A381781 a(n) = k where k*Pi is the solution to sin(x) = 0 obtained using Newton's method starting from x = n.

0, 0, 1, 1, 1, 3, 2, 2, 5, 3, 3, 74, 4, 4, 2, 5, 5, 4, 6, 6, 6, 7, 7, 7, 8, 8, 8, 10, 9, 9, 13, 10, 10, 30, 11, 11, 9, 12, 12, 11, 13, 13, 13, 14, 14, 14, 15, 15, 15, 25, 16, 16, 14, 17, 17, 32, 18, 18, 16, 19, 19, 18, 20, 20, 17, 21, 21, 21, 22, 22, 22, 25, 23, 23, 26

Offset: 0

Author

Simcha Z. Katzoff, Mar 07 2025

Keywords

• sign
• easy

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.

Examples

			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.
		

Links

• Simcha Z. Katzoff, Table of n, a(n) for n = 0..1000

Crossrefs

Cf. A381892, A381893, A082964, A381473.

Programs

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]

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).