OFFSET
1,3
COMMENTS
Beginning with a(4) = 3, these are the numbers m such that if r = golden ratio and the fractional parts {r}, {2 r}, ..., {mr} are arranged in increasing order, then the set of differences {kr} - {(k - 1)r}, for k = 2..m, consists of exactly two numbers.
LINKS
Clark Kimberling, Table of n, a(n) for n = 1..1000
Index entries for linear recurrences with constant coefficients, signature (0,2,0,0,0,-1).
FORMULA
From Colin Barker, Jun 28 2015: (Start)
a(n) = 2*a(n-2) - a(n-6) for n>8.
G.f.: -x^2*(x^6+x^5+x^4-x^2-2*x-1) / ((x-1)*(x+1)*(x^4+x^2-1)).
(End)
EXAMPLE
F = (1,1,2,3,5,8,13,...); F-1 = (0,0,1,2,4,7,12,...), so that the ordered list of F and F-1 is (0,1,2,3,4,5,7,8,...).
Regarding the fractional parts in Comment, for r = golden ratio and m = 7, the fractional parts are ordered as follows: -8+r, -3+2r, -11+7r, -6+4 r,-1+r, -9+6r, -4+3r. The set of differences is {5-3r, -8+5r}, so that 7 is a term in A258085.
MAPLE
map((t->(t-1, t)) @ combinat:-fibonacci, [1, $4..100]); # Robert Israel, Jun 29 2015
MATHEMATICA
f = Fibonacci[Range[60]]; u = Union[f, f - 1]
PROG
(Magma) [0, 1] cat &cat[[Fibonacci(n)-1, Fibonacci(n)]: n in [4..40]]; // Vincenzo Librandi, Jun 28 2015
(PARI) a(n)=if(n<6, n-1, fibonacci((n+5)\2)-n%2) \\ Charles R Greathouse IV, Jun 28 2015
(PARI) concat(0, Vec(-x^2*(x^6+x^5+x^4-x^2-2*x-1)/((x-1)*(x+1)*(x^4+x^2-1)) + O(x^100))) \\ Colin Barker, Feb 16 2017
CROSSREFS
KEYWORD
nonn,easy
AUTHOR
Clark Kimberling, Jun 27 2015
STATUS
approved