[seqfan] Re: Earn $500 for the OEIS

[Sascha Kurz]

Am So, 12.01.2014, 01:10, schrieb William Keith:
> I have sent him a proof but I haven't heard back from him.  Perhaps it
> got dropped into his spam folder?

>From his website:
Note: all E-mail to me must contain the word MathIsFun  (no spaces!, i.e.
: "Math Is Fun" is no good),
or else I (might) never get it.

> 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.

Maybe I did not get your explanation right, but I obtain a small problem
here for d=4 (using your method or at thr very least my interpretation of
it).

But I still think, that should be able to prove the conjecture in a
similar way. I have implemented a small C++ programm, which first swaps
every pairs of rows so that the diagonal entries equal 1. Then I
recursively eleminate the -1's left from the diagonal by adding the
suitable row above. If this procedure finishes, a right upper triangular
matrix is obtained. Checking that all diagonal entries still equal 1
settles the conjecture for a specific d.

Computationally checking 1<=d<=5000 took less than 15 minutes. For larger
d things naturally slow down. (Currently my computer is at d=7000). So it
seems to be more money for the OEIS.

If this problem is still open tomorrow, I might try to look for a general
scheme of the elemination steps.

Best,
Sascha Kurz