[seqfan] Re: Season's present to OEIS

[Charles Greathouse]

Interesting -- I'd certainly like to see the file!

A023110 is just the solutions to ten (well, actually just five have
solutions) quadratic Diophantine equations -- it shouldn't be hard to
list their members in several explicit families.  I wonder if that's
related or unrelated to what you show.

Charles Greathouse
Analyst/Programmer
Case Western Reserve University

On Fri, Dec 23, 2011 at 5:55 PM, Richard Guy <rkg at cpsc.ucalgary.ca> wrote:
> 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.
>
>
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