Observation on A000058

[David Wilson]

Believe it or not, I figured this out on the way to work.  Shows you how
exciting my drive is...

It appears that if a = A000058, then a(k)^2 + 1 divides a(k+1)^2 + 1.

I arrived at this observation by trying to calculate an increasing sequence
of numbers of the form k^2+1 such that each element divides the next.
After observing that the initial k were 0,1,2,3,7,43,... I was led to the
observation.  If we continue choosing the k greedily, however, we obtain
the sequence 0,1,2,3,7,43,857,95343,... which diverges from Sylvester's
sequence.