[seqfan] Season's present to OEIS

[Richard Guy]

Dear all,
          I wanted to send a suitable season's present, but it's
in a fairly large and incoherent file.  For those who are interested
I can send the file (when I'm back in my office on Tuesday).  I'll send
the file to Neil in any case, and he can touch the D key or pass it on
as he deems fit.  For other, to whet your appetite, here are a few snippets:

Sequence A023110 in OEIS begins

1, 4, 9, 16, 49, 169, 256, 361, 1444, 3249, 18496, 64009,

[snip]

The sequence of remaining squares,

0, 0, 0, 1, 4, 16, 25, 36, 144, 324, 1849,

does not, at the time of writing, appear to be in OEIS,
but their roots,

0, 0, 0, 1, 2, 4, 5, 6, 12, 18, 43, 80, 154, 191,

appear as A031150.  The roots of the original sequence ...
are not in OEIS, and show some remarkable connexions ...
which will be explained below.

[snip]


Neg to the rescue!
$$\sqrt{10} = [\bar4; \overbrace{\bar2, \bar2, \bar2, \bar2, \bar2, 
\bar8}],$$
where the overbrace indicates the periodic part, has convergents
$$\left(\frac{0}{1}\right), \frac{1}{0}, \frac{4}{1}, \frac{7}{2}, ...

[snip]

We need to make two adjustments and then we have all the solutions.

[snip]

We are now in a position to confirm the connexions noticed ...

In fact,
\begin{eqnarray*}
N_{7n+3}^2 & = & \left\{(4+\sqrt{10})(3+\sqrt{10})^{2n}+
(4-\sqrt{10})(3-\sqrt{10})^{2n}\right\}^2\left/4\right. \\

[snip]

   & = & N_{14n+5} + 3 \\

[snip]

One can pose the same questions in bases other than 10.

[snip]

and the squares of the numerators satisfy the relations
$$2N_{3n}^2=N_{6n}+1, \ N_{3n+1}^2=N_{6n+1}+2, \quad
N_{3n+2}^2=N_{6n+3}+1$$

All the best for 2012.      R.