Maybe binary numeral system could help...

We are going to prove the answer by showing that a bijection exists between the possible arrangements in $a(n)$ and $b(n)$ . For this, let us encode every construction in $a(n)$ in the following way:

Firstly, convert every $x_i$ to its binary form and write them into $i$ rows (the $i$ th row containing $x_i$ ) such that their first digits are below each other. Since all $x_i$ 's are in the form of $2^{k_i}-1$ for positive integers $k_i$ , their binary form consists only of 1's. Moreover, from the condition $x_i \leq x_{i+1}$ it is evident that the $i+1$ th row must contain at least as much digits as the $i$ th. Finally, let us assign a unique value to every digit which represents the amount it contributes to the value of its corresponding $x_i$ ; this can be achieved by defining the value of a certain digit to be $2^l$ if the number of digits in the same row which are on the right hand side of the digit appears to be $l$ . A clear consequence of this assigment is the fact that $x_i$ is equal to the sum of the values of digits in the $i$ th row.

Now we seek to show that if we add together the digit's values in every column instead of row, we obtain a proper construction in $b(n)$ . Indeed, denoting the sum of the values in the $i$ th row from the right by $y_i$ it is apparent that the condition $y_{i+1} \geq 2y_i$ is met, since every digit corresponding to $y_i$ has a neighbouring digit from the left with twice as much value that is contained in $y_{i+1}$ . Moreover, as the sums $n=x_1+x_2+\ldots$ and $y_1+y_2+\ldots$ both can be calculated by summing up the values of every digit in the construction, it follows that $n=y_1+y_2+\ldots$ . Henceforth, we have found an injection from $a(n)$ to $b(n)$ .

It suffices to show that this injection is a one-to-one mapping. We'll prove it by correlating every $n=y_1+y_2+\ldots$ arrangement to a unique $n=x_1+x_2+\ldots$ arrangement such that the latter is corresponding to the former according to our injective reasoning. We proceed by the following:

Place $y_1$ number of 1's in a column. If $i$ number of columns have already been constructed, we add the $(i+1)$ st by writing 1's on the left side of every digit in the $i$ th column, and $y_{i+1}-2y_i$ number of 1's over them in addition (note that this can always be accomplished in view of $y_{i+1} \geq 2y_i$ ). After completing the construction, let all of the digits be given the same defined value as formerly. Finally, observe that the sum of values of 1's in the $i$ th column from the right equals to $y_i$ for every $i$ , and if the values are summed up in every row we end up with a proper $n=x_1+x_2+\ldots$ arrangement which corresponds to the $b(n)$ construction we started with. That's exactly what we wanted to show.

Hence, our proof is complete.