A300078 Number of steps of iterating 0 under z^2 + c before escaping, i.e., abs(z^2 + c) > 2, with c = -5/4 - epsilon^2 + epsilon*i, where epsilon = 10^(-n) and i^2 = -1.

1, 18, 159, 1586, 15731, 157085, 1570800, 15707976

Offset: 0

Author

Martin Renner, Feb 24 2018

Keywords

• nonn
• more

Comments

A relation between Pi and the Mandelbrot set: 2*a(n)*epsilon converges to Pi.

c = -5/4 - epsilon^2 + epsilon*i is a parabolic route into the point c = -5/4, the second neck of the Mandelbrot set.

The difference between the terms of a(n) and A300077(n) = floor(1/2*Pi*10^n) is d(n) = 0, 3, 2, 16, 24, 6, 4, 13, ...

Links

• Gerald Edgar, Pi and the Mandelbrot set. (The Ohio State University.)
• Boris Gourévitch, Pi and fractal sets. The Mandelbrot set -- Dave Boll -- Gerald Edgar. (The World of Pi.)
• Aaron Klebanoff, Pi in the Mandelbrot Set. In: Fractals 9 (2001), nr. 4, p. 393-402.

Crossrefs

Cf. A097486, A299415, A300077.

Programs

Maple

Digits:=2^8:
f:=proc(z, c, k) option remember;
  f(z, c, k-1)^2+c;
end;
a:=proc(n)
local epsilon, c, k;
  epsilon:=10.^(-n):
  c:=-1.25-epsilon^2+epsilon*I:
  f(0, c, 0):=0:
  for k do
    if abs(f(0, c, k))>2 then
      break;
    fi;
  od:
  return(k);
end;
seq(a(n), n=0..7);

Python

from fractions import Fraction
def A300078(n):
    zr, zc, c = Fraction(0,1), Fraction(0,1), 0
    cr, cc = Fraction(-5,4)-Fraction(1,10**(2*n)), Fraction(1,10**n)
    zr2, zc2 = zr**2, zc**2
    while zr2 + zc2 <= 4:
        zr, zc = zr2 - zc2 + cr, 2*zr*zc + cc
        zr2, zc2 = zr**2, zc**2
        c += 1
    return c # Chai Wah Wu, Mar 03 2018