OFFSET
0,6
COMMENTS
Each step of Newton's method for this is x -> f(x) = x - tan(x).
If |x - k*Pi| < A381473 = 1.16556... for some k then each step takes x closer to k*Pi (by absolute difference) and so converges on the multiple k*Pi.
If n itself is within |n - k*Pi| < A381473 then that convergence to k*Pi begins with the initial x = n and in that case a(n) = round(n/Pi) = A082964(n).
The width of this convergence region around each k*Pi is 2*A381473 = A257451 = 2.331... so contains at least 2 integers n and so every k >= 0 occurs at least twice in the sequence.
When x (or n) is further from its nearest multiple of Pi, the slope of sin(x) sends a step off to possibly much bigger or smaller places on the real line and may converge on some k*Pi far from the initial n.
Those steps can be large enough to reach negative k, as for example first at a(99) = -810 (see A381892 and A381893).
Conjecture: Infinitely many negative integers appear.
LINKS
Simcha Z. Katzoff, Table of n, a(n) for n = 0..1000
FORMULA
a(n) = (1/Pi)*lim_{k->oo} f_k(n), where f_k(n) denotes the k-th iteration of the function f(x) = x - tan(x).
EXAMPLE
a(11) = 74: applying Newton's method to f(x) = sin(x) with initial guess x_0 = 11 gives x_1 = 236.9508, x_2 = 232.8538, etc., eventually converging to x = 232.477856365644 with a(11) = x/Pi = 74.
MATHEMATICA
delta = 1.1655611852072113068339179779585606691;
a[n_]:=(x=n; While[Abs[x-Round[x, Pi]]>delta, x=x-Tan[x]]; Round[x, Pi]/Pi);
Array[a, 100]
CROSSREFS
KEYWORD
sign,easy
AUTHOR
Simcha Z. Katzoff, Mar 07 2025
STATUS
approved