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 A087799 a(n) = 10*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 10. 6
 2, 10, 98, 970, 9602, 95050, 940898, 9313930, 92198402, 912670090, 9034502498, 89432354890, 885289046402, 8763458109130, 86749292044898, 858729462339850, 8500545331353602, 84146723851196170, 832966693180608098, 8245520207954884810 (list; graph; refs; listen; history; text; internal format)
 OFFSET 0,1 COMMENTS a(n+1)/a(n) converges to (5+sqrt(24)) = 9.8989794... a(0)/a(1)=2/10; a(1)/a(2)=10/98; a(2)/a(3)=98/970; a(3)/a(4)=970/9602; ... etc. Lim a(n)/a(n+1) as n approaches infinity = 0.10102051... = 1/(5+sqrt(24)) = (5-sqrt(24)). Except for the first term, positive values of x (or y) satisfying x^2 - 10xy + y^2 + 96 = 0. - Colin Barker, Feb 25 2014 A triangle whose sides are a(n) - 1, a(n) and a(n) + 1 is nearly Fleenor-Heronian since its area is the product of an integer and the square root of 2. See A003500. - Charlie Marion, Dec 18 2020 LINKS T. D. Noe, Table of n, a(n) for n = 0..200 Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5. Tanya Khovanova, Recursive Sequences Index entries for linear recurrences with constant coefficients, signature (10,-1). FORMULA a(n) = (5+sqrt(24))^n + (5-sqrt(24))^n. G.f.: (2-10*x)/(1-10*x+x^2). - Philippe Deléham, Nov 02 2008 From Peter Bala, Jan 06 2013: (Start) Let F(x) = Product_{n = 0..inf} (1 + x^(4*n+1))/(1 + x^(4*n+3)). Let alpha = 5 - sqrt(24). This sequence gives the simple continued fraction expansion of 1 + F(alpha) = 2.09989 80642 72052 68138 ... = 2 + 1/(10 + 1/(98 + 1/(970 + ...))). Also F(-alpha) = 0.89989 78538 78393 34715 ... has the continued fraction representation 1 - 1/(10 - 1/(98 - 1/(970 - ...))) and the simple continued fraction expansion 1/(1 + 1/((10-2) + 1/(1 + 1/((98-2) + 1/(1 + 1/((970-2) + 1/(1 + ...))))))). F(alpha)*F(-alpha) has the simple continued fraction expansion 1/(1 + 1/((10^2-4) + 1/(1 + 1/((98^2-4) + 1/(1 + 1/((970^2-4) + 1/(1 + ...))))))). Cf. A174503 and A005248. (End) a(-n) = a(n). - Michael Somos, Feb 25 2014 From Peter Bala, Oct 16 2019: (Start) 8*Sum_{n >= 1} 1/(a(n) - 12/a(n)) = 1. 12*Sum_{n >= 1} (-1)^(n+1)/(a(n) + 8/a(n)) = 1. Series acceleration formulas for sums of reciprocals: Sum_{n >= 1} 1/a(n) = 1/8 - 12*Sum_{n >= 1} 1/(a(n)*(a(n)^2 - 12))  and Sum_{n >= 1} (-1)^(n+1)/a(n) = 1/12 + 8*Sum_{n >= 1} (-1)^(n+1)/(a(n)*(a(n)^2 + 8)). Sum_{n >= 1} 1/a(n) = ( (theta_3(5-sqrt(24)))^2 - 1 )/4 and Sum_{n >= 1} (-1)^(n+1)/a(n) = ( 1 - (theta_3(sqrt(24)-5))^2 )/4, where theta_3(x) = 1 + 2*Sum_{n >= 1} x^(n^2) (see A000122). Cf. A153415 and A003499. (End) E.g.f.: 2*exp(5*x)*cosh(2*sqrt(6)*x). - Stefano Spezia, Oct 18 2019 From Peter Bala, Mar 29 2022: (Start) a(n) = 2*T(n,5), where T(n,x) denotes the n-th Chebyshev polynomial of the first kind. a(2^n) = A135927(n+1) and a(3^n) = A006242(n+1), both for n >= 0. (End) EXAMPLE a(4) = 9602 = 10*a(3) - a(2) = 10*970 - 98 = (5+sqrt(24))^4 + (5-sqrt(24))^4. MATHEMATICA a = 2; a = 10; a[n_] := 10a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 17}] (* Robert G. Wilson v, Jan 30 2004 *) LinearRecurrence[{10, -1}, {2, 10}, 30] (* G. C. Greubel, Nov 07 2018 *) PROG (Sage) [lucas_number2(n, 10, 1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008 (PARI) polsym(x^2 - 10*x + 1, 20) \\ Charles R Greathouse IV, Jun 11 2011 (PARI) {a(n) = 2 * real( (5 + 2 * quadgen(24))^n )}; /* Michael Somos, Feb 25 2014 */ (MAGMA) I:=[2, 10]; [n le 2 select I[n] else 10*Self(n-1) - Self(n-2): n in [1..30]]; // G. C. Greubel, Nov 07 2018 CROSSREFS Cf. A005248, A006242, A036336, A086927, A135927, A174503. Sequence in context: A193435 A132572 A069247 * A124214 A098279 A345258 Adjacent sequences:  A087796 A087797 A087798 * A087800 A087801 A087802 KEYWORD easy,nonn AUTHOR Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Oct 11 2003 EXTENSIONS More terms from Colin Barker, Feb 25 2014 STATUS approved

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Last modified May 20 06:18 EDT 2022. Contains 353852 sequences. (Running on oeis4.)