OFFSET
0,3
COMMENTS
a(n) = floor(lim_{k->oo} Fibonacci(k)/Fibonacci(k-n)). - Jon Perry, Jun 10 2003
For n > 1, a(n) is the maximum element in the continued fraction for A000045(n)*phi. - Benoit Cloitre, Jun 19 2005
a(n) is also the curvature (rounded down) of the circle inscribed in the n-th kite arranged in a spiral, starting with a unit circle, as shown in the illustration in the links section. - Kival Ngaokrajang, Aug 29 2013
a(n) is the n-th Lucas number (A000032) if n is odd, and a(n) is the n-th Lucas number minus 1 if n is even. (Mario Catalani's formula below expresses this fact.) This is related to the fact that the powers of phi approach the values of the Lucas numbers, the odd powers from above and the even powers from below. - Geoffrey Caveney, Apr 18 2014
a(n) is the sum of the last summands over all Arndt compositions of n (see the Checa link). - Daniel Checa, Dec 25 2023
LINKS
Alois P. Heinz, Table of n, a(n) for n = 0..4784 (first 301 terms from T. D. Noe)
Mohammad K. Azarian, Problem 123, Missouri Journal of Mathematical Sciences, Vol. 10, No. 3, Fall 1998, p. 176. Solution published in Vol. 12, No. 1, Winter 2000, pp. 61-62.
Daniel F. Checa, Arndt Compositions: Connections with Fibonacci Numbers, Statistics, and Generalizations, 2023. p. 29.
Daniel F. Checa and José L. Ramírez, Arndt Compositions: A Generating Functions Approach, Integers (2024) Vol. 24, A35. See p. 14.
G. Harman, One hundred years of normal numbers
Kival Ngaokrajang, Illustration for n = 0..7
Stefan Schuster and Tatjana Malycheva, Enumeration of saturated and unsaturated substituted N-heterocycles, arXiv:2309.02343 [q-bio.BM], 2023.
Index entries for linear recurrences with constant coefficients, signature (1,2,-1,-1).
FORMULA
a(n) = a(n-1) + 2*a(n-2) - a(n-3) - a(n-4).
a(n) = a(n-1) + a(n-2) + (1-(-1)^n)/2 = a(n-1) + a(n-2) + A000035(n).
a(n) = A000032(n) - (1 + (-1)^n)/2. - Mario Catalani (mario.catalani(AT)unito.it), Jan 17 2003
G.f.: (1-x^2+x^3)/((1+x)*(1-x)*(1-x-x^2)). - R. J. Mathar, Sep 06 2008
a(2n-1) = (Fibonacci(4n+1)-2)/Fibonacci(2n+2). - Gary Detlefs, Feb 16 2011
a(n) = floor(Fibonacci(2n+3)/Fibonacci(n+3)). - Gary Detlefs, Feb 28 2011
a(2n) = Fibonacci(2*n-1) + Fibonacci(2*n+1) - 1. - Gary Detlefs, Mar 10 2011
a(n+2*k) - a(n) = A203976(k)*A000032(n+k) if k odd, a(n+2*k) - a(n) = A203976(k)*A000045(n+k) if k even, for k > 0. - Paul Curtz, Jun 05 2013
a(n+6) - a(n-6) = 40*A000045(n), case k=6 of my formula above. - Paul Curtz, Jun 13 2013
From Paul Curtz, Jun 17 2013: (Start)
a(n-3) + a(n+3) = A153382(n).
a(n-1) + a(n+2) = A022319(n). (End)
For k > 0, a(2k) = A169985(2k)-1 and a(2k+1) = A169985(2k+1) (which is equivalent to Catalani's 2003 formula). - Danny Rorabaugh, Apr 15 2015
a(n) = ((-1)^(1+n)-1)/2 + ((1-sqrt(5))/2)^n + ((1+sqrt(5))/2)^n. - Colin Barker, Nov 05 2017
a(n) = floor(2*sinh(n*arccsch(2))). - Federico Provvedi, Feb 23 2022
E.g.f.: 2*exp(x/2)*cosh(sqrt(5)*x/2) - cosh(x). - Stefano Spezia, Jul 26 2022
MAPLE
A014217 := proc(n)
option remember;
if n <= 3 then
op(n+1, [1, 1, 2, 4]) ;
else
procname(n-1)+2*procname(n-2)-procname(n-3)-procname(n-4) ;
end if;
end proc: # R. J. Mathar, Jun 23 2013
#
a:= n-> (<<0|1|0|0>, <0|0|1|0>, <0|0|0|1>, <-1|-1|2|1>>^n. <<1, 1, 2, 4>>)[1, 1]:
seq(a(n), n=0..40); # Alois P. Heinz, Oct 12 2017
MATHEMATICA
Table[Floor[GoldenRatio^n], {n, 0, 36}] (* Vladimir Joseph Stephan Orlovsky, Dec 12 2008 *)
LinearRecurrence[{1, 2, -1, -1}, {1, 1, 2, 4}, 40] (* Jean-François Alcover, Nov 05 2017 *)
PROG
(PARI) my(x='x+O('x^44)); Vec((1-x^2+x^3)/((1+x)*(1-x)*(1-x-x^2))) \\ Joerg Arndt, Jul 10 2023
(Magma) [Floor( ((1+Sqrt(5))/2)^n ): n in [0..100]]; // Vincenzo Librandi, Apr 16 2011
(Haskell)
a014217 n = a014217_list !! n
a014217_list = 1 : 1 : zipWith (+)
a000035_list (zipWith (+) a014217_list $ tail a014217_list)
-- Reinhard Zumkeller, Jan 06 2012
(Sage) [floor(golden_ratio^n) for n in range(37)] # Danny Rorabaugh, Apr 19 2015
(Python)
from sympy import floor, sqrt
def A014217(n): return floor(((1+sqrt(5))/2)**n) # Chai Wah Wu, Dec 17 2021
CROSSREFS
KEYWORD
nonn,easy,nice
AUTHOR
EXTENSIONS
Corrected by T. D. Noe, Nov 09 2006
Edited by N. J. A. Sloane, Aug 29 2008 at the suggestion of R. J. Mathar
STATUS
approved