Sequence conjectures
Gerald McGarvey
Gerald.McGarvey at comcast.net
Sat May 20 23:27:27 CEST 2006
A060867, which is 1,9,49,225,961,... with formula a(n) = (2^n - 1)^2
starting at n=1 is in the OEIS,
- Gerry
At 12:18 AM 5/20/2006, Gene Smith wrote:
>Here's a sequence definition and a conjecture about it, which makes me
>ask, what do you do with such a thing (beyond proving the conjecture, of
>course, and even then there's a problem.)
>
>Let f_1 = z, and f_n = f_(n-1)^2 + z. Now define the nth Mandelbrot curve,
>M_n, as the curve in the real plane defined by the complex equation
>|f_n(z)| = 1. That is, substututing both x+iy for z, and x-iy for z, and
>multiplying, obtain a polynomial in x and y G_n of degree 2^{n+1), and
>then the Mandelbrot curve is G_n = 1. The reason for bringing it up is
>that it converges to the boundry of the Mandelbrot set.
>
>The genus of a nonsingular plane curve of degree 2^(n+1) is
>(2^n-1)(2^(n+1)-1). I'm conjecuring, based on the first five values,that
>the correct genus for the Mandelbrot curve is (2^n-1)^2, for a sequence
>0,1,9,49,225, ... which isn't in the sequence directory, but could
>be--except that its pretty short!
>
>However, I have a fix in mind for it being short.I think I can prove this
>(the curve seems to have an ordinary
>2^n-fold point, and no other singularities.) In that case the sequence
>could go into the sequence database, but now--do you rely on my proof?
>What am I supposed to do here? This *does* strike me as an interesting
>sequence!
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