A087799 a(n) = 10*a(n-1) - a(n-2), starting with a(0) = 2 and a(1) = 10.

2, 10, 98, 970, 9602, 95050, 940898, 9313930, 92198402, 912670090, 9034502498, 89432354890, 885289046402, 8763458109130, 86749292044898, 858729462339850, 8500545331353602, 84146723851196170, 832966693180608098, 8245520207954884810

Offset: 0

Author

Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Oct 11 2003

Keywords

• easy
• nonn

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

Examples

			a(4) = 9602 = 10*a(3) - a(2) = 10*970 - 98 = (5+sqrt(24))^4 + (5-sqrt(24))^4.
		

Links

• T. D. Noe, Table of n, a(n) for n = 0..200
• Peter Bala, Some simple continued fraction expansions for an infinite product, Part 1
• 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 recurrences a(n) = k*a(n - 1) +/- a(n - 2)
• Index entries for linear recurrences with constant coefficients, signature (10,-1).

Crossrefs

Cf. A005248, A006242, A036336, A086927, A135927, A174503.

Programs

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

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 *)

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 */

Sage

[lucas_number2(n,10,1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008

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)

Extensions

More terms from Colin Barker, Feb 25 2014