[seqfan] Re: Earn $500 for the OEIS

[William Keith]

I have sent him a proof but I haven't heard back from him.  Perhaps it got
dropped into his spam folder?

Anyway, if you look at it, the matrix consists of a series of 2x2 blocks of
{{0,1,{1,0}} down the main diagonal, a "pseudo-diagonal" of -1s at slope
-1/2 above the diagonal, and a shallower "pseudo-diagonal" of -1s below the
blocks.  You can use row addition operations to use the 1s in the blocks to
clear the -1s below the diagonal by working from the bottom up, and since
the pseudo-diagonal of -1s above the main diagonal is steeper, this will
never produce additional entries below the main diagonal.  Then you use row
additions upward (again, bottom up) to clear above the main diagonal.

Interchange the two rows of each block, and you get the identity matrix.
There are d blocks and each row interchange gives a -1 to the determinant
while the row operations didn't change it, so the determinant of the
original matrix is (-1)^d.  QED.

It seems straightforward to me.  Did I miss something?

I hope he got it, anyway.  Maybe he monitors this list?

William Keith