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User: Simcha Z. Katzoff

Simcha Z. Katzoff has authored 3 sequences.

A381893 Negative values of A381781.

-810, -39184, -4396135, -124970, -61325522, -64927344, -439288021

Offset: 1

Simcha Z. Katzoff, Mar 09 2025

			A381781(99) = -810.

Cf. A381781, A381892.

a(n) = A381781(A381892(n)).

A381892 Numbers k such that A381781(k) is negative.

Original entry on oeis.org

99, 260515, 18997153, 37362253, 50601157, 122925461, 534483448

Offset: 1

Simcha Z. Katzoff, Mar 09 2025

			A381781(99) = -810.

Cf. A381781, A381893.

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

Original entry on oeis.org

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

Simcha Z. Katzoff, Mar 07 2025

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.

			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.

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

Cf. A381892, A381893, A082964, A381473.

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]

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).

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User: Simcha Z. Katzoff

A381893 Negative values of A381781.

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A381892 Numbers k such that A381781(k) is negative.

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

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