OFFSET
0,1
COMMENTS
From Klaus Purath, Sep 02 2025: (Start)
Let (t) be any recurrence of the form (24,-1), including this sequence itself. Then the following always applies: a(n) = (t(k-n) + t(k+n))/t(k) for any k, provided that t(k) != 0 and n >= 1. In general, this relationship applies to every Lucas sequence V(P,1). As a special case for V(3,1) see A005248.
This sequence has the discriminant D = 572 = a(1)*a(3) - (a(0) + a(4)) = x*z - y^2 = 24*x*y - x^2 - y2 where (x,y,z) and (x,y) are any consecutive terms of the sequence. Similarly, this relationship applies to every Lucas sequence V(P,1). (End)
LINKS
Indranil Ghosh, Table of n, a(n) for n = 0..723
Tanya Khovanova, Recursive Sequences
Index entries for linear recurrences with constant coefficients, signature (24,-1).
FORMULA
a(n) = p^n + q^n, where p = 12 + sqrt(143) and q = 12 - sqrt(143). - Tanya Khovanova, Feb 06 2007
G.f.: (2-24*x)/(1-24*x+x^2). - Philippe Deléham, Nov 02 2008
a(n) = 2*A077424(n). - R. J. Mathar, Sep 27 2014
E.g.f.: 2*exp(12*x)*cosh(sqrt(143)*x). - Stefano Spezia, Sep 09 2025
MATHEMATICA
a[0] = 2; a[1] = 24; a[n_] := 24a[n - 1] - a[n - 2]; Table[ a[n], {n, 0, 15}] (* Robert G. Wilson v, Jan 30 2004 *)
LinearRecurrence[{24, -1}, {2, 24}, 30] (* Harvey P. Dale, Sep 19 2011 *)
PROG
(SageMath) [lucas_number2(n, 24, 1) for n in range(0, 20)] # Zerinvary Lajos, Jun 26 2008
(PARI) a(n)=([0, 1; -1, 24]^n*[2; 24])[1, 1] \\ Charles R Greathouse IV, Feb 07 2017
CROSSREFS
KEYWORD
easy,nonn
AUTHOR
Nikolay V. Kosinov (kosinov(AT)unitron.com.ua), Jan 18 2004
STATUS
approved