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 A090249 a(n) = 28a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 28. 3
 2, 28, 782, 21868, 611522, 17100748, 478209422, 13372763068, 373959156482, 10457483618428, 292435582159502, 8177738816847628, 228684251289574082, 6394981297291226668, 178830792072864772622, 5000867196742922406748 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS 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). LINKS Indranil Ghosh, Table of n, a(n) for n = 0..689 Tanya Khovanova, Recursive Sequences Index entries for linear recurrences with constant coefficients, signature (28, -1). FORMULA 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. G.f.: (2-28*x)/(1-28*x+x^2). - Philippe Deléham, Nov 02 2008 EXAMPLE a(4) = 611522 = 28a(3) - a(2) = 28*21868 - 782 =(14+sqrt(195))^4 + (14-sqrt(195))^4 =611521.999998364 + 0.000001635 =611522. MATHEMATICA 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 *) 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 *) PROG (Sage) [lucas_number2(n, 28, 1) for n in xrange(0, 16)] # Zerinvary Lajos, Jun 27 2008 CROSSREFS Cf. A053204, A063872. Sequence in context: A151332 A098631 A089836 * A264411 A009256 A012725 Adjacent sequences:  A090246 A090247 A090248 * A090250 A090251 A090252 KEYWORD easy,nonn AUTHOR Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Jan 24 2004 EXTENSIONS More terms from Robert G. Wilson v, Jan 30 2004 STATUS approved

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Last modified December 18 08:16 EST 2018. Contains 318219 sequences. (Running on oeis4.)