A258085 Strictly increasing list of F and F - 1, where F = A000045, the Fibonacci numbers.

0, 1, 2, 3, 4, 5, 7, 8, 12, 13, 20, 21, 33, 34, 54, 55, 88, 89, 143, 144, 232, 233, 376, 377, 609, 610, 986, 987, 1596, 1597, 2583, 2584, 4180, 4181, 6764, 6765, 10945, 10946, 17710, 17711, 28656, 28657, 46367, 46368, 75024, 75025, 121392, 121393, 196417

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)

For n >= 3, a(n) = A000045(A004526(n+5)) - A000035(n). - Robert Israel, Jun 29 2015

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

Cf. A000035, A000045, A004526.

Sequence in context: A306203 A111795 A046098 * A247350 A057484 A091997

Adjacent sequences: A258082 A258083 A258084 * A258086 A258087 A258088

Keyword

nonn,easy

Author

Clark Kimberling, Jun 27 2015

Status

approved