[math-fun] Is 312 a black hole for N?
Hello Math-fun, a little puzzle slightly adapted from here (in French), thanks to my friend Alexandre Wajnberg: http://dev.ulb.ac.be/urem/IMG/pdf/PRO12102010.pdf For any natural number "n" (base-10-written) we build "next-n" like this: - write the quantity of digits of "n" - concatenate the quantity of even digits in "n" - concatenate to the former concatenation the quantity of odd digits in "n" Example for "n" = 80322057626942 - there are 14 digits in "n": --> 14 - there are 10 even digits in "n": --> 1410 - there are 4 odd digits in "n": --> 14104 If we iterate the construction, we get the sequence: 80322057626942, 14104, 532, 312, 312, 312, ... Another example: 5771, 404, 330, 312, 312, 312, ... Is 312 the end of all such sequences? Best, É.
On Fri, Oct 15, 2010 at 10:06 AM, Eric Angelini <Eric.Angelini@kntv.be>wrote:
Hello Math-fun,
a little puzzle slightly adapted from here (in French),
thanks to my friend Alexandre Wajnberg:
http://dev.ulb.ac.be/urem/IMG/pdf/PRO12102010.pdf
For any natural number "n" (base-10-written) we build
"next-n" like this:
- write the quantity of digits of "n"
- concatenate the quantity of even digits in "n"
- concatenate to the former concatenation the quantity
of odd digits in "n"
Example for "n" = 80322057626942
- there are 14 digits in "n": --> 14
- there are 10 even digits in "n": --> 1410
- there are 4 odd digits in "n": --> 14104
If we iterate the construction, we get the sequence:
80322057626942, 14104, 532, 312, 312, 312, ...
Another example:
5771, 404, 330, 312, 312, 312, ...
Is 312 the end of all such sequences?
Yes, certainly it is. 312 is a fixed point. If the input is 3 digits, then the output is one of 330, 321, 312, 303. So any 3-digit number becomes 312 in <=2 steps. Any input with up to 9 digits must have an output that's 3 digits. So any number up to 9 digits becomes 312 in <=3 steps. Any larger input has an output that's fewer digits than the input, so eventually reaches something 9 digits or less, and so reaches 312. (If you want bounds better than "eventually", though, you can continue the above logic: any input up to 999 digits produces an output with at most 9 digits, so becomes 312 in <=4 steps.) --Michael -- Forewarned is worth an octopus in the bush.
Michael, You beat me too it. However, to be complete you need to handle 1 and 2 digit numbers: 1 digit number --> 110 or 101 2 digit number ---> 202, 211 or 220 and now we're in the 3 digit realm Victor On Fri, Oct 15, 2010 at 10:31 AM, Michael Kleber <michael.kleber@gmail.com>wrote:
On Fri, Oct 15, 2010 at 10:06 AM, Eric Angelini <Eric.Angelini@kntv.be
wrote:
Hello Math-fun,
a little puzzle slightly adapted from here (in French),
thanks to my friend Alexandre Wajnberg:
http://dev.ulb.ac.be/urem/IMG/pdf/PRO12102010.pdf
For any natural number "n" (base-10-written) we build
"next-n" like this:
- write the quantity of digits of "n"
- concatenate the quantity of even digits in "n"
- concatenate to the former concatenation the quantity
of odd digits in "n"
Example for "n" = 80322057626942
- there are 14 digits in "n": --> 14
- there are 10 even digits in "n": --> 1410
- there are 4 odd digits in "n": --> 14104
If we iterate the construction, we get the sequence:
80322057626942, 14104, 532, 312, 312, 312, ...
Another example:
5771, 404, 330, 312, 312, 312, ...
Is 312 the end of all such sequences?
Yes, certainly it is.
312 is a fixed point.
If the input is 3 digits, then the output is one of 330, 321, 312, 303. So any 3-digit number becomes 312 in <=2 steps.
Any input with up to 9 digits must have an output that's 3 digits. So any number up to 9 digits becomes 312 in <=3 steps.
Any larger input has an output that's fewer digits than the input, so eventually reaches something 9 digits or less, and so reaches 312. (If you want bounds better than "eventually", though, you can continue the above logic: any input up to 999 digits produces an output with at most 9 digits, so becomes 312 in <=4 steps.)
--Michael
-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
On Fri, Oct 15, 2010 at 10:36 AM, Victor Miller <victorsmiller@gmail.com>wrote:
Michael, You beat me too it. However, to be complete you need to handle 1 and 2 digit numbers:
I don't think this is a special case. Any number with up to 9 digits produces a 3-digit output (since you're concatenating three single digits) -- this applies to 1 or 2 digit numbers just as well as 4 or 9. --Michael
1 digit number --> 110 or 101 2 digit number ---> 202, 211 or 220
and now we're in the 3 digit realm
Victor
On Fri, Oct 15, 2010 at 10:31 AM, Michael Kleber <michael.kleber@gmail.com>wrote:
On Fri, Oct 15, 2010 at 10:06 AM, Eric Angelini <Eric.Angelini@kntv.be
wrote:
Hello Math-fun,
a little puzzle slightly adapted from here (in French),
thanks to my friend Alexandre Wajnberg:
http://dev.ulb.ac.be/urem/IMG/pdf/PRO12102010.pdf
For any natural number "n" (base-10-written) we build
"next-n" like this:
- write the quantity of digits of "n"
- concatenate the quantity of even digits in "n"
- concatenate to the former concatenation the quantity
of odd digits in "n"
Example for "n" = 80322057626942
- there are 14 digits in "n": --> 14
- there are 10 even digits in "n": --> 1410
- there are 4 odd digits in "n": --> 14104
If we iterate the construction, we get the sequence:
80322057626942, 14104, 532, 312, 312, 312, ...
Another example:
5771, 404, 330, 312, 312, 312, ...
Is 312 the end of all such sequences?
Yes, certainly it is.
312 is a fixed point.
If the input is 3 digits, then the output is one of 330, 321, 312, 303. So any 3-digit number becomes 312 in <=2 steps.
Any input with up to 9 digits must have an output that's 3 digits. So any number up to 9 digits becomes 312 in <=3 steps.
Any larger input has an output that's fewer digits than the input, so eventually reaches something 9 digits or less, and so reaches 312. (If you want bounds better than "eventually", though, you can continue the above logic: any input up to 999 digits produces an output with at most 9 digits, so becomes 312 in <=4 steps.)
--Michael
-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Forewarned is worth an octopus in the bush.
Oops, I guess that you're right. Victor On Fri, Oct 15, 2010 at 10:43 AM, Michael Kleber <michael.kleber@gmail.com>wrote:
On Fri, Oct 15, 2010 at 10:36 AM, Victor Miller <victorsmiller@gmail.com
wrote:
Michael, You beat me too it. However, to be complete you need to handle 1 and 2 digit numbers:
I don't think this is a special case. Any number with up to 9 digits produces a 3-digit output (since you're concatenating three single digits) -- this applies to 1 or 2 digit numbers just as well as 4 or 9.
--Michael
1 digit number --> 110 or 101 2 digit number ---> 202, 211 or 220
and now we're in the 3 digit realm
Victor
On Fri, Oct 15, 2010 at 10:31 AM, Michael Kleber <michael.kleber@gmail.com>wrote:
On Fri, Oct 15, 2010 at 10:06 AM, Eric Angelini <Eric.Angelini@kntv.be
wrote:
Hello Math-fun,
a little puzzle slightly adapted from here (in French),
thanks to my friend Alexandre Wajnberg:
http://dev.ulb.ac.be/urem/IMG/pdf/PRO12102010.pdf
For any natural number "n" (base-10-written) we build
"next-n" like this:
- write the quantity of digits of "n"
- concatenate the quantity of even digits in "n"
- concatenate to the former concatenation the quantity
of odd digits in "n"
Example for "n" = 80322057626942
- there are 14 digits in "n": --> 14
- there are 10 even digits in "n": --> 1410
- there are 4 odd digits in "n": --> 14104
If we iterate the construction, we get the sequence:
80322057626942, 14104, 532, 312, 312, 312, ...
Another example:
5771, 404, 330, 312, 312, 312, ...
Is 312 the end of all such sequences?
Yes, certainly it is.
312 is a fixed point.
If the input is 3 digits, then the output is one of 330, 321, 312, 303. So any 3-digit number becomes 312 in <=2 steps.
Any input with up to 9 digits must have an output that's 3 digits. So any number up to 9 digits becomes 312 in <=3 steps.
Any larger input has an output that's fewer digits than the input, so eventually reaches something 9 digits or less, and so reaches 312. (If you want bounds better than "eventually", though, you can continue the
above
logic: any input up to 999 digits produces an output with at most 9 digits, so becomes 312 in <=4 steps.)
--Michael
-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
So how about the following generalization? We're working in base B, we have some fixed partition of the digit set {0,...,B-1} into r pieces, numbered from 1 to r. For any base B number we make a new number the same way: first write down the base B representation of the number of digits, concatenating the base B counts of the digits for the r pieces given above. It's clear by the same reasoning that eventually we'll reach a cycle. So the challenge is to describe other situations where the cycle is length 1. Victor On Fri, Oct 15, 2010 at 10:44 AM, Victor Miller <victorsmiller@gmail.com>wrote:
Oops, I guess that you're right.
Victor
On Fri, Oct 15, 2010 at 10:43 AM, Michael Kleber <michael.kleber@gmail.com
wrote:
On Fri, Oct 15, 2010 at 10:36 AM, Victor Miller <victorsmiller@gmail.com
wrote:
Michael, You beat me too it. However, to be complete you need to handle 1 and 2 digit numbers:
I don't think this is a special case. Any number with up to 9 digits produces a 3-digit output (since you're concatenating three single digits) -- this applies to 1 or 2 digit numbers just as well as 4 or 9.
--Michael
1 digit number --> 110 or 101 2 digit number ---> 202, 211 or 220
and now we're in the 3 digit realm
Victor
On Fri, Oct 15, 2010 at 10:31 AM, Michael Kleber <michael.kleber@gmail.com>wrote:
On Fri, Oct 15, 2010 at 10:06 AM, Eric Angelini <
wrote:
Hello Math-fun,
a little puzzle slightly adapted from here (in French),
thanks to my friend Alexandre Wajnberg:
http://dev.ulb.ac.be/urem/IMG/pdf/PRO12102010.pdf
For any natural number "n" (base-10-written) we build
"next-n" like this:
- write the quantity of digits of "n"
- concatenate the quantity of even digits in "n"
- concatenate to the former concatenation the quantity
of odd digits in "n"
Example for "n" = 80322057626942
- there are 14 digits in "n": --> 14
- there are 10 even digits in "n": --> 1410
- there are 4 odd digits in "n": --> 14104
If we iterate the construction, we get the sequence:
80322057626942, 14104, 532, 312, 312, 312, ...
Another example:
5771, 404, 330, 312, 312, 312, ...
Is 312 the end of all such sequences?
Yes, certainly it is.
312 is a fixed point.
If the input is 3 digits, then the output is one of 330, 321, 312,
Eric.Angelini@kntv.be 303.
So
any 3-digit number becomes 312 in <=2 steps.
Any input with up to 9 digits must have an output that's 3 digits. So any number up to 9 digits becomes 312 in <=3 steps.
Any larger input has an output that's fewer digits than the input, so eventually reaches something 9 digits or less, and so reaches 312. (If you want bounds better than "eventually", though, you can continue the above logic: any input up to 999 digits produces an output with at most 9 digits, so becomes 312 in <=4 steps.)
--Michael
-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Forewarned is worth an octopus in the bush. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
One possibility: the second digit is the number of non-zero digits, the third digit is the number of zeroes. Then by the same argument as above we always end up with a three digit number, except that in this case we alternate between 330 and 321 (cycle of length 2). Seb On Fri, Oct 15, 2010 at 16:49, Victor Miller <victorsmiller@gmail.com> wrote:
So how about the following generalization?
We're working in base B, we have some fixed partition of the digit set {0,...,B-1} into r pieces, numbered from 1 to r.
For any base B number we make a new number the same way:
first write down the base B representation of the number of digits, concatenating the base B counts of the digits for the r pieces given above. It's clear by the same reasoning that eventually we'll reach a cycle. So the challenge is to describe other situations where the cycle is length 1.
Victor
On Friday 15 October 2010 15:49:35 Victor Miller wrote:
So how about the following generalization?
We're working in base B, we have some fixed partition of the digit set {0,...,B-1} into r pieces, numbered from 1 to r.
For any base B number we make a new number the same way:
first write down the base B representation of the number of digits, concatenating the base B counts of the digits for the r pieces given above. It's clear by the same reasoning that eventually we'll reach a cycle. So the challenge is to describe other situations where the cycle is length 1.
*Two* things are special in the 10/even/odd case: we always reach the same cycle, and that cycle has length 1. I think the best generalization is "describe other situations where there is a unique such cycle and it has length 1". -- g
participants (5)
-
Eric Angelini -
Gareth McCaughan -
Michael Kleber -
Seb Perez-D -
Victor Miller