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Thursday, September 2, 2010

3x+1 look alike

Number theory fascinates me. I studied elementary number theory, the next step,  analytical number theory requires knowledge of group and representatuon theory ( characters ) and complex analysis. I am working hard on that. In the meanwhile I like to study the famous but accessible problems of number theory. An interesting problem is the following.

It has been conjectured that the function $x_n = f(x_{n-1})$ with $x_0 \in \mathbf{N}$ always ends in $1,4,2,1,4,2,1 \cdots$ where

$$f: x \mapsto \begin{cases}
\frac{x}{2} \text{ if } x \equiv 0 \mod{2} \\
3x+1 \text{ if } x \equiv 1 \mod{2}
\end{cases}$$

Example:
27
82
41
124
62
31
94
47
142
71
214
107
322
161
484
242
121
364
182
91
274
137
412
206
103
310
155
466
233
700
350
175
526
263
790
395
1186
593
1780
890
445
1336
668
334
167
502
251
754
377
1132
566
283
850
425
1276
638
319
958
479
1438
719
2158
1079
3238
1619
4858
2429
7288
3644
1822
911
2734
1367
4102
2051
6154
3077
9232
4616
2308
1154
577
1732
866
433
1300
650
325
976
488
244
122
61
184
92
46
23
70
35
106
53
160
80
40
20
10
5
16
8
4
2
1

A fornal proof that this sequence ends like this is unknown. According to Conway it is even undecidable. Many papers have been written on the problem and it has several dedicated websites.


Very similar, but nevertheless generally generating smaller sequences is the function
$$f: x \mapsto \begin{cases}

1 \text{ if } x = 2 \\
\text{ else } x \mapsto \begin{cases}
\frac{x}{4} \text{ if } x \equiv 0 \mod{4} \\
6x+2 \text{ if } x \equiv 1 \text{ or } 3 \mod{4} \\
\frac{x-2}{4} \text{ if } x \equiv 2 \mod{4}
\end{cases}
\end{cases}$$

 27
164
41
248
62
15
92
23
140
35
212
53
320
80
20
5
32
8
2
1

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