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%I #26 Dec 07 2019 12:18:24
%S 2,28,782,21868,611522,17100748,478209422,13372763068,373959156482,
%T 10457483618428,292435582159502,8177738816847628,228684251289574082,
%U 6394981297291226668,178830792072864772622,5000867196742922406748
%N a(n) = 28a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 28.
%C a(n+1)/a(n) converges to (14+sqrt(195)) =27.96424004... Lim a(n)/a(n+1) as n approaches infinity = 0.03575995... = 1/(14+sqrt(195)) = (14-sqrt(195)). Lim a(n+1)/a(n) as n approaches infinity = 27.96424004... = (14+sqrt(195)) = 1/(14-sqrt(195)). Lim a(n)/a(n+1) = 28 - Lim a(n+1)/a(n).
%H Indranil Ghosh, <a href="/A090249/b090249.txt">Table of n, a(n) for n = 0..689</a>
%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/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (28, -1).
%F a(n) = 28a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 28. a(n) = (14+sqrt(195))^n + (14-sqrt(195))^n. (a(n))^2 =a(2n)+2.
%F G.f.: (2-28*x)/(1-28*x+x^2). - _Philippe Deléham_, Nov 02 2008
%e a(4) = 611522 = 28a(3) - a(2) = 28*21868 - 782 =(14+sqrt(195))^4 + (14-sqrt(195))^4 =611521.999998364 + 0.000001635 =611522.
%t a[0] = 2; a[1] = 28; a[n_] := 28a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 15}] (* _Robert G. Wilson v_, Jan 30 2004 *)
%t LinearRecurrence[{28,-1},{2,28},20] (* or *) CoefficientList[ Series[ (2-28x)/(x^2-28x+1),{x,0,20}],x] (* _Harvey P. Dale_, Jun 25 2011 *)
%o (Sage) [lucas_number2(n,28,1) for n in range(0,16)] # _Zerinvary Lajos_, Jun 27 2008
%Y Cf. A053204, A063872.
%K easy,nonn
%O 0,1
%A Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Jan 24 2004
%E More terms from _Robert G. Wilson v_, Jan 30 2004
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