06 2017 14:58:28
%S 2,9,79,702,6239,55449,492802,4379769,38925119,345946302,3074591599,
%T 27325378089,242853811202,2158358922729,19182376493359,
%U 170483029517502,1515164889164159,13466000972959929,119678843867475202
%N a(n) = 9*a(n-1)-a(n-2); a(0)=2, a(1)=9.
%C All nonnegative integer solutions of Pell equation a(n)^2 - 77*b(n)^2 = +4 together with b(n)=A018913(n), n>=0. - _Wolfdieter Lang_, Aug 31 2004
%C Except for the first term, positive values of x (or y) satisfying x^2 - 9xy + y^2 + 77 = 0. - _Colin Barker_, Feb 13 2014
%H Reinhard Zumkeller, <a href="/A056918/b056918.txt">Table of n, a(n) for n = 0..1000</a>
%H P. Bala, <a href="/A174500/a174500_2.pdf">Some simple continued fraction expansions for an infinite product, Part 1</a>
%H E. I. Emerson, <a href="http://www.fq.math.ca/Scanned/7-3/emerson.pdf">Recurrent Sequences in the Equation DQ^2=R^2+N</a>, Fib. Quart., 7 (1969), pp. 231-242.
%H A. F. Horadam, <a href="http://www.fq.math.ca/Scanned/5-5/horadam.pdf">Special Properties of the Sequence W(n){a,b; p,q}</a>, Fib. Quart., Vol. 5, No. 5 (1967), pp. 424-434.
%H Tanya Khovanova, <a href="http://www.tanyakhovanova.com/RecursiveSequences/RecursiveSequences.html">Recursive Sequences</a>
%H <a href="/index/Rea#recur1">Index entries for recurrences a(n) = k*a(n - 1) +/- a(n - 2)</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 (9,-1).
%F a(n) = 9*S(n-1, 9) - 2*S(n-2, 9) = S(n, 9) - S(n-2, 9) = 2*T(n, 9/2), with S(n, x) := U(n, x/2) (see A049310), S(-1, x) := 0, S(-2, x) := -1. S(n-1, 9)=A018913(n). U-, resp. T-, are Chebyshev's polynomials of the second, resp. first, kind.
%F a(n) = {9*[((9+sqrt(77))/2)^n - ((9-sqrt(77))/2)^n] - 2*[((9+sqrt(77))/2)^(n-1) - ((9-sqrt(77))/2)^(n-1)]}/sqrt(77).
%F G.f.: (2-9*x)/(1-9*x+x^2).
%F a(n) = ap^n + am^n, with ap := (9+sqrt(77))/2 and am := (9-sqrt(77))/2.
%F G.f.: (2-9*x)/(1-9*x+x^2). - _Philippe Deléham_, Nov 03 2008
%F From _Peter Bala_, Jan 06 2013: (Start)
%F Let F(x) = product {n = 0..inf} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 1/2*(9 - sqrt(77)). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.11095 50589 89701 91909 ... = 2 + 1/(9 + 1/(79 + 1/(702 + ...))).
%F Also F(-alpha) = 0.88873 23915 40314 47623 ... has the continued fraction representation 1 - 1/(9 - 1/(79 - 1/(702 - ...))) and the simple continued fraction expansion 1/(1 + 1/((9-2) + 1/(1 + 1/((79-2) + 1/(1 + 1/((702-2) + 1/(1 + ...))))))). F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((9^2-4) + 1/(1 + 1/((79^2-4) + 1/(1 + 1/((702^2-4) + 1/(1 + ...))))))). Cf. A005248.
%F (End)
%t a[0] = 2; a[1] = 9; a[n_] := 9a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 17}] (* _Robert G. Wilson v_, Jan 30 2004 *)
%o (Sage) [lucas_number2(n,9,1) for n in range(23)] # _Zerinvary Lajos_, Jun 25 2008
%o (Haskell)
%o a056918 n = a056918_list !! n
%o a056918_list = 2 : 9 :
%o zipWith (-) (map (* 9) $ tail a056918_list) a056918_list
%o -- _Reinhard Zumkeller_, Jan 06 2013
%Y Cf. A018913. a(n)=sqrt(77*A018913(n)^2 + 4). A005248.
%K easy,nonn
%O 0,1
%A _Barry E. Williams_, Aug 21 2000
%E More terms from _James A. Sellers_, Sep 07 2000
%E Chebyshev comments from _Wolfdieter Lang_, Oct 31 2002
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