OFFSET
1,4
COMMENTS
a(n) = k = n - either ceiling or floor of n/phi, according to which minimizes abs((n-k)/k - phi).
Each term is equal to or one greater than the previous term.
The average run length approaches 1+phi.
The 4 following statements are equivalent for any positive integer n and any function f(x) such that for any real x, f(x) equals a integer within the range (x-1,x+1):
a(n) != A371627(n);
a(n) != n-f(n/phi) xor A371627(n) != n-f(n/phi);
Let s(n) = (phi*n - 1 - sqrt(1+(n^2)*(phi^-4)))/2.
Floor(s(n)) equals the number of times that a(n) swapped from being equal to n-floor(n/phi) to being equal to n-ceiling(n/phi) when n is extended to the reals.
This is true because s(n) is the solution to the equation n = (phi/4) * (phi(2r+1) + sqrt((2r+1)^2 / phi^4 + 4/phi)) solved for w. The equation gives the n-value of w-th swap from a(n) = n-floor(n/phi) to a(n) = n-ceiling(n/phi).
s(n) is asymptotic to n/phi - 1/2.
Floor(s(n)) != floor(n/phi - 1/2) <-> a(n) != round(n).
Floor(n/phi) equals the number of that a(n) swapped from being equal to n-ceiling(n/phi) to being equal to n-floor(n/phi) when n is extended to the reals.
FORMULA
a(n) = n - A371625(n).
Let s(n) = (phi*n - 1 - sqrt(1+(n^2) / phi^4))/2.
Floor(s(n))+floor(n/phi) is even -> a(n) = n-ceiling(n/phi) = (n mod 1) + floor(n/phi^2).
Floor(s(n))+floor(n/phi) is odd -> a(n) = n-floor(n/phi) = (n mod 1) + ceiling(n/phi^2).
a(n) = -a(-n).
EXAMPLE
For n=5, the possibilities are (0,5), (1,4), (2,3), (3,2), & (4,1). Of those, 3/2 is the closest to phi, so a(5)=3.
CROSSREFS
KEYWORD
nonn,frac
AUTHOR
Colin Linzer, Mar 29 2024
STATUS
approved