# [seqfan] Re: Wajnberg variations (100 terms and percentages)

Eric Angelini Eric.Angelini at kntv.be
Thu Oct 2 18:11:33 CEST 2014

```Hello Allan,
my idea (badly exposed, I'm afraid) is that
we must have 100 elements in the set -- and
every single element "speaks" and "says":
« There are fourteen '14's in this set »
or:
« There is one '1' in this set »

-----Message d'origine-----
De : SeqFan [mailto:seqfan-bounces at list.seqfan.eu] De la part de Allan Wechsler
Envoyé : jeudi 2 octobre 2014 17:27
À : Sequence Fanatics Discussion list
Objet : [seqfan] Re: Wajnberg variations (100 terms and percentages)

I think I am not really understanding the rules.

Let us abbreviate a sequence of n n's, then m m's, and so on, as [n,m...]; for example, "2,2,3,3,3" would be written "[2,3]".

Then Eric's S% is [1,2,3,4,6,7,8,9,10,11,12,13,14]; T% = [1,4,5,6,7,8,9,10,11,12,13,14]; U% = [2,3,5,6,7,8,9,10,11,12,13,14]. Then he says that these are the only such 100-sequences. What is wrong with, say, [1,99]? Or [4,5,6,12,30,43]? What additional constraint is there besides that it must be a partition of 100 into blocks of distinct sizes?

On Thu, Oct 2, 2014 at 6:16 AM, Eric Angelini <Eric.Angelini at kntv.be> wrote:

> Hello SeqFans,
> here are 3 seq I've discussed with my friend Alexandre W.
> S%, T% and U% are 3 finite seq of 100 terms each.
> They self-describe their content in a "percentage way":
>
> S% =
> 1,2,2,3,3,3,4,4,4,4,6,6,6,6,6,6,7,7,7,7,7,7,7,8,8,8,8,8,8,8,8,9,9,9,9,9,9,9,9,9,10,10,10,10,10,10,10,10,10,10,11,11,11,11,11,11,11,11,11,11,11,12,12,12,12,12,12,12,12,12,12,12,12,13,13,13,13,13,13,13,13,13,13,13,13,13,14,14,14,14,14,14,14,14,14,14,14,14,14,14.
>
> T% =
>
> 1,4,4,4,4,5,5,5,5,5,6,6,6,6,6,6,7,7,7,7,7,7,7,8,8,8,8,8,8,8,8,9,9,9,9,9,9,9,9,9,10,10,10,10,10,10,10,10,10,10,11,11,11,11,11,11,11,11,11,11,11,12,12,12,12,12,12,12,12,12,12,12,12,13,13,13,13,13,13,13,13,13,13,13,13,13,14,14,14,14,14,14,14,14,14,14,14,14,14,14.
>
> U% =
>
> 2,2,3,3,3,5,5,5,5,5,6,6,6,6,6,6,7,7,7,7,7,7,7,8,8,8,8,8,8,8,8,9,9,9,9,9,9,9,9,9,10,10,10,10,10,10,10,10,10,10,11,11,11,11,11,11,11,11,11,11,11,12,12,12,12,12,12,12,12,12,12,12,12,13,13,13,13,13,13,13,13,13,13,13,13,13,14,14,14,14,14,14,14,14,14,14,14,14,14,14.
>
> Explanation (for U%, immediately above):
>
> Among the 100 terms of U%, only two are '2' --> there is 2% of '2'
> Among the 100 terms of U%, only three are '3' --> there is 3% of '3'
> Among the 100 terms of U%, only five are '5' --> there is 5% of '5'
> Among the 100 terms of U%, only six are '6' --> there is 6% of '6'
> ...
> Among the 100 terms of U%, only fourteen are '14' --> there is 14% of '14'
> END.
>
> Those are the only such 100-term possible seq -- and S% is the
> lexically first (then T%, then U%).
> A 1000-term such seq is easy to build. And a 10,000 one also, etc.
> But is it possible to build an infinite seq like that? Well, there is
> at least this one : H% = 100,100,100,100,100,100,100,100,...
> ;-D
> Best,
> É.
>
>
> _______________________________________________
>
> Seqfan Mailing list - http://list.seqfan.eu/
>

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