Derek's sequence

Transition from $a_k$ to $a_{k+1}$ is given by the Mobius transformation with a matrix: $M=\left(\begin{array}{cc} 1 & -i\\ 1 & i\end{array}\right)$ The composition of Mobius transformations is, again, Mobius transformation, with a matrix given by the matrix product of initial transformation matrices. Therefore, $a_k$ to $a_{k+3}$ transformation is given by: $M_3=M^3=\left(\begin{array}{cc} 1 & -i\\ 1 & i\end{array}\right)\times\left(\begin{array}{cc} 1 & -i\\ 1 & i\end{array}\right)\times\left(\begin{array}{cc} 1 & -i\\ 1 & i\end{array}\right)=\left(\begin{array}{cc} 1-i & 1-i\\ 1+i & -1-i\end{array}\right)\times\left(\begin{array}{cc} 1 & -i\\ 1 & i\end{array}\right)=\left(\begin{array}{cc} 2-2i & 0\\ 0 & 2-2i\end{array}\right)=(2-2i)\left(\begin{array}{cc} 1 & 0\\ 0 & 1\end{array}\right)$ Since overall multiplier is not important for Mobius transforms, we see, that the composition of three transforms is an identity, and $a_k$ is periodic with a period of a divisor of three, i.e. 1 or 3. The alternating series partial sums, used to compute members of S can therefore be written using just 3 terms: $a_1,a_1-a_2,a_1-a_2+a_3,-a_2+a_3,a_3,0,a_1,...$ i.e. it's periodic with a period of a divisor of 6. Quick check with $a_1=2i$ gives $a_2=1/3$ and $a_3=-0.8-0.6i$ , which gives partial sums: $2i,-1/3+2i,-17/15+7/5i,-17/15-3/5i,-4/5-3/5i,0,2i,...$ i.e. for at least one $a_1$ these series of partial sums is periodic with a period of 6, and all the terms within a period are different.

For a given prime $p$ the partial sum would be determined by a remainder after division of p by 6. For $p<6$ the reminders are: $2, 3, 5$ . For $p>6$ the remainders have to be coprime with 6, otherwise, according to the equality $p=6q+r$ , $p$ has to divide by their common divisor, which is less then 6, and, therefore, not equal to p. Therefore the remainders 0, 2, 3 and 4 are impossible for $p>6$ , or, there are no remainders 0 and 4 for any $p$ . Writing these remainders down for the first prime numbers $2, 3, 5, 1, ...$ we see that other four possibilities are met for some primes. Answer: 4

Let $f(z) = \dfrac{z-i}{z+i}$ for all $z \neq -i$ . So, $a_{k+1} = f(a_k)$ for all $k \ge 1$ . Then, $f(f(z)) = -i\dfrac{z+1}{z-1}$ and $f(f(f(z))) = z$ hold for all $z \neq -i,1$ .

If $a_k = -i, 1$ then $a_{k+1} = f(a_k)$ or $a_{k+2} = f(f(a_k))$ will be undefined.

Thus, $a_k \neq -i, 1$ , and so, $a_{k+3} = f(f(f(a_k))) = a_k$ for all $k \ge 1$ .

Let $s_n = \displaystyle\sum_{k = 1}^{n}(-1)^k a_k$ . So, $S(a_1) = \{s_p \ | \ p \ \text{is prime}\}$ . Note that: $s_{n+6} - s_n = (-1)^n(-a_{n+1} + a_{n+2} - a_{n+3} + a_{n+4} - a_{n+5} + a_{n+6})$

$= (-1)^n\left(-a_{n+1} + a_{n+2} - a_{n+3} + a_{n+1} - a_{n+2} + a_{n+3}\right) = 0$ .

Hence, $s_{n+6} = s_n$ for all $n \ge 1$ . So, if $n \equiv r \pmod{6}$ then $s_n = s_r$ .

For all primes $p$ we have $p = 2,3$ or $p \equiv 1,5 \pmod{6}$ .

Therefore, $s_p \in \{s_1,s_2,s_3,s_5\}$ for all primes $p$ .

Hence, $S(a_1) \subseteq \{s_1,s_2,s_3,s_5\}$ , and thus, $|S(a)| \le 4$ for all $a$ .

Now, we must check that $|S(a)| = 4$ is in fact attainable for some $a$ .

For $a_1 = 2$ we have $a_2 = \tfrac{3}{5} - \tfrac{4}{5}i$ , and $a_3 = -3i$ . Hence,

$s_2 = -a_1+a_2 = -\tfrac{7}{5} - \tfrac{4}{5}i$ , $s_3 = -a_1+a_2-a_3 = -\tfrac{7}{5} + \tfrac{11}{5}i$ ,

$s_5 = -a_3 = 3i$ , and $s_7 = s_1 = -a_1 = -2$ . So, $|S(2)| = 4$ .

Since $|S(a)| \le 4$ for all $a$ and $|S(2)| = 4$ , we have $\displaystyle\max_{a} |S(a)| = \boxed{4}$ .