A049670 a(n) = Fibonacci(10*n)/55.

0, 1, 123, 15128, 1860621, 228841255, 28145613744, 3461681649257, 425758697244867, 52364858079469384, 6440451785077489365, 792123204706451722511, 97424713727108484379488, 11982447665229637126954513, 1473743638109518258131025611, 181258485039805516112989195640

Offset: 0

Author

Clark Kimberling

Keywords

• nonn
• easy

Comments

Chebyshev polynomials S(n-1,123).

Used for all positive integer solutions of Pell equation x^2 - 5*(5*y)^2 = -4. See A097842 with A097843.

This is the k = 10 member of the k-family of sequences {F(k*n)/F(k)}, n >= 0 for k >= 1, with o.g.f. x/(1 - L(k)*x + (-1)^k*x^2). Proof: Binet-de Moivre formula for F and L. See also A028412. - Wolfdieter Lang, Aug 26 2012

Links

• Robert Israel, Table of n, a(n) for n = 0..383
• 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 sequences related to Chebyshev polynomials.
• Index entries for linear recurrences with constant coefficients, signature (123,-1).

Crossrefs

A column of array A028412.

Cf. A000045.

Programs

Magma

[ Fibonacci(10*n)/55: n in [0..30]]; // G. C. Greubel, Dec 02 2017

Maple

seq(combinat:-fibonacci(10*n)/55, n=0..20); # Robert Israel, Apr 03 2015

Mathematica

Table[Fibonacci[10 n]/55, {n, 12}] (* Michael De Vlieger, Apr 03 2015 *)
LinearRecurrence[{123,-1},{0,1},20] (* Harvey P. Dale, Dec 03 2019 *)

MuPAD

numlib::fibonacci(10*n)/55 $ n = 0..25; // Zerinvary Lajos, May 09 2008

PARI

a(n)=fibonacci(10*n)/55 \\ Charles R Greathouse IV, Oct 07 2016

Formula

G.f.: x/(1-123*x+x^2), 123=L(10)=A000032(10) (Lucas).

a(n+1) = S(n, 123) = U(n, 123/2) = S(2*n+1, 5*sqrt(5))/(5*sqrt(5)), n>=0, with S(n, x) = U(n, x/2) Chebyshev's polynomials of the second kind, A049310. S(-1, x)= 0 = U(-1, x).

a(n) = 123*a(n-1) - a(n-2), n >= 2; a(0)=0, a(1)=1.

a(n) = (ap^n - am^n)/(ap-am) with ap := (123+55*sqrt(5))/2 and am := (123-55*sqrt(5))/2 = 1/ap.

From Peter Bala, Nov 29 2013: (Start)

a(n) = 1/(11*55)*(F(10*n + 5) - F(10*n - 5)).

Sum_{n >= 1} 1/( 11*a(n) + 1/(11*a(n)) ) = 1/11. Compare with A001906 and A049660. (End)

From Peter Bala, Apr 03 2015: (Start)

For integer k, 1 + k*(22 - k)*Sum_{n >= 1} a(n)*x^(2*n) = ( 1 + k/5*Sum_{n >= 1} Fibonacci(5*n)*x^n )*( 1 + k/5*Sum_{n >= 1} Fibonacci(5*n)*(-x)^n ).

1 + 4*Sum_{n >= 1} a(n)*x^(2*n) = ( 1 + 2/5*Sum_{n >= 1} Fibonacci(5*n+5)*x^n )*( 1 + 2/5*Sum_{n >= 1} Fibonacci(5*n+5)*(-x)^n ) = ( 1 + 2/5*Sum_{n >= 1} Fibonacci(5*n-5)*x^n )*( 1 + 2/5*Sum_{n >= 1} Fibonacci(5*n-5)*(-x)^n ).

1 + 25*Sum_{n >= 1} a(n)*x^(2*n) = ( 1 + Sum_{n >= 1} Fibonacci(5*n+3)*x^n )*( 1 + Sum_{n >= 1} Fibonacci(5*n+3)*(-x)^n ) = ( 1 + Sum_{n >= 1} Fibonacci(5*n-3)*x^n )*( 1 + Sum_{n >= 1} Fibonacci(5*n-3)*(-x)^n ).

1 + 100*Sum_{n >= 1} a(n)*x^(2*n) = ( 1 + 2*Sum_{n >= 1} Fibonacci(5*n+1)*x^n )*( 1 + 2*Sum_{n >= 1} Fibonacci(5*n+1)*(-x)^n ) = ( 1 + 2*Sum_{n >= 1} Fibonacci(5*n-1)*x^n )*( 1 + 2*Sum_{n >= 1} Fibonacci(5*n-1)*(-x)^n ).

1 + 125*Sum_{n >= 1} a(n)*x^(2*n) = ( 1 + Sum_{n >= 1} Lucas(5*n)*x^n )*( 1 + Sum_{n >= 1} Lucas(5*n)*(-x)^n ). (End)

Extensions

More terms from James Sellers, Jan 20 2000

Chebyshev and Pell comments from Wolfdieter Lang, Sep 10 2004