2-dimensional version of Fibonacci numbers

[benoit]

>> It seems there is another possible constant for this sequence a(n) :
>>
>> letting b(n)=a(n+2)*a(n)/a(n+1)^2
>>
>>  limit n-->infty b(n)=c2= 2.2591...
>>
>> with  b(2n)<c2<b(2n-1)
>>
>> Any more accurate value for this limit c2?
>
> It is almost certainly the square of c1.  Perhaps this can be proved
> from the known asymptotic behaviour of a(n), but I'm not sure.
>
> My experimental observation is that a(n) appears to behave like
>   A * c3^n * c1^(n^2)
> where c1 is as above,
>   c3 = 1.143519129587 approx
>    A = 1.0660826 approx
>
> I base this conjecture on numerical analysis of a(n) for n up to 19.
>
> Brendan.
>




This seems very natural since a similar asymptotic should hold for :

A067962(n) = F(n+2)*prod(i=1,n+1,F(i))^2 where F(i)=A000045(i) is the 
i-th Fibonacci number

S. Kitaev and T. Mansour, Problem of the pawns, 
http://arxiv.org/abs/math.CO/0305253

They give the general formula for the number of binary mxn matrix 
avoiding some patterns (attacking pawns) as U(m,n) the number of binary 
matrix avoiding adjacent 1's nw-se.

Benoit



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