[seqfan] Re: A130011 and the definition of "slowest increasing".

[Frank Adams-Watters]

Note, by the way, that neither of these always exists. One possible problem is that the greedy algorithm takes you to a dead end, while an infinite extenuation is possible.

Flor slowest increasing, there may be two possible sequences that keep switching off which is the smallest at a given index.

Franklin T. Adams-Watters

-----Original Message-----
From: Lars Blomberg <lars.blomberg at visit.se>
To: 'Sequence Fanatics Discussion list' <seqfan at list.seqfan.eu>
Sent: Mon, Jul 13, 2015 4:28 pm
Subject: [seqfan] Re: A130011 and the definition of "slowest increasing".


Alois, 

Thank you for the clarification.

/Lars

-----Ursprungligt
meddelande-----
Från: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] För Heinz,
Alois
Skickat: den 13 juli 2015 15:23
Till: Sequence Fanatics Discussion
list
Ämne: [seqfan] Re: A130011 and the definition of "slowest increasing".

Am
13.07.2015 um 08:19 schrieb Lars Blomberg:

>
> Could someone please define what
"slowest increasing" means?
>
> And what is the difference between "slowest
increasing" and 
> "lexicographically first"?
>

"lexicographically first" ist
the greedy approach.  Use the smallest a(n) that satisfies the condition given
a(1), ..., a(n-1).
And do not change it later.  This is easy
algorithmically.

"slowest increasing" here means that you accept a larger than
greedy
a(n) if it is possible to get a smaller a(m) for a larger m>n.
This is
more complicated algorithmically.

Please do not change the definition of
A130011.

If you want to have a new sequence with the greedy approach, please
use a new A-number.

Best,
Alois






















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