OFFSET
0,3
COMMENTS
Let b(1)=1, b(k) = floor(b(k-1)) + 3/b(k-1); then for n>1, b(n) = a(n)/a(n-1). - Benoit Cloitre, Sep 09 2002
In general, x/(1 - a*x - b*x^2) has a(n) = Sum_{k=0..floor((n-1)/2)} C(n-k-1,k)*b^k*a^(n-2k-1). - Paul Barry, Apr 23 2005
Pisano period lengths: 1, 2, 1, 4, 24, 2, 21, 4, 3, 24, 40, 4, 84, 42, 24, 8, 288, 6, 18, 24, ... . - R. J. Mathar, Aug 10 2012
This is the Lucas sequence U(4,-3). - Bruno Berselli, Jan 09 2013
LINKS
Vincenzo Librandi, Table of n, a(n) for n = 0..1000
Lucyna Trojnar-Spelina and Iwona Włoch, On Generalized Pell and Pell-Lucas Numbers, Iranian Journal of Science and Technology, Transactions A: Science (2019), 1-7.
Wikipedia, Lucas sequence.
Index entries for linear recurrences with constant coefficients, signature (4,3).
FORMULA
a(n) = 4*a(n-1) + 3*a(n-2).
a(n) = Sum_{k=0..floor((n-1)/2)} C(n-k-1, k)*3^k*4^(n-2k-1). - Paul Barry, Apr 23 2005
a(n) = ((2+sqrt(7))^n - (2-sqrt(7))^n)/sqrt(28). Offset 1. a(3)=19. - Al Hakanson (hawkuu(AT)gmail.com), Jan 05 2009
From Johannes W. Meijer, Aug 01 2010: (Start)
Limit(a(n+k)/a(k), k=infinity) = A108851(n)+a(n)*sqrt(7).
Limit(A108851(n)/a(n), n=infinity) = sqrt(7). (End)
G.f.: x*G(0) where G(k)= 1 + (4*x+3*x^2)/(1 - (4*x+3*x^2)/(4*x + 3*x^2 + 1/G(k+1))); (continued fraction, 3rd kind, 3-step). - Sergei N. Gladkovskii, Jul 28 2012
G.f.: G(0)*x/(2-4*x), where G(k)= 1 + 1/(1 - x*(7*k-4)/(x*(7*k+3) - 2/G(k+1))); (continued fraction). - Sergei N. Gladkovskii, Jun 16 2013
MATHEMATICA
LinearRecurrence[{4, 3}, {0, 1}, 30] (* Vincenzo Librandi, Jun 19 2012 *)
PROG
(Sage) [lucas_number1(n, 4, -3) for n in range(0, 23)]# Zerinvary Lajos, Apr 23 2009
(Magma) I:=[0, 1]; [n le 2 select I[n] else 4*Self(n-1)+3*Self(n-2): n in [1..30]]; // Vincenzo Librandi, Jun 19 2012
(PARI) x='x+O('x^30); concat([0], Vec(x/(1-4*x-3*x^2))) \\ G. C. Greubel, Jan 24 2018
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
STATUS
approved