A258085 - OEIS

%I #28 Sep 08 2022 08:46:12

%S 0,1,2,3,4,5,7,8,12,13,20,21,33,34,54,55,88,89,143,144,232,233,376,

%T 377,609,610,986,987,1596,1597,2583,2584,4180,4181,6764,6765,10945,

%U 10946,17710,17711,28656,28657,46367,46368,75024,75025,121392,121393,196417

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

%C 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.

%H Clark Kimberling, <a href="/A258085/b258085.txt">Table of n, a(n) for n = 1..1000</a>

%H <a href="/index/Rec#order_06">Index entries for linear recurrences with constant coefficients</a>, signature (0,2,0,0,0,-1).

%F From _Colin Barker_, Jun 28 2015: (Start)

%F a(n) = 2*a(n-2) - a(n-6) for n>8.

%F G.f.: -x^2*(x^6+x^5+x^4-x^2-2*x-1) / ((x-1)*(x+1)*(x^4+x^2-1)).

%F (End)

%F For n >= 3, a(n) = A000045(A004526(n+5)) - A000035(n). - _Robert Israel_, Jun 29 2015

%e 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,...).

%e 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.

%p map((t->(t-1,t)) @ combinat:-fibonacci,[1,$4..100]); # _Robert Israel_, Jun 29 2015

%t f = Fibonacci[Range[60]]; u = Union[f, f - 1]

%o (Magma) [0,1] cat &cat[[Fibonacci(n)-1, Fibonacci(n)]: n in [4..40]]; // _Vincenzo Librandi_, Jun 28 2015

%o (PARI) a(n)=if(n<6,n-1,fibonacci((n+5)\2)-n%2) \\ _Charles R Greathouse IV_, Jun 28 2015

%o (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

%Y Cf. A000035, A000045, A004526.

%K nonn,easy

%O 1,3

%A _Clark Kimberling_, Jun 27 2015