A384509 a(n) = number of iterations of z -> z^2 + c(n) with c(n) = ((5/n+1) + (5/n-1)*i)/(n*sqrt(2)) + 1/4 + (1/2)*i to reach |z| > 2, starting with z = 0.

1, 2, 2, 3, 4, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 16, 17, 19, 19, 21, 22, 23, 24, 25, 26, 27, 28, 29, 31, 32, 33, 34, 35, 36, 37, 38, 39, 40, 42, 43, 44, 45, 46, 47, 48, 49, 51, 51, 53, 54, 55, 56, 57, 58, 59, 60, 62, 63, 64, 65, 66, 67, 68, 69, 71, 71, 73, 74, 75, 76, 77, 78, 79, 80

Offset: 1

Author

Luke Bennet, May 31 2025

Keywords

• nonn

Comments

a(n)/n seems to converge to Pi/(2*sqrt(2)).

a(n) counts the escape time of points outside the Mandelbrot set that converge to the Mandelbrot set's 1/4 period bulb.

Links

• Luke Bennet, Table of n, a(n) for n = 1..10001
• Thies Brockmöller, Oscar Scherz, and Nedim Srkalović, Pi in the Mandelbrot set everywhere, arXiv preprint arXiv:2505.07138 [math.DS], 2025.
• Aaron Klebanoff, Pi in the Mandelbrot Set, Fractals 9 (2001), nr. 4, p. 393-402.

Crossrefs

Cf. A093954, A097486, A383750, A384513.

Programs

PARI

c(n) = ((5/n+1) + (5/n-1)*I)/(n*sqrt(2)) + 1/4 + (1/2)*I;
a(n) = my(z=0, k=0, c=c(n)); while(norml2(z)<=4, z = z^2 + c; k++); k; \\ Michel Marcus, Jun 01 2025

Python

import mpmath
from mpmath import iv
def a(n):
    dps = 1
    while True:
        mpmath.iv.dps = dps
        c = iv.mpc(iv.mpf(5) / n + 1, iv.mpf(5) / n - 1)
        c = c / (n * iv.sqrt(2)) + 0.25 + 0.5j
        z = iv.mpc(0, 0)
        counter = 0
        while (z.real**2 + z.imag**2).b <= 4:
            z = z ** 2 + c
            counter += 1
        if (z.real**2 + z.imag**2).a > 4:
            return counter
        dps *= 2