Let's analyze the given sequence.

a_{n + 1} = ⎩ ⎪ ⎪ ⎨ ⎪ ⎪ ⎧ \frac{1}{2} a_{n}, 3 a_{n} + 1, if a_{n} is even if a_{n} is odd

We are asked to write terms of that sequence for

a_{1} = 34

and for

a_{1} = 25 .

$a_{1} = 34$

Let $a_{1} = 34 .$ Since $a_{1} = 34$ is even, the second term is $a_{2} = \frac{1}{2} a_{1} = \frac{1}{2} (34) = 17 .$ Since $a_{2} = 17$ is odd, the third term is $a_{3} = 3 a_{2} + 1 = 3 (17) + 1 = 52 .$ We will continue to do the same until we come across a pattern.

$n$	$a_{n - 1}$	Is $a_{n - 1}$ even?	$a_{n}$
$1$	$-$	$-$	$a_{1} = 34$
$2$	$a_{1} = 34$	Yes $✓$	$a_{2} = \frac{1}{2} (34) = 17$
$3$	$a_{2} = 17$	No $\times$	$a_{3} = 3 (17) + 1 = 52$
$4$	$a_{3} = 52$	Yes $✓$	$a_{4} = \frac{1}{2} (52) = 26$
$5$	$a_{4} = 26$	Yes $✓$	$a_{5} = \frac{1}{2} (26) = 13$
$6$	$a_{5} = 13$	No $\times$	$a_{6} = 3 (13) + 1 = 40$
$7$	$a_{6} = 40$	Yes $✓$	$a_{7} = \frac{1}{2} (40) = 20$
$8$	$a_{7} = 20$	Yes $✓$	$a_{8} = \frac{1}{2} (20) = 10$
$9$	$a_{8} = 10$	Yes $✓$	$a_{9} = \frac{1}{2} (10) = 5$
$10$	$a_{9} = 5$	No $\times$	$a_{10} = 3 (5) + 1 = 16$
$11$	$a_{10} = 16$	Yes $✓$	$a_{11} = \frac{1}{2} (16) = 8$
$12$	$a_{11} = 8$	Yes $✓$	$a_{12} = \frac{1}{2} (8) = 4$
$13$	$a_{12} = 4$	Yes $✓$	$a_{13} = \frac{1}{2} (4) = 2$
$14$	$a_{13} = 2$	Yes $✓$	$a_{14} = \frac{1}{2} (2) = 1$
$15$	$a_{14} = 1$	No $\times$	$a_{15} = 3 (1) + 1 = 4$
$16$	$a_{15} = 4$	Yes $✓$	$a_{16} = \frac{1}{2} (4) = 2$
$17$	$a_{16} = 2$	Yes $✓$	$a_{17} = \frac{1}{2} (2) = 1$

We got the term $1$ again. This tells us that the sequence $4,$ $2,$ $1$ will repeat indefinitely.

$a_{1} = 25$

We will continue to do the same method as the first sequence.

$n$	$a_{n - 1}$	Is $a_{n - 1}$ even?	$a_{n}$
$1$	$-$	$-$	$a_{1} = 25$
$2$	$a_{1} = 25$	No $\times$	$a_{2} = 3 (25) + 1 = 76$
$3$	$a_{2} = 76$	Yes $✓$	$a_{3} = \frac{1}{2} (76) = 38$
$4$	$a_{3} = 38$	Yes $✓$	$a_{4} = \frac{1}{2} (38) = 19$
$5$	$a_{4} = 19$	No $\times$	$a_{5} = 3 (19) + 1 = 58$
$6$	$a_{5} = 58$	Yes $✓$	$a_{6} = \frac{1}{2} (58) = 29$
$7$	$a_{6} = 29$	No $\times$	$a_{7} = 3 (29) + 1 = 88$
$8$	$a_{7} = 88$	Yes $✓$	$a_{8} = \frac{1}{2} (88) = 44$
$9$	$a_{8} = 44$	Yes $✓$	$a_{9} = \frac{1}{2} (44) = 22$
$10$	$a_{9} = 22$	Yes $✓$	$a_{10} = \frac{1}{2} (22) = 11$
$11$	$a_{10} = 11$	No $\times$	$a_{11} = 3 (11) + 1 = 34$

Notice that we get $a_{11} = 34 .$ Therefore, after this term we will get the same sequence as in the first sequence, where $a_{1} = 34 .$

Conclusion

Notice that that each time an even number is reached, it is divided until it reaches an odd number. Then, it grows again, but it appears that at some point we reach the number $4,$ from which the sequence has the subsequence $4,$ $2,$ $1,$ that repeats indefinitely.

Extra

Collatz Conjecture

The Collatz conjecture is a mathematical conjecture related to the sequence in this exercise. The conjecture is that no matter what value of $a_{1}$ we start with, the given sequence will always reach the subsequence $4,$ $2,$ $1,$ which repeats indefinitely.

Paul Erdos, a famous mathematician, said about the Collatz conjecture: Mathematicians may not be ready for such problems.
Jeffrey Lagarias, another famous mathematician, said: This is an extraordinarily difficult problem, completely out of reach of present day mathematics.

Even today, mathematicians do not know if the Collatz conjecture is true or false.

a1​=34

a1​=25

Conclusion

Extra

$a_{1} = 34$

$a_{1} = 25$

Exercise