For any number N the parity vector v(N) is defined as $v_i(N) = S_i \pmod 2$
If N is a positive integer of the form $a\cdot 2^k + b (b < 2^k)$ then the first k elements of the parity vector are dependent on b only
The proof is by induction on k. For k = 1 we have $N = 2\cdot a + b$ with b either 0 or 1 and the lemma is true by the definition of the 3x + 1 algorithm Now assume the lemma is true for some k Let $N = a\cdot 2^{k+1} + b (b < 2^{k+1})$ Then if $S_0 = N$ is even $S_1 = a\cdot 2^k + b / 2$ and if $S_0 = N$ is odd $S_1 = 3\cdot a\cdot 2^k + (3b+1) / 2$ In either case the first k elements of v(S1) depend on b only Therefore the first k+1 elements of N are independent of a as well This concludes the proof of the lemma.
I am trying to understand the proof. Is the inductive hypotesis used when it is said that the first k elements of $v(s_1)$ depend only on b?
Is correct that the First k elements of v(S1) depend on b only because in every case $a\cdot 2^k$ is even for k steps and so the party Vector depends only on b?
Presumably $S_i$ is a "Collatz sequence" given by $S_0=N$ and $S_{i+1}=\begin{cases} \frac12S_i & S_i\text{ even}\\ \frac12(3S_i+1)& S_i\text{ odd}.\end{cases}$
Yes, that is where the inductive hypothesis is used.