number of 1's in "Feigenbaum" matrix

[benoit]

Bonjour,

Vaguely inspired by the construction of Hadamard matrix, I was looking 
to the (2^n)x(2^n) (0,1)-matrix built using a rule defined in first 
Feigenbaum symbolic sequence A035263. Let M(0)=1 and M(n) be the (2^n) 
x( 2^n) symetric matrix  defined as follows :

M(0)=1 and if M(n) is known, to get M(n+1) ( r(i,n) = i-th row of M(n) 
c(i,n)= i-th column ) :

first step :

for i=1 up to 2^n :

r(i,n+1)=r(i,n)r'(i,n) where r'(i,n) is obtained from r(i,n) by 
changing last term by 0 if 1, by 1 if 0.
c(i,n+1)=c(i,n)c'(i,n) where c'(i,n) is obtained from c(i,n) by 
changing last term by 0 if 1, by 1 if 0.

secund step :

The rest of the matrix is filled with same rule, using new created 
partial rows or columns .

So, M(0), M(1), M(2) are :

1

1,0
0,1

1,0,1,1
0,1,0,0
1,0,1,1
1,0,1,1

To get M(3) :

First step : applying the rule on M(2) :

1,0,1,1,1,0,1,0
0,1,0,0,0,1,0,1
1,0,1,1,1,0,1,0,
1,0,1,1,1,0,1,0,
1,0,1,1,
0,1,0,0,
1,0,1,1,
0,1,0,0,

Second step : to fill the matrix, applying the "Feigenbaum" rule to new 
created rows or columns of length 4 gives M(3) :

1,0,1,1,1,0,1,0
0,1,0,0,0,1,0,1
1,0,1,1,1,0,1,0,
1,0,1,1,1,0,1,0,
1,0,1,1,1,0,1,0
0,1,0,0,0,1,0,1
1,0,1,1,1,0,1,0
0,1,0,0,0,1,0,1

In 1-D,  let b(n) be the number of 1's from A035263(1) up to 
A035263(2^n). Then b(n) = Jacobsthal sequence (A001045)  and  
b(n)=b(n-1)+2*b(n-2)

In 2-D, let a(n) be the number of 1's in M(n) : 1,2,10,34,134,... seems 
to me  : a(0)=1,a(1)=2,a(2)=10,  a(n)=3*a(n-1)+4*a(n-2)-4*a(n-3)  which 
is  A052965.

Is it possible to extend the process to 3-D starting with 1 : if 
working, what could be the number of 1's in the successive 
(2^n)x(2^n)x(2^n) cubic boxes?

I'm also curious ot see what is the sequence of determinants of M(n).

Thanks
BC









 


 

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