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<font face="Courier New">Heinz,<br>
<br>
It isn't clear what you need, I don't underastand the relationship
between EE and E.<br>
<br>
Except for the first 0 in E, it seems that each consecutive number
n is repeated n+1 times up to the number indicated by the first
component of EE. Is this what you are looking for?<br>
<br>
This is a simple (though not very efficient) way of attaining
that:<br>
<br>
m = EE(1);<br>
E = []; //or E = 0 if the first 0 is really needed<br>
for k=0:m<br>
E = [E, k*ones(1, k+1)];<br>
end<br>
<br>
Regards.<br>
<br>
Federico Miyara <br>
</font><br>
<div class="moz-cite-prefix">On 07/02/2022 14:35, Heinz Nabielek
wrote:<br>
</div>
<blockquote type="cite"
cite="mid:33C03EA0-D7FF-470F-83CE-55FA7E7DA733@me.com">
<pre class="moz-quote-pre" wrap="">Sorry, colleagues - please help, I have just a mental blockade.
Given vector EE= [3 5 8]
I want to create a stepwise increasing vector E= [0 0 1 1 2 2 2 3 3 3 3]
And would need a system that works for much larger numbers....
Probably dead easy?
Heinz
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