[seqfan] Re: Continued Fractions

[Neil Sloane]

Melvin, Please add a comment to A058182, and also submit the denominator
sequence.  Nice question!

I added a ref to the "Proofs" book, but maybe you could fill in the page
number.

Best regards
Neil

Neil J. A. Sloane, President, OEIS Foundation.
11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
Phone: 732 828 6098; home page: http://NeilSloane.com
Email: njasloane at gmail.com


On Sun, Feb 21, 2016 at 2:04 AM, Melvin M Peralta <melvinmperalta at gmail.com>
wrote:

> Hello SeqFans,
>
> a(0) = 1
> a(n+1) = numerator of the simplified continued fraction resulting from
> [a(0), a(1), ..., a(n)]
>
> 1, 1, 2, 5, 27, 734, 538783, ...
>
> a(n) is also the number of ways to tile an n-board with dominos and
> stackable squares, where nothing can be stacked on a domino but
> otherwise the i-th cell may be stacked by as many as a_i squares. (a nice
> proof of this is given in "Proofs That Really Count").
>
> *My question*: From n>=2, the sequence appears identical to
> https://oeis.org/A058182. Does it always coincide? If so is the connection
> already implicit in the entry? Or should a comment be made?
>
> Note a similar sequence does not appear in OEIS:
> a(0) = 1
> a(n+1) = denominator of the simplified continued fraction resulting from
> [a(0), a(1), ..., a(n)]
>
> 1, 1, 2, 3, 10, 103, 10619, ...
>
> This is also nice because a(n+1) is the number of ways to tile an n-board
> in the same way described, except the first cell should be ignored
> entirely.
>
> Best Regards,
> Melvin
>
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