Question

- so let n=1 when is odd than multiply by 3 and adds 1 -- 1x3+1=4 -- when is even divide by 2 so -- 4/2=2 -- 2/2=1 -- so always the result will be 1
- n=2 -- 2/2=1
- n=3 -- 3x3+1=10 -- 10/2=5 -- 5x3+1=16 -- 16/2=8 -- 8/2=4 -- 4/2=2 -- 2/2=1
- n=4 -- 4/2=2 -- 2/2=1
- n=5 -- 5x3+1=16 -- 16/2=8 -- 8/2=4 -- 4/2=2 -- 2/2=1
- n=6 -- 6/2=3 -- 3x3+1=10 -- 10/2=5 -- 5x3+1=16 -- 16/2=8 -- 8/2=4 -- 4/2=2 -- 2/2=1
- n=7 -- 7x3+1=22 -- 22/2=11 -- 11x3+1=34 -- 34/2=17 -- 17x3+1=52 -- 52/2=26 -- 26/2=13 -- 13x3+1=40 -- 40/2=20 -- 20/2=10 -- 10/2=5 -- 5x3+1=16 -- 16/2=8 -- 8/2=4 -- 4/2=2 -- 2/2=1
- de

Answer 1

- deductible can be concluded that all odd and even numbers will result in the end those the same terms ?

Answer 2

We take any number n. If n is even, we halve it (n/2), else we do ”triple
plus one” and get 3n + 1. The conjecture is that for all numbers this process
converges to 1. Hence it has been called ’Half Or Triple Plus One’, sometimes
called HOTPO.
Paul Erd˝os said about the Collatz conjecture: ’Mathematics is not yet ready
for such problems.’ He offered $500 for its solution. ”
In this paper we present a generalization of the Collatz map, and prove the
corresponding conjecture for a set of initial values; this proof has no relevance
for the original Collatz conjecture, although one might hope that it provide
some hint for solving that problem.
1.1 The Collatz problem
Consider the following operation on an arbitrary positive integer:
* If the number is even, divide it by two.
* If the number is odd, triple it and add one.
For example, if this operation is performed on 3, the result is 10; if it is

Answer 3

One possible Proof of Collatz’s Conjecture


Andrew John Gábor
Gizella út 6/b
1143 Budapest - Hungary gbrndrs1968@yahoo.com



,,...Take any natural number n. If n is even, divide it by 2 to get n / 2. If n is odd, multiply it by 3 and add 1 to obtain 3n + 1. Repeat the process (which has been called "Half Or Triple Plus One", or HOTPO) indefinitely. The conjecture is that no matter what number you start with, you will always eventually reach 1. The property has also been called oneness.


Paul Erdős said, allegedly, about the Collatz conjecture: "Mathematics is not yet ripe for such problems" and also offered $500 for its solution....”


n1=1 --- 3x1+1 -- 4/2 -- 2/2 = 1
n2=2 --- 2/2 = 1
n3=3 --- 3x3+1 = 10/2 -- 5x3+1 -- 16/2 -- 8/2 -- 4/2 -- 2/2 = 1
n4=4 --- 4/2 -- 2/2 = 1
n5=5 --- 5x3+1 -- 16/2 -- 8/2 -- 4/2 -- 2/2 = 1
n6=6 --- 6/2 -- 3x3+1 -- 10/2 -- 5x3+1 -- 16/2 -- 8/2 -- 4/2 -- 2/2 = 1
n7=7 --- 7x3+1 -- 22/2 -- 11x3+1 -- 34/2 -- 17x3+1 -- 52/2 -- 26/2 -- 13x3+1 -
- 40/2 -- 20/2 -- 10/2 -- 5x3+1 -- 16/2 -- 8/2 -- 4/2 -- 2/2 = 1
n8=8 --- 8/2 -- 4/2 -- 2/2 = 1
n9=9 --- 9x3+1 -- 28/2 -- 14/2 -- 7x3+1 -- 22/2 -- 11x3+1 -- 34/2 -- 17x3+1 --
52/2 -- 26/2 -- 13x3+1 -- 40/2 -- 20/2 -- 10/2 -- 5x3+1 -- 16/2 -- 8/2 -- 4/2 --
-- 2/2 = 1
n10=10 --- 10/2 -- 5x3+1 -- 16/2 -- 8/2 -- 4/2 -- 2/2 = 1
…....................................................................................................................
.....
2.





for n=(n2)n
(n2)n=2n --- ... -- 5n2/2 -- (2n2 + 1)x3 + 1 -- 6n2 + 4 -- 8n2/2 -- 4n2/2 -- 2n2/2 --
2/2 = 1
(n2)n+1=2n+1 --- ... -- 5n2/2 -- (2n2 + 1)x3 + 1 -- 6n2 + 4 -- 8n2/2 -- 4n2/2 --
2n2/2 -- 2/2 = 1

for n = k suppose that is true
(n2)n=(k2)n=2k --- ... -- 5k2/2 -- (2k2 + 1)x3 + 1 -- 6k2 + 4 -- 8k2/2 -- 4k2/2 --
2k2/2 -- 2/2 = 1
(n2)n+1=(k2)n+1=2k+1 --- ... -- 5k2/2 -- (2k2 + 1)x3 + 1 -- 6k2 + 2k2 -- 8k2/2 --
4k2/2 -- 2k2/2 -- 2/2 = 1

for k = k + 1
k2n=(k2n +1) --- … -- (5(k2 +1))x3 +1 -- 15k2 +15 +1 -- 15k2 +16 -- 15k2 +8k2
-- 23k2/2 -- (11k2 +1)x3 +1 -- 33k2 +2k2 -- 35k2/2 -- (17k2 +1)x3 +1 -- 51k2
+ 2k2 -- 53k2/2 -- (26k2 +1)x3 +1 -- 78k2 +2k2 -- 80k2/2 -- 40k2/2 -- 20k2/2 --
10k2/2 -- 5k2/2 -- (2k2 +1)x3 +1 -- 6k2 +2k2 -- 8k2/2 -- 4k2/2 -- 2k2/2 -- 2/2 =
1
k2n+1=(k2n+1 +1) --- … -- (5(k2 +1) +1) -- 5k2 +6 -- 5k2 +3k2 -- 8k2/2 -- 4k2/2 --
2k2/2 -- 2/2 = 1


q.e.d.