[seqfan] Wajnberg variations (100 terms and percentages)
Eric Angelini
Eric.Angelini at kntv.be
Thu Oct 2 12:16:47 CEST 2014
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.
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