[seqfan] Re: Connection between Galton-Watson branching process and Julia sets?

israel at math.ubc.ca israel at math.ubc.ca
Thu Jun 2 08:58:13 CEST 2016


If u(n) = z(n) - 1/2, this recurrence becomes u(n) = u(n-1)^2 - u(n-1). 
With u(0) = t, u(n) is a polynomial in t of degree 2^n with integer 
coefficients: u(1) = t^2-t u(2) = t^4-2*t^3+t u(3) = 
t^8-4*t^7+4*t^6+2*t^5-5*t^4+2*t^3+t^2-t u(4) = 
t^16-8*t^15+24*t^14-28*t^13-10*t^12+60*t^11-50*t^10-14*t^9+48*t^8-20*t^7-14*t^6+12*t^5+2*t^4-4*t^3+t 
u(5) = 
t^32-16*t^31+112*t^30-440*t^29+1004*t^28-1064*t^27-756*t^26+4212*t^25-5340*t^24+120*t^23+7980*t^22-9120*t^21+120*t^20+8880*t^19-7236*t^18-1566*t^17+5871*t^16-2352*t^15-2040*t^14+2064*t^13+150*t^12-960*t^11+270*t^10+270*t^9-180*t^8-24*t^7+54*t^6-8*t^5-10*t^4+4*t^3+t^2-t 
The table of coefficients of this doesn't seem to be in the OEIS. Maybe it 
should be.

Cheers,
Robert


On Jun 1 2016, Alonso Del Arte wrote:

>Given z_0 = i = sqrt(-1), z_n = z_(n - 1)^2 + 1/4, we get a sequence of all
>real numbers (aside from z_0) and all positive numbers (aside from z_0 and
>z_1): i, -3/4, 13/16, 233/256, 70673/65536, etc. Obviously the denominators
>are powers of 4. My first thought for the numerators was Fibonacci numbers,
>but the coincidence is broken by 70673. Just the first four real numerators
>are enough to bring up A015701, which involves iteration in the
>Galton-Watson branching process.
>
>The mention of iteration suggests I got the right match, but I had never
>heard of Galton-Watson before. From what I've read in the second Google
>result http://galton.uchicago.edu/~lalley/Courses/312/Branching.pdf , it
>makes sense that the denominators would be powers of 4 as well. But the
>OEIS entry says nothing about fractals or at least fractions. I've been
>scrutinizing the Julia sets for quadratic functions with c purely real.
>
>Any thoughts?
>
>Al
>
>


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