cp's OEIS Frontend

A331814 Fourier coefficients of the boundary of the Mandelbrot set (normalized by a power of two).

%I A331814 #19 Apr 07 2020 02:08:13
%S A331814 -1,1,-8,15,0,-94,-512,987,0,-7346,65536,-122058,0,-2757820,-22020096,
%T A331814 59250963,0,-329425898,2617245696,-4805611678,-34359738368,
%U A331814 -19403249316,498216206336,-36302282082,0,14136557849100,-71399536328704,-88183884706356
%N A331814 Fourier coefficients of the boundary of the Mandelbrot set (normalized by a power of two).
%C A331814 a(n) = 2^(2*n+1)*b(n) where b(n) is the unique sequence of complex numbers such that f(z) := z + Sum_{n>=0} (b(n)*z^-n) defines an analytic homeomorphism (biholomorphic bijection) between the complement of the unit disk and the complement of the Mandelbrot set, sometimes known as the "Jungreis function".  (The b(n) are rationals, so we multiply them by the appropriate power of two to make them integers; this is equivalent to a simple rescaling of the complex plane.)  It is conjectured that |b(n)| <= 1/n, so |a(n)| <= 2^(2*n+1)/n.
%C A331814 Note that the table given in Ewing and Schober (1992) gives the coefficients of the inverse series (contrary to what the text itself says): it's not wrong, it's just mislabeled.
%H A331814 John H. Ewing and Glenn Schober, <a href="https://doi.org/10.1307/mmj/1029004138">On the Coefficients of the Mapping to the Exterior of the Mandelbrot set</a>, Michigan Math. J. 37 (1990), 315-320.
%H A331814 John H. Ewing & Glenn Schober, <a href="https://doi.org/10.1007/BF01385497">The area of the Mandelbrot set</a>, Numer. Math. 61 (1992) 59-72 (note that table 1 gives the coefficients of the INVERSE series).
%H A331814 Irwin Jungreis, <a href="https://projecteuclid.org/euclid.dmj/1077304731">The uniformization of the complement of the Mandelbrot set</a>, Duke Math. J. 4 (1985), 935-938.
%H A331814 David A. Madore, <a href="https://gist.github.com/Gro-Tsen/4c5434b23c5a7e95e9c1cab345cdbf5f">Sage code</a>
%F A331814 a(m)=B(0,m+1) where B(0,0)=1/2 and by downwards induction on k we have B(k-1,m) = 2^(2^(k+1)-1)*B(k,m) - 2^(2^(k+1)-4)*Sum_{j=2^k-1..m-2^k+1} (B(k-1,j)*B(k-1,m-j) - 2*B(0,m-2^k+1)) if m >= 2^k-1, 0 otherwise.
%e A331814 a(0)=-1 because B(1,1)=0 and B(0,1) = 8*B(1,1) - 2*B(0,0) = -1; then a(1)=1 because B(1,2)=0 and B(0,2) = 8*B(1,2) - B(0,1)^2 - 2*B(0,1) = 1.
%K A331814 sign
%O A331814 0,3
%A A331814 _David A. Madore_, Jan 27 2020