I have a little confusion regarding inductive method.
Notation and Definition: Given, $n!=2^{n- s_2(n)}\times a \times r$ ($a, r$ are odd numbers), here, $s_{2}(n) $ is the number of $1$'s in the binary representation of $n$. We use logarithm of base $2$ and ignore any positive number less than $1$.
Context: By induction I would like to show -
$ \log(a)-\log(r)=n- s_2(n)-2, \forall n\geq 7 $.
The basis step would be for $n=7$. In inductive step, we assume that- $ \log(a)-\log(r)=n- s_2(n)-2, \forall n\geq 7 $ is true, this is the inductive hypothesis
Now, we need to show that, $ \log(a_1)-\log(r_1)=(n+1)- s_2(n+1)-2$ is also true. where $(n+1)!=2^{(n+1)- s_2(n+1)}\times a_1 \times r_1$ ($a_1, r_1$ are odd numbers).
If I am not wrong it can be shown that if $n$ is even then $ \log(a_1)-\log(r_1) \neq (n+1)- s_2(n+1)-2$.
Question: Since the proposition is false if $n$ is even, and $n, (n+1)$ can not be odd number at the same time, does it imply that-
$ \log(a)-\log(r)\neq n- s_2(n)-2, \forall n > 7 $.
If not, why above implication is false? please, write your explanation, thanks.