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%I #17 Nov 22 2017 02:36:31
%S 1,1,0,1,1,0,1,1,0,0,1,1,2,0,0,1,1,2,1,0,0,1,1,2,5,0,0,0,1,1,2,5,6,0,
%T 0,0,1,1,2,5,14,6,0,0,0,1,1,2,5,14,26,4,0,0,0,1,1,2,5,14,42,44,1,0,0,
%U 0,1,1,2,5,14,42,100,69,0,0,0,0
%N Array read by antidiagonals: a(n,k)= coefficient of z^n of p_k(z), where p_k+1(z)=(p_k(z))^2+z, p_1(z)=z.
%C a(n,k+1)=a(n,k), n<=k; a(n,n)=A000108. Note that the set {z: limit(p_k(z),k->infinity) not=infinity} of complex numbers defines the Mandelbrot set.
%H R. P. Munafo, <a href="http://www.mrob.com/pub/muency.html">Mu-Ency - The Encyclopedia of the Mandelbrot Set</a>
%H Eric Weisstein's World of Mathematics, <a href="http://mathworld.wolfram.com/MandelbrotSet.html">Mandelbrot Set</a>
%H R. Munafo, <a href="http://www.mrob.com/pub/math/seq-a052154.html">Coefficients of Lemniscates for Mandelbrot Set</a>
%F a(n, k+1)=sum(a(i, k)*a(n-i, k), i=1..n-1) for n=2..2^k, a(1, k)=1, a(n, k+1)=0 for n>2^k.
%e p_1(z)=z: coefficient = 1 = a(1,1); p_2(z)=z^2+z: coefficients = 1, 1 = a(1,2), a(2,2); p_3(z)=(z^2+z)^2+z=z+z^2+2z^3+z^4: coefficients = 1,1,2,1 = (1,3), a(2,3), a(3,3), a(4,3); ...
%e Triangle starts:
%e 1,
%e 1, 0,
%e 1, 1, 0,
%e 1, 1, 0, 0,
%e 1, 1, 2, 0, 0,
%e 1, 1, 2, 1, 0, 0,
%e 1, 1, 2, 5, 0, 0, 0,
%e 1, 1, 2, 5, 6, 0, 0, 0,
%e 1, 1, 2, 5, 14, 6, 0, 0, 0,
%e 1, 1, 2, 5, 14, 26, 4, 0, 0, 0,
%e 1, 1, 2, 5, 14, 42, 44, 1, 0, 0, 0,
%e 1, 1, 2, 5, 14, 42, 100, 69, 0, 0, 0, 0,
%e ...
%t p[1, z_] := z; p[k_, z_] := p[k, z] = p[k-1, z]^2 + z; a[n_, k_] := Coefficient[p[k, z], z, n]; Flatten[ Table[a[n-k, k], {n, 1, 13}, {k, n-1, 1, -1}]] (* _Jean-François Alcover_, Jun 13 2012 *)
%Y Cf. A000108.
%Y Cf. A137560, which gives the same array read by rows. [From _Robert Munafo_, Dec 12 2009]
%K nice,nonn,tabl,easy
%O 1,13
%A Klaus Strassburger (strass(AT)ddfi.uni-duesseldorf.de), Jan 24 2000
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