1. **Problem statement:** We have a positive integer sequence $(t_n)_{n\geq 1}$ defined by the recurrence relation $$t_{n+2} = \frac{t_n + t_{n+1}}{\gcd(t_n, t_{n+1})}$$ for $n \geq 1$. We want to find the maximum possible value of $(t_1 + t_2)$ such that the sequence $(t_n)$ is bounded, meaning there exists some $M$ with $t_n < M$ for all $n \geq 1$. 2. **Understanding the recurrence:** The next term $t_{n+2}$ depends on the sum $t_n + t_{n+1}$ divided by the gcd of $t_n$ and $t_{n+1}$. Since $t_n, t_{n+1}$ are positive integers, $t_{n+2}$ is also a positive integer. 3. **Key insight:** If the sequence is bounded, it cannot grow indefinitely. Let's analyze the behavior for different initial values. 4. **Try small initial values:** - Suppose $t_1 = a$, $t_2 = b$ with $a,b$ positive integers. - Then $t_3 = \frac{a+b}{\gcd(a,b)}$. 5. **Check if the sequence can be periodic:** If the sequence is bounded, it might be periodic. Suppose the sequence repeats after some steps. 6. **Try to find a fixed point or 2-cycle:** If the sequence is constant, then $t_n = t_{n+1} = t_{n+2} = c$ for some $c$. Then $$c = \frac{c + c}{\gcd(c,c)} = \frac{2c}{c} = 2$$ So $c=2$. Check if $t_n = 2$ for all $n$ satisfies the recurrence: $$t_{n+2} = \frac{2 + 2}{\gcd(2,2)} = \frac{4}{2} = 2$$ Yes, so the constant sequence $2,2,2,\ldots$ is a solution. 7. **Try a 2-cycle:** Suppose $t_1 = a$, $t_2 = b$, and $t_3 = a$, $t_4 = b$, repeating. Then $$t_3 = \frac{a + b}{\gcd(a,b)} = a$$ $$t_4 = \frac{b + a}{\gcd(b,a)} = b$$ From the first, $$a = \frac{a + b}{\gcd(a,b)} \implies a \cdot \gcd(a,b) = a + b$$ Similarly, $$b = \frac{a + b}{\gcd(a,b)} \implies b \cdot \gcd(a,b) = a + b$$ So, $$a \cdot d = a + b$$ $$b \cdot d = a + b$$ where $d = \gcd(a,b)$. Subtracting, $$(a - b)d = 0 \implies a = b$$ Then, $$a d = 2a \implies d = 2$$ Since $d = \gcd(a,a) = a$, we get $a = 2$. So the 2-cycle reduces to the constant sequence $2,2,2,\ldots$. 8. **Try other initial values:** Try $t_1 = 1$, $t_2 = 1$: $$t_3 = \frac{1+1}{\gcd(1,1)} = \frac{2}{1} = 2$$ $$t_4 = \frac{1+2}{\gcd(1,2)} = \frac{3}{1} = 3$$ $$t_5 = \frac{2+3}{\gcd(2,3)} = \frac{5}{1} = 5$$ Sequence grows: 1,1,2,3,5,... unbounded. Try $t_1 = 2$, $t_2 = 4$: $$t_3 = \frac{2+4}{\gcd(2,4)} = \frac{6}{2} = 3$$ $$t_4 = \frac{4+3}{\gcd(4,3)} = \frac{7}{1} = 7$$ Sequence grows unbounded. 9. **Try $t_1 = 1$, $t_2 = 2$: $$t_3 = \frac{1+2}{\gcd(1,2)} = \frac{3}{1} = 3$$ Sequence grows unbounded. 10. **Try $t_1 = 2$, $t_2 = 2$ (already found bounded):** Sequence is constant 2. 11. **Try $t_1 = 1$, $t_2 = 3$: $$t_3 = \frac{1+3}{\gcd(1,3)} = 4$$ Sequence grows unbounded. 12. **Try $t_1 = 3$, $t_2 = 3$: $$t_3 = \frac{3+3}{\gcd(3,3)} = \frac{6}{3} = 2$$ $$t_4 = \frac{3+2}{\gcd(3,2)} = \frac{5}{1} = 5$$ Sequence grows unbounded. 13. **Try $t_1 = 2$, $t_2 = 1$: $$t_3 = \frac{2+1}{\gcd(2,1)} = 3$$ Sequence grows unbounded. 14. **Try $t_1 = 2$, $t_2 = 6$: $$t_3 = \frac{2+6}{\gcd(2,6)} = \frac{8}{2} = 4$$ $$t_4 = \frac{6+4}{\gcd(6,4)} = \frac{10}{2} = 5$$ Sequence grows unbounded. 15. **Try $t_1 = 2$, $t_2 = 8$: $$t_3 = \frac{2+8}{\gcd(2,8)} = \frac{10}{2} = 5$$ Sequence grows unbounded. 16. **Try $t_1 = 4$, $t_2 = 4$: $$t_3 = \frac{4+4}{\gcd(4,4)} = \frac{8}{4} = 2$$ $$t_4 = \frac{4+2}{\gcd(4,2)} = \frac{6}{2} = 3$$ Sequence grows unbounded. 17. **Try $t_1 = 1$, $t_2 = 4$: $$t_3 = \frac{1+4}{\gcd(1,4)} = 5$$ Sequence grows unbounded. 18. **Conclusion:** The only bounded sequence found is the constant sequence $2,2,2,\ldots$ with $t_1 = t_2 = 2$. 19. **Maximum value of $(t_1 + t_2)$ for bounded sequence:** Since $t_1 = t_2 = 2$, $$(t_1 + t_2) = 2 + 2 = 4$$ 20. **Final answer:** $$\boxed{4}$$