| Title: | Mathematics at DEC |
| Moderator: | RUSURE::EDP |
| Created: | Mon Feb 03 1986 |
| Last Modified: | Fri Jun 06 1997 |
| Last Successful Update: | Fri Jun 06 1997 |
| Number of topics: | 2083 |
| Total number of notes: | 14613 |
Please see note 249 for background of the following math problem:
Consider a hailstorm cycle which takes an odd positive integer
to another odd positive integer like this:
NEW = (OLD*3+1)/2/2/2/2/2/2/2/2...
That is, you divide by 2 as much as necessary to make NEW be odd.
Prove that for all positive integers N, there exists a set
S(N)
of numbers that require at least N cycles to reach the integer 1.
For example, try 7. One cycle is (3*7+1)/2, or 11, and two cycles
is (3*11+1)/2 or 17, and three cycles is (3*17+1)/4 or 13, and four
cycles is (3*13+1)/8 or 5, and five cycles is (3*5+1)/16 or 1.
So 7 takes 5 cycles. Hence
s(5)={7,...}
So for N=5, we've proved the conjecture.
Another way of stating the conjecture, is that although the
grand conjecture of claiming that all numbers lead to 1 is not
proved, perhaps we can prove this simpler conjecture that shows
that numbers can be found that lead to 1 but take arbitrarily as
long as we like to get there.
/Eric
| T.R | Title | User | Personal Name | Date | Lines |
|---|---|---|---|---|---|
| 998.1 | KOBAL::GILBERT | Ownership Obligates | Tue Dec 27 1988 11:10 | 14 | |
Let S(N) be the set of odd numbers that require exactly N cycles (as
described in .0) to reach the integer 1.
Choose an element p of S(N). Suppose p = 3*s+2. Now 2*s+1, 8*s+5,
32*s+21, ..., (2*4^k)*s + (4*4^k-1)/3, ... are all members of S(N+1),
because one cycle leads them to p. Suppose p = 3*s+1; now 4*s+1,
16*s+5, ..., (4*4^k)*s + (4*4^k-1)/3, ... are all members of S(N+1).
If p = 3*s, there is no value that leads to p in one cycle. However,
the above construction shows that for every N, S(N) contains an infinite
number of values that aren't multiples of 3. Thus, for any non-empty S(N),
the above construction produces an infinite number of elements in S(N+1).
By induction, every S(N) contains an infinite number of elements.
| |||||
| 998.2 | New book sheds light | CIVAGE::LYNN | Lynn Yarbrough @WNP DTN 427-5663 | Wed Jun 12 1991 17:13 | 4 |
There's a lot of recent ideas about this problem - and a 'Mathematica' program to investigate it - in the recently published 'Computational Recreations in Mathematica' (Author: Vardi) by Addison-Wesley. Anyone for rewriting it in MAPLE? | |||||