The OEIS Foundation is supported by donations from users of the OEIS and by a grant from the Simons Foundation.

 Hints (Greetings from The On-Line Encyclopedia of Integer Sequences!)
 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 LINKS T. D. Noe, Table of n, a(n) for n = 0..200 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 EXAMPLE a(4) = 9602 = 10a(3) - a(2) = 10*970 - 98 =(5+sqrt(24))^4 + (5-sqrt(24))^4 = 9601.999895855 + 0.000104144 = 9602. MATHEMATICA a[0] = 2; a[1] = 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. A086927, A036336. A005248, A174503. Sequence in context: A193435 A132572 A069247 * A124214 A098279 A248615 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

Lookup | Welcome | Wiki | Register | Music | Plot 2 | Demos | Index | Browse | More | WebCam
Contribute new seq. or comment | Format | Style Sheet | Transforms | Superseeker | Recent
The OEIS Community | Maintained by The OEIS Foundation Inc.

Last modified February 25 22:55 EST 2020. Contains 332270 sequences. (Running on oeis4.)