Easter supplement
Tim S Roberts
their English names, so1,2,3,4,5,6,....->
3,3,5,4,4,3,....
Suppose we do the same with this new sequence, so we now have
5,5,4,4,4,5,....
Suppose we do the same again, so we now have
4,4,4,4,4,4,....
What is the first point in this sequence to have a number other than 4?
Tim
Tim
Martin DeMello
11 (-> 6 -> 3 -> 5)
martin
Tim S Roberts
"Martin DeMello" <martin...@yahoo.com> wrote in message
news:QxX6e.984643$8l.982176@pd7tw1no...
And if we do the sequence just once more, then the answer is.....?
Tim
Tim S Roberts
r.e.s.
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There is no such point.
Consider the mapping n -> L(n), where L(n) is the length of the
'name' of the natural number n. For the usual English names,
the mapping looks like
n name L(n)
- ---- ----
0 zero < 4
1 one < 3
2 two < 3
3 three < 5
4 four = 4
5 five > 4
6 six > 3
7 seven > 5
etc., with
n < 4 ==> n < L(n)
n = 4 ==> n = L(n)
n > 4 ==> n > L(n).
It follows that when started at any natural n, iteration of
the mapping n -> L(n) eventually converges to 4. (Every n
is "attracted" to the orbit 4 -> 4 -> 4 -> ...)
(This particular attractor is a peculiarity of English --
note that if 'five' were spelled 'phaive', say, then there
would be two attracting orbits instead of one.)
It might interesting to find the similar attractors for
languages other than English.
--r.e.s.
Carl G.
"r.e.s." <r...@ZZmindspring.com> wrote in message
news:gKZ6e.5061$An2....@newsread2.news.pas.earthlink.net...
odd ==> odd
even ==> even
longer ==> shorter
shorter ==> longer
;-)
Carl G.
r.e.s.
> "Tim S Roberts" <t.ro...@cqu.edu.au> wrote ...
>> Suppose we take the integers, and replace them with the number of letters
>> Suppose we do the same with this new sequence, so we now have
>> 5,5,4,4,4,5,....
>> Suppose we do the same again, so we now have
>> 4,4,4,4,4,4,....
>> What is the first point in this sequence to have a number other than 4?
> There is no such point.
>
> Consider the mapping n -> L(n), where L(n) is the length of the
> 'name' of the natural number n. For the usual English names,
> the mapping looks like
>
> n name L(n)
> - ---- ----
> 0 zero < 4
> 1 one < 3
> 2 two < 3
> 3 three < 5
> 4 four = 4
> 5 five > 4
> 6 six > 3
> 7 seven > 5
> etc., with
> n < 4 ==> n < L(n)
> n = 4 ==> n = L(n)
> n > 4 ==> n > L(n).
>
> It follows that when started at any natural n, iteration of
> the mapping n -> L(n) eventually converges to 4. (Every n
> is "attracted" to the orbit 4 -> 4 -> 4 -> ...)
>
> (This particular attractor is a peculiarity of English --
> note that if 'five' were spelled 'phaive', say, then there
> would be two attracting orbits instead of one.)
>
> It might [be] interesting to find the similar attractors for
> languages other than English.
For the perversely curious, here's what I find for a few
languages, with the orbits enclosed in parentheses ...
Language #Orbits Orbits
-------- ------- ----------------------------------------
English 1 (4)
German 1 (4)
French 1 (4 -> 6 -> 3 -> 5)
Spanish 2 (4 -> 6),(5)
Zulu* 7 (11),(15),(16),(22),(23),(19 -> 24),(27)
* http://www.sf.airnet.ne.jp/~ts/language/number/zulu.html
It's clear, I think, that if there exists a number k such that
n > k ==> n > L(k), then there will be finitely-many orbits,
and every number will eventually fall into one of those orbits.
(And such a number k will exist for most number-naming systems.
An exception would be unary 'names', where the name of n
consists of n or more marks, so n <= L(n) for every n.)
--r.e.s.
Martin DeMello
>
> "Martin DeMello" <martin...@yahoo.com> wrote in message
> news:QxX6e.984643$8l.982176@pd7tw1no...
> > Tim S Roberts <t.ro...@cqu.edu.au> wrote:
> >> Suppose we take the integers, and replace them with the number of letters
> >> in
> >> their English names, so1,2,3,4,5,6,....->
> >> 3,3,5,4,4,3,....
> >> Suppose we do the same with this new sequence, so we now have
> >> 5,5,4,4,4,5,....
> >> Suppose we do the same again, so we now have
> >> 4,4,4,4,4,4,....
> >> What is the first point in this sequence to have a number other than 4?
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> >
> > 11 (-> 6 -> 3 -> 5)
> >
> > martin
>
> And if we do the sequence just once more, then the answer is.....?
Twenty three, I'd guess. Hm - I wonder if we back-project it a few more
times, if we'll get an IESable sequence.
martin
r.e.s.
> "Tim S Roberts" <t.ro...@cqu.edu.au> wrote ...
>> Suppose we take the integers, and replace them with the number of letters
>> in their English names, so 1,2,3,4,5,6,.... -> 3,3,5,4,4,3,....
>> Suppose we do the same with this new sequence, so we now have
>> 5,5,4,4,4,5,....
>> Suppose we do the same again, so we now have
>> 4,4,4,4,4,4,....
>> What is the first point in this sequence to have a number other than 4?
Please ignore my earlier remark about there being no such point,
since I misread your question -- sorry about the confusion. In
the notation defined below, your sequences are as follows:
n 1,2,3,4,5,6,7,8,9,10,11,12,....
L^1(n) 3,3,5,4,4,3,5,5,4, 3, 6, 6,....
L^2(n) 5,5,4,4,4,5,4,4,4, 5, 3, 3,....
L^3(n) 4,4,4,4,4,4,4,4,4, 4, 5, 4,....
... ^
and you were looking at n=11. What I say below concerns the
*columns* of the above display extended indefinitely -- each
column eventually converges to 4 in the case of English number-
names (but not in the case of various other languages).
r.e.s.
> "r.e.s." <r...@ZZmindspring.com> wrote ...
>> "Tim S Roberts" <t.ro...@cqu.edu.au> wrote ...
>> > Suppose we take the integers, and replace them with the number of
> letters
>> > in their English names, so1,2,3,4,5,6,....->
>> > 3,3,5,4,4,3,....
>> > Suppose we do the same with this new sequence, so we now have
>> > 5,5,4,4,4,5,....
>> > Suppose we do the same again, so we now have
>> > 4,4,4,4,4,4,....
>> > What is the first point in this sequence to have a number other than 4?
>> There is no such point.
Sorry to have misinterpreted Tim's puzzle, which considers only
the sequence L^3(n), n=1,2,3,..., (using notation defined below):
n 1,2,3,4,5,6,7,8,9,10,11,12,....
L^1(n) 3,3,5,4,4,3,5,5,4, 3, 6, 6,....
L^2(n) 5,5,4,4,4,5,4,4,4, 5, 3, 3,....
L^3(n) 4,4,4,4,4,4,4,4,4, 4, 5, 4,....
... ^
so he was of course looking for n=11. Having misread the question,
what I say below concerns the behaviour of the *columns* of the
above display -- each column eventually converges to 4 in the
case of English number-names (but not in the case of various other
languages).
>> Consider the mapping n -> L(n), where L(n) is the length of the
> odd ==> odd
> even ==> even
>
> longer ==> shorter
> shorter ==> longer
>
> ;-)
You've lost me ;o(
Ed Murphy
>> odd ==> odd
>> even ==> even
>>
>> longer ==> shorter
>> shorter ==> longer
>>
>> ;-)
>
> You've lost me ;o(
"odd" has an odd number of letters.
"even" has an even number of letters.
"longer" is a shorter word (than "shorter").
"shorter" is a longer word (than "longer").
r.e.s.
> r.e.s. wrote:
> [Carl G. wrote:]
>>> odd ==> odd
>>> even ==> even
>>>
>>> longer ==> shorter
>>> shorter ==> longer
>>>
>>> ;-)
>>
>> You've lost me ;o(
>
> "odd" has an odd number of letters.
> "even" has an even number of letters.
>
> "longer" is a shorter word (than "shorter").
> "shorter" is a longer word (than "longer").
>
> See also http://mathworld.wolfram.com/GrellingsParadox.html
Hmmm.... OK ... but Grelling's paradox doesn't arise in
this puzzle. (Also, Carl's comment looked like a reply
to my suggestion about languages other than English).
--r.e.s.
Danny Kodicek
So:
L(n) is the length of the English word for n
L'(m,n) is L(L(L(...(n)...))), with m L's
f(m) is the smallest integer n>0 such that L'(m,n)<>4
It looks to me like the sequence f(m) starts (beginning with f(0)):
1,1,11,23,124
That was as high as my computer could manage in normal time - the next
answer would have to have at least 124 letters. Can anyone beat
1043373373373 ("one trillion, forty three billion, three hundred and seventy
three million, three hundred and seventy three thousand, three hundred and
seventy three")?
Danny
Danny Kodicek
"Danny Kodicek" <use...@well-spring.co.uk> wrote in message
news:kJe7e.60$v8...@newsfe5-gui.ntli.net...
> It looks to me like the sequence f(m) starts (beginning with f(0)):
> 1,1,11,23,124
>
> That was as high as my computer could manage in normal time - the next
> answer would have to have at least 124 letters. Can anyone beat
> 1043373373373 ("one trillion, forty three billion, three hundred and
seventy
> three million, three hundred and seventy three thousand, three hundred and
> seventy three")?
1024373373373 is slightly better, of course. By the way, my algorithm is
only counting letters, not commas or spaces, and as you can see uses British
'and's between words (although it uses US-style billions, which are
increasingly standard these days)
Danny
Danny Kodicek
I know, I know, bad form to keep replying to myself, but I thought I'd point
out that by using 'milliard' and British billions, the length of the number
remains the same in this case: "one billion, forty three milliard, three
hundred and seventy three million, three hundred and seventy three thousand,
three hundred and seventy three". But no one uses 'milliard' today.
Danny
Ted Schuerzinger
news:vQ%6e.5218$An2...@newsread2.news.pas.earthlink.net:
> Language #Orbits Orbits
> -------- ------- ----------------------------------------
> English 1 (4)
> German 1 (4)
> French 1 (4 -> 6 -> 3 -> 5)
> Spanish 2 (4 -> 6),(5)
> Zulu* 7 (11),(15),(16),(22),(23),(19 -> 24),(27)
In Russian, there are three orbits:
(3), (4 -> 6 -> 5), and (11)
--
Ted <fedya at bestweb dot net>
TV Announcer: It's 11:00. Do you know where your children are?
Homer: I told you last night, *no*!
<http://www.snpp.com/episodes/4F06.html>
Ted S.
wrote in news:kJe7e.60$v8...@newsfe5-gui.ntli.net:
> That was as high as my computer could manage in normal time - the
> next answer would have to have at least 124 letters. Can anyone beat
> 1043373373373 ("one trillion, forty three billion, three hundred and
> seventy three million, three hundred and seventy three thousand,
> three hundred and seventy three")?
373,373,373,373
Three hundred and seventy three billion,
three hundred and seventy three million,
three hundred and seventy three thousand
three hundred and seventy three.
If I count correctly, it's got 130 letters.
You can get rid of three letters by going to 123,373,373,373.
r.e.s.
> "r.e.s." <r...@ZZmindspring.com> wrote:
>
>> Language #Orbits Orbits
>> -------- ------- ----------------------------------------
>> English 1 (4)
>> German 1 (4)
>> French 1 (4 -> 6 -> 3 -> 5)
>> Spanish 2 (4 -> 6),(5)
>> Zulu* 7 (11),(15),(16),(22),(23),(19 -> 24),(27)
>
> In Russian, there are three orbits:
>
> (3), (4 -> 6 -> 5), and (11)
I'm also curious about Hebrew ... Maybe someone will confirm
or correct my impression (from googling) that the Hebrew
number-names depend on gender, there being one orbit in
each case: (4) if feminine, and (4 -> 5) if masculine. (?)
Which names are used for numbers as abstract objects?
Arabic would be interesting too, but I couldn't find a site
that gives the Arabic number-names (in Arabic, of course).
--r.e.s.
Danny Kodicek
"Ted S." <fe...@bestweb.spam> wrote in message
news:Xns9638E2DB9A9CF8j...@ID-121946.user.dfncis.de...
> Somebody claiming to be "Danny Kodicek" <use...@well-spring.co.uk>
> wrote in news:kJe7e.60$v8...@newsfe5-gui.ntli.net:
>
> > That was as high as my computer could manage in normal time - the
> > next answer would have to have at least 124 letters. Can anyone beat
> > 1043373373373 ("one trillion, forty three billion, three hundred and
> > seventy three million, three hundred and seventy three thousand,
> > three hundred and seventy three")?
>
> 373,373,373,373
>
> Three hundred and seventy three billion,
> three hundred and seventy three million,
> three hundred and seventy three thousand
> three hundred and seventy three.
>
> If I count correctly, it's got 130 letters.
It's longer than mine, but takes fewer steps to reach 4:
L(130)=19 ("one hundred and thirty")
L(19)=8
L(8)=5
L(5)=4
> You can get rid of three letters by going to 123,373,373,373.
Better but still not as good.:
L(127)=24
L(24)=10
L(10)=3
L(3)=5
L(5)=4
But now that I see it's possible to use numbers below the trillions (don't
know how I managed to miss that), I suggest 113,373,373,373, which is back
to 124 letters. I doubt we could go lower than that.
Danny