Writing a program to illustrate a fractal
On Sunday, August 26, 2018 at 2:35:13 PM UTC-5, Richard Damon wrote:
> On 8/26/18 1:58 PM, Musatov wrote:
> > On Sunday, August 26, 2018 at 12:49:16 PM UTC-5, Richard Damon wrote:
> >> On 8/26/18 12:48 PM, Dennis Lee Bieber wrote:
> >>>> The sequence is defined by:
> >>>> For 1 <= n <= 3, a(n) = n; thereafter, a(2n) = a(n) + a(n+1), a(2n-1) = a(n) + a(n-2).
> >>> I am not sure what 'fractal' property this sequence has that he
> >>> wants to
> >> display.
> > I'm sorry, let me try to explain:
> > Here is my output:
> > 1, 2, 3, 5, 4, 8, 7, 9, 7, 12, 13, 15, 11, 16, 17, 16, 14, 19, 21, 25, 20, 28, 27, 26, 24, 27, 31, 33, 28, 33, 32, 30, 31, 33, 35, 40, 35, 46, 44, 45, 41, 48, 53, 55, 47, 53, 54, 50, 51, 51, 53,
> > It is an OEIS sequence.
> > I was told this image of the scatterplot emphasizes the 'fractal nature' of my sequence:
> > https://oeis.org/A292575/a292575.png
> Something is wrong with that image compared to the sequence, as the
> sequence is always positive, and in fact the lowest the sequence can get
> to is always increasing (as it starts always positive, and each term is
> the sum of two previous terms),while the graph is going negative.
> (actually going to the definition of the sequence, the plot isn't of
> a(n) but a(n)-n, which can go negative)
> I normally think for fractals as a sequence of patterns of increasing
> complexity, or a pattern looked at with increasing resolution revealing
> the growth pattern. This sequence isn't quite like that, but I suppose
> if you think of the sequence a(n) in the interval m <= n <= 2*m, and
> then the interval 2*m <= n <= 4*m, that second interval is somewhat like
> the first with some recursively added pattern (especially if you include
> the -n in the sequence).
> That graph is probably the best way to show that pattern.
> One thing that might help, is to clean up the definition of a(n) to be
> more directly computable, and? maybe even include the subtraction of n.
> A rewriting of your rules would be:
> n=1,2,3:??? a(n) = n
> n>3, and even: a(n) = a(n/2) + a(n/2+1)
> n>3 and odd: a(n) = a((n+1)/2) + a(n-3)/2)
> If I have done my math right, this is the same sequence definition, but
> always defining what a(n) is equal to.
> If we want to define the sequence b(n) = a(n) - n, we can transform the
> above by substitution
> n=1,2,3: b(n) = 0
> n>3 and even: b(n) = a(n/2)+a(n/2+1)-n
> ??????? = b(n/2)+b(n/2+1) + n/2 + n/2+1 -n
> ??????? = b(n/2) + b(n/2+1) + 1
> n>3 and odd: b(n) = a((n+1)/2) + a((n-3)/2) - n
> ??????? = b((n+1)/2) + b((n-3)/2) + (n+1)/2 + (n-3)/2 -n
> ??????? = b((n+1)/2) + b((n-3)/2) -1
> Richard Damon
Thank you, Richard. If anyone is interested further, even in writing a Python code to generate the sequence or further preparing of an animation I would be delighted.