A puzzling sequence

[Roland Bacher]

Dear sequence fans,

in relation with the Pascal triangle I stumbled upon 3 intriguing 
sequences (denoted alpha(n),beta(n) and tau(n) below). They
are not yet in the database since all I know about them are entirely
empirical observations. I hope that somebody of you will have some
new idea putting part of it on a firmer ground.

The story starts with the symmetric matrix P(n) with entries
P_{i,j}={i+j\choose i}, 0<=i,j<n, the binomial coefficients.

First observation:

  det( (P(n))^3- x Identity(n) ) == (1-x)^n modulo 2  ,

seemingly for all n (it holds for the first 100 or so values).

This (if true) implies the factorisation

  det( P(n) - x Identity(n) ) == 
     ==(1+x)^{alpha(n)} (1+x+x^2)^{beta(n)} modulo 2  

with    alpha(n)+2 beta(n)=n.
  
The first values alpha(0),alpha(1),.. are

0,1,0,3,2,5,0,3,2,5,0,11,6,9,4,7,6,9,4,15,10,21,0,11,6,9,..  .

The sequence alpha(n) is seemingly best computed using an 
auxiliary sequence tau(n) defined recursively by  tau(1)=0 and 

tau(n)= 4 tau(r)+1  if n=2^k+r, 1 <= r < 2^(k-1)  ,

tau(n)= tau(r)      if n=2^k+r, 2^(k-1) <= r <= 2^k .

(examples: tau(2)=tau(2^0+1)=tau(1)              since 1 = 2^0,
           tau(5)=tau(2^2+1)= 4 tau(1) + 1 = 1   since 1 < 2^1 .)

Its first terms tau(1),tau(2),.. are

0,0,0,0,1,0,0,0,1,1,1,0,1,0,0,0,1,1,1,1,5,1,1,0,1,... .

The sequence alpha(n) seems now to be recursively defined by
alpha(0)=0 , alpha(1)=1 and

alpha(2n)   = n-alpha(n)                ,
alpha(2n+1) = n+3-alpha(n)+8 tau(n)

for n>=1.

Up to 10^4, one has 0<=alpha(n)<=n   (which is not obvious to me from 
the recursive definition).

Even more puzzling, the sequence alpha(n) satisfies also the identity

alpha(2^k+r)=1+alpha(2^k+r-1)+2 alpha(2^k-r)-2 alpha(2^k+1-r),  1<=r<=2^k

(which, together with some initial values yields a different
recursive definition of the sequence alpha(n)).

Does anyone have an idea or proof of some of these facts? 

Thanks   Roland