# More on Pell numbers "term doubling"

Creighton Dement crowdog at crowdog.de
Thu Nov 24 22:26:01 CET 2005

```Hello,

In reference to the comment for a = A000129:

Define c(2n) = -A001108(n), c(2n+1) = -A001108(n+1) and d(2n) =
d(2n+1) = A001652(n), then ((-1)^n)*(c(n) + d(n)) = a(n).
- Proof given
by Max Alekseyev (maxal(AT)cs.ucsd.edu) - Creighton Dement
(crowdog(AT)crowdog.de), Jul 21 2005

Here's another which is similar in several respects:

Define c(2n) = -2*A001109(n), c(2n+1) = -2*A001109(n+1) and d(2n) =
d(2n+1) = A000129(n), then ((-1)^n)*(c(n) + d(n)) =
A001333(n)

In other words, c = 2*(0, -1, -1, -6, -6, -35, -35, -204)
d = (1, 1, 5, 5, 29, 29, 169, 169, 985, 985)

This time, I don't believe a "standard proof" using generating functions
is necessary. Exactly as was used before in the first case, the identity
jes(pos) + jes(neg) = jes holds. But this is nothing other than the
equation (c(n) + d(n)) = ((-1)^n)*A001333(n) provided one assumes that
the sequence given by FAMP as 1, -1, 3, -7, 17, -41, 99, -239, 577
really is ((-1)^n)*A001333(n)), etc. More specifically, one must assume
that any sequence generated by the floretion E*x  (where E =  (1/4)(-'i
- i' - 'ii' + 'jj' + 'kk' + 'jk' + 'kj' - e) and x is any floretion)
satisfies a 2nd order linear recurrence relation (a proof of a similar
case has already been found).

FAMP's output is given at:
http://www.crowdog.de/Pell/pellbeforesplit.html
http://www.crowdog.de/Pell/pellaftersplit.html

Side note: I've added several new ray-traced pictures of
floretion-generated integer sequences to my homepage at
http://www.crowdog.de  If you have time, you might enjoy having a look.

Sincerely,
Creighton

-It's a shame when the girl of your dreams would still rather be with
someone else when you're actually in a dream.

```

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