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A371625 The x-coordinate of the point (x,y) where x + y = n, x is an integer, and x/y is as close as possible to phi (by absolute difference).

%I A371625 #91 Aug 04 2025 14:21:29
%S A371625 0,1,2,2,3,4,4,5,5,6,7,7,8,9,9,10,10,11,12,12,13,14,14,15,15,16,17,17,
%T A371625 18,19,19,20,20,21,22,22,23,23,24,25,25,26,27,27,28,28,29,30,30,31,32,
%U A371625 32,33,33,34,35,35,36,36,37,38,38,39,40,40,41,41,42,43
%N A371625 The x-coordinate of the point (x,y) where x + y = n, x is an integer, and x/y is as close as possible to phi (by absolute difference).
%C A371625 a(n) = x = either ceiling or floor of n/phi, according to which minimizes abs(x/(n-x) - phi).
%C A371625 Each term is equal to or one greater than the previous term.
%C A371625 The average run length approaches phi.
%C A371625 The 2 following statements are equivalent for any real n and any function f(x) such that for any real x, f(x) equals an integer within the range (x-1,x+1) (e.g., round(x), ceiling(x), floor(x)):
%C A371625   A371626(n) != A371627(n);
%C A371625   A371626(n) != n-f(n/phi) xor A371627(n) != n-f(n/phi).
%C A371625 Let s(n) = (phi*n - 1 - sqrt(1+(n^2)*(phi^-4)))/2.
%C A371625 Floor(s(n)) equals the number of times that a(n) swapped from being equal to floor(n/phi) to being equal to ceiling(n/phi) when n is extended to the reals.
%C A371625 This is true because s(n) is the solution to the equation n = (phi/4)(phi(2w+1)+sqrt((2w+1)^2 * phi^-4 + 4/phi)) solved for w. The equation gives the n-value of w-th swap from a(n) = floor(n/phi) to a(n) = ceiling(n/phi).
%C A371625 s(n) is asymptotic to n/phi - 1/2.
%C A371625 floor(s(n)) != floor(n/phi - 1/2) <-> a(n) != round(n).
%C A371625 Floor(n/phi) equals the number of times that a(n) swapped from being equal to ceiling(n/phi) to being equal to floor(n/phi) when n is extended to the reals.
%F A371625 a(n) = n - A371626(n).
%F A371625 a(n) = ceiling(n/phi) if floor(s(n)) + floor(n/phi) is even.
%F A371625 a(n) = floor(n/phi) if floor(s(n)) + floor(n/phi) is odd.
%F A371625 a(n) = ceiling(n/phi) - (floor(s(n))+floor(n/phi) mod 2).
%F A371625 a(n) = round(n/phi) + floor(s(n)) - floor(n/phi+1/2).
%e A371625 For n=5, the possibilities are (0,5), (1,4), (2,3), (3,2) and (4,1). Of those, 3/2 is the closest to phi, so a(5)=3.
%Y A371625 Cf. A001622 (phi), A371626 (y_coordinate), A371627 (with 1/phi), A002163 (sqrt(5)).
%K A371625 nonn
%O A371625 1,3
%A A371625 _Colin Linzer_, Mar 29 2024