Re: Find (cyclic) Sequence

• To: mathgroup at smc.vnet.net
• Subject: [mg109375] Re: Find (cyclic) Sequence
• From: Albert Retey <awnl at gmx-topmail.de>
• Date: Sat, 24 Apr 2010 04:00:56 -0400 (EDT)
• References: <hqou0k\$9fp\$1@smc.vnet.net> <hqp99h\$grb\$1@smc.vnet.net> <hqrjar\$m0u\$1@smc.vnet.net>

```Hi,

>
> Ok, but
>
> FindSequenceFunction[{0, 3, 2, 1, 0, 3, 2, 1}]
>
> really gives:
>
> Mod[1 + 3 #1 + 2 #1^2 + 2 #1^3, 4] &
>
> because of 1-based indexing. The alternative is to specify
>
> FindSequenceFunction[Table[{k, Mod[4 - k, 4]}, {k, 0, 7}]]
>
> resulting in
>
> Mod[#1 (13 + 2 #1^2), 4] &
>
> Now, how is this reduced to the cleaner 4-k?

Honestly, I have no idea, and I have heard about the function only from
your first post. I think it is obvious that FindSequenceFunction can
only find one of infinitely many solutions to the problem. Of course it
will not always find exactly the same solution that you have in mind.

Unfortunatly, Simplify and FullSimplify don't help with proving that the
two are the same:

FullSimplify[
FindSequenceFunction[{0, 3, 2, 1, 0, 3, 2, 1}, x] - Mod[4 - x, 4]]

I really don't know what you try to achieve, but I think that
FindSequenceFunction is probably just not the right tool. It is like
Simplify or FindInstance, they will both also find _a_ solution but are
no good to find a exactly one or one form of solution.

hth,

albert

```

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