A078039 revisited

[Dion Gijswijt]

If b(n),c(n),d(n),e(n) are the numbers of zeros, ones, twos and threes
after n iterations, then:

[b(n),c(n),d(n),e(n)]=[1,0,0,0]A^n, there the matrix A equals:

A=[[1,1,1,0],[1,1,1,1],[1,1,0,0],[1,1,1,0]].

(the rows of A correspond to the replacements of 0,..,3 respectively.)

Of course a(n)=b(n)+c(n)+d(n)+e(n)=
[1,0,0,0]A^n [1,1,1,1]^T.

So the GF for a(n) equals:
 sum_n=0^infty x^n [1,0,0,0]A^n [1,1,1,1]^T =
 [1,0,0,0] sum_n=0^infty (xA)^n [1,1,1,1]^T =
 [1,0,0,0] 1/(1-xA) [1,1,1,1]^T=
 (1+x)/(-1+2x+3x^2+x^3).

Hope this is of use.

Dion Gijswijt.

> the repeated substitution
> 0  ->  {0,1,2}
> 1  ->  {0,1,2,3}
> 2  ->  {0,1}
> 3  ->  {0,1,2}
> starting on list {0}
> and flattening at each step
> { {sequence1}, {sequence2},..}
> to {sequence1,sequence2,..}
> generates a list after n steps
> with length =a(n)=A078039
>
> with GF=       (1 - x)/(1 + x - 2*x^2 + x^3)
> or, unsigned : (1 + x)/(1 - x - 2*x^2 - x^3)
>
> Who sees the link between the substitutions
> and the GF?
>
>
> Wouter Meeussen
>
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>
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