2-dimensional version of Fibonacci numbers

[Marc LeBrun]

>=Yuval Dekel <dekelyuval at hotmail.com>
> Let a(m,n) = number of mXn (0,1) matrices that have no consecutive
> horizontal or vertical 0's.
> So a(1,n)=a(n,1) are the Fibonacci numbers - A000045.
> Is there a formula for a(n,n)?
> I guess that for fixed m, a(m,n) has a linear recurrence formula
similar
> to A000045.

Interesting idea.  Here's the initial 7x7 corner of this table:

  2   3     5      8      13       21         34  
  3   7    17     41      99      239        577  
  5  17    63    227     827     2999      10897  
  8  41   227   1234    6743    36787     200798  
 13  99   827   6743   55447   454385    3729091  
 21 239  2999  36787  454385  5598861   69050253  
 34 577 10897 200798 3729091 69050253 1280128950

which is as far as I've computed it.  The diagonal values check with
A006506 (thanks for the ref, Ed).

Due to the symmetry I've got enough values to submit a full OEIS entry
(which I'll do soon), since extending the rows is relatively cheap.

Row 2 is A001333 and Rows 3 & 4 are A051736 and A015737.  These have an
interesting looking link to Calkin & Wilf, "The number of independent
sets in a grid" but unfortunately I wasn't able to download the reprint.

Should I bother adding Rows 5, 6 and 7?

The orders of the first seven row recurrences are 2 2 4 5 9 11 21.  From
this I don't expect a very easy formula for the diagonal values, but
perhaps the linked paper contains something magic...

PS: Row 2 coincidentally happens to be one of those continued fraction
sequences I was just recently ranting about!