Hilbert's fully occupied Grand Hotel

Question

Q: Suppose that a countably infinite number of buses, each containing a countably infinite number of guests, arrive at Hilbert's fully occupied Grand Hotel. Show that all the arriving guests can be accommodated without evicting any current guest.

The answer in the book:

Move the guest currently in Room $i$ to Room $2i+1$ for $i=1, 2, 3, ... .$ Put the $jth$ guest from the $kth$ bus into Room $2^k(2j+1)$

If we follow the way the textbook tells us, then the Room 1, 2, and 4 are empty.

This is because the 1st guest from the 1st bus is provided the Room $2^1(2*1+1)=6$, and the guest currently in Room 1 is moved to the Room $2*1+1=3.$

Hence, I think it will be more efficient if the answer is

Move the guest currently in Room $i$ to Room $2i-1$ for $i=1, 2, 3, ... .$ Put the $jth$ guest from the $kth$ bus into Room $2^k(2j-1)$

This answer fulfills all the rooms in the hotel, I think.

Is my idea correct?

Answer 1

(This is a very mildly elaborated version of Amritanshu Prasad's comment)

Yes, this method completely fills the hotel.

Every number $N$ has a (unique) prime factorization $N=2^{e_1}p_2^{e_2}\cdots p_k^{e_k}$ where each $p_i$ is odd, and hence $P=p_2^{e_2}p_3^{e_3}\cdots p_k^{e_k}$ is odd. Thus for all $N$ we have the $N^\text{th}$ room occupied, containing (only) the $(P+1)/2\,^\text{th}$ person in the $e_1\,\!^\text{th}$ bus.