OFFSET
0,1
COMMENTS
From Klaus Purath, Sep 25 2024: (Start)
a(n) = (t(i+3n) - t(i))/(t(i+2n) - t(i+n)) - 1 for n >= 1, where (t) is any sequence satisfying t(i) = 12t(i-1) - 12t(i-2) + t(i-3) or t(i) = 11t(i-1) - t(i-2) without regard to initial values and including this sequence itself, as long as t(i+2n) - t(i+n) != 0 for integer i.
a(n) = (t(i+3n) + t(i))/(t(i+2n) + t(i+n)) + 1 for i >= 0, n >= 1, where (t) is any sequence satisfying t(i) = 10t(i-1) + 10t(i-2) - t(i-3) or t(i) = 11t(i-1) - t(i-2) without regard to initial values and including this sequence itself, as long as t(i+2n) + t(i+n) != 0.
a(n) = (t(i-n) + t(i+n))/t(i) for i >= n >= 0, where (t) is any recurrence of the form (11,-1) including this sequence itself, as long as t(i) != 0.
a(n) = t(n) - t(n-1) = (t(n+1) - t(n-2))/12, where (t) is any third order recurrence with constant coefficients (12,-12,1) and initial values t(0) = x, t(1) = x + 2, t(2) = x + 13 for integer x.
a(n) = t(n-1) + t(n) = (t(n-2) + t(n+1))/10, where (t) is any third order recurrence with constant coefficients (10,10,-1) and initial values t(0) = x, t(1) = 2 - x, t(2) = x + 9 for integer x. (End)
LINKS
Michael De Vlieger, Table of n, a(n) for n = 0..963
Hacène Belbachir, Soumeya Merwa Tebtoub, and László Németh, Ellipse Chains and Associated Sequences, J. Int. Seq., Vol. 23 (2020), Article 20.8.5.
P. Bhadouria, D. Jhala, and B. Singh, Binomial Transforms of the k-Lucas Sequences and its Properties, The Journal of Mathematics and Computer Science (JMCS), Volume 8, Issue 1, Pages 81-92; sequence R_3.
S. Falcon, Relationships between Some k-Fibonacci Sequences, Applied Mathematics 5 (2014), 2226-2234.
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (11,-1).
FORMULA
a(n) = S(n, 11) - S(n-2, 11) = 2*T(n, 11/2) with S(n, x) := U(n, x/2), S(-1, x) := 0, S(-2, x) := -1. S(n, 11)=A004190(n). U-, resp. T-, are Chebyshev's polynomials of the second, resp. first, case. See A049310 and A053120.
G.f.: (2-11*x)/(1-11*x+x^2).
a(n) = ap^n + am^n, with ap := (11+sqrt(117))/2 and am := (11-sqrt(117))/2.
a(n) = sqrt(4+117*A004190(n-1)^2), n>=1.
a(n) = a(-n). - Michael Somos, Apr 25 2003
E.g.f.: 2*exp(11*x/2)*cosh(3*sqrt(13)*x/2). - Stefano Spezia, Aug 07 2024
From Klaus Purath, Sep 25 2024: (Start)
a(n) = (a(n-1)*a(n-2) + 1287)/a(n-3) for integer n.
a(n+1)^2 - a(n)*a(n+2) = -117 for integer n. (End)
EXAMPLE
G.f. = 2 + 11*x +119*x^2 + 1298*x^3 + 14159*x^4 + 154451*x^5 + ...
MATHEMATICA
a[0] = 2; a[1] = 11; a[n_] := 11a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 17}] (* Robert G. Wilson v, Jan 30 2004 *)
a[ n_] := 2 ChebyshevT[ n, 11/2]; (* Michael Somos, May 28 2014 *)
LinearRecurrence[{11, -1}, {2, 11}, 30] (* Harvey P. Dale, Sep 13 2024 *)
PROG
(PARI) {a(n) = subst( poltchebi(n), x, 11/2) * 2};
(PARI) {a(n) = 2 * poltchebyshev(n, 1, 11/2)}; /* Michael Somos, May 28 2014 */
(PARI) Vec((2-11*x)/(1-11*x+x^2) + O(x^40)) \\ Michel Marcus, Feb 18 2016
(Sage) [lucas_number2(n, 11, 1) for n in range(27)] # Zerinvary Lajos, Jun 25 2008
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Wolfdieter Lang, Oct 31 2002
STATUS
approved