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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.
3
1, 18, 159, 1586, 15731, 157085, 1570800, 15707976
OFFSET
0,2
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.)
Aaron Klebanoff, Pi in the Mandelbrot Set. In: Fractals 9 (2001), nr. 4, p. 393-402.
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);
PROG
(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
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Martin Renner, Feb 24 2018
STATUS
approved