Choose N numbers in order to produce maximum equidistant numbers by summing the chosen numbers Preliminaries
I have a device, an attenuator which is made up of resistive cells. To set the desired level of attenuation there are 7 switches on its front panel with the following values of attenuation (in dBs):
$$
{0.5, 1, 2, 3, 6, 12, 20}
$$
These switches allow to choose the attenuation between 0 and 44.5 dB with the step of 0.5 dB.
I was wondering how the designers of this device have chosen those values. I was thinking that maybe they wanted to produce the maximum possible attenuation range with a step of 0.5 dB by means of 7 resistive cells. But then I realized that it was not the case because the following values produces a wider range:
$$
{0.5, 1, 2, 4, 8, 16, 32}
$$
The range in this case is from 0 to 63.5 dB (and the step is still 0.5dB). There the problem came.
Problem
Let's consider an $N$-tuple
$$
a(N) \in \mathbb{N}^N
$$
and a set
$$
S(a(N)) = \bigg\{\sum_{i\in p,\, p \in \mathcal{P}(\overline{[1,N]})} a_i \bigg | a_i \text{ is the $i$th coordinate of $a(N)$}\bigg\},
$$
where
$\mathcal{P}(A)$ denotes the power set of $A$ and $\overline{[1,N]}$  is $\{i|0 < i < N, i\in\mathbb{N}\}$
So $S(a(N))$ is the set of all possible sums of coordinates of $a(N)$ (including the empty sum).
Let's call an $N$-tuple $a(N)$ good if
$$S(a(N)) = \overline{[1,M]},\text{ for some } M\in\mathbb{N}$$
Let's denote the set of all good $N$-tuples as $G(N)$.
For example, the tuple $a=(1,2,3)$ is good because $S(a) = \{1, 2,3,4,5,6\}$. On the other hand the tuple $b=(1,2,5)$ is not good because $S(b) = \{1,2,3,5,6,7,8\}$ and 4 is missing.

Given $N \in \mathbb{N}$ prove that
$$\bigg [a(N)\in G(N) \text{ and } \sum_i a_i  = \max_{a(N)\in G(N)}({\max_{s_i \in S(a(N))}s_i})\bigg] \iff \bigg[a(N) = (1, 2, 4, ..., 2^{N-1})\bigg]$$
In other words, a good $N$-tuple with the maximum possible sum of its coordinates is $(1,2,...,2^{N-1})$.

Proof
I will prove the statement by induction. Without loss of generality, it is assumed that the coordinates of tuples is sorted in the ascending order.
The base case $N=1$ is obvious.
Let's suppose that the statement is true for all $k\leq N$. Then consider the case for $N+1$. Suppose that in this case the desired tuple $a\neq (1,2,4,...,2^{N})$. Then take the tuple $a' = (a_1, a_2, ..., a_N)$ i.e. $a$ without the last coordinate. There are two possibilities here:
Case $a' = (1,2,4, ..., 2^{N-1})$. In this case $S(a)$ looks like this:
$$
S(a) = \{ \color{red}{0, 1, 2, 3, ..., (2^N-1)}, \color{blue}{(a_{N+1} + 0), (a_{N+1} + 1), (a_{N+1} + 2), ..., (a_{N+1} + 2^N - 1)}\},
$$
where red elements are those obtained only from $a'$ and blue - are those obtained from the rest ($a_{N+1}$ and $a'$). Due to our convention we know that $a_{N+1} \geq a_{N}$. Therefore, taking into account the fact that $a$ is a good tuple, we conclude that $\color{blue}{a_{N+1}+0}$ must be equal to $\color{red}{2^N-1} + 1$. Of course, the reds and blues might intersect but in this case the sum of all coordinates wouldn't be the maximum sum. I.e.
$$
a_{N+1} = 2^N
$$
This contradicts the initial suggestion that $a$ is not $(1,2,4, ..., 2^N)$.
Case $a'\neq (1,2,4, ..., 2^{N-1})$. In this case $S(a)$ is the following (if $a\in G(N+1)$ then $a'\in G(N)$, see Update):
$$
S(a) = \{\color{red}{0, 1, 2, 3, ..., M}, \color{blue}{(a_{N+1} + 0), (a_{N+1} + 1), ..., (a_{N+1} + M)}\},
$$
where the colors have the same meaning as in the previous case and
$$
M < 2^N - 1\quad \text{by induction}
$$
Using the same reasoning as in the previous case we conclude that
$$
a_{N+1} = M + 1
$$
So
$$
\sum_i a_i = 2M + 1 < 2^{N+1} - 1
$$

Update
If $a\in G(N+1)$ then $a'\in G(N)$. I thought it was true but now I cannot prove it :). But this doesn't really violate the proof I provided: suppose that $a'\notin G(N)$ then there is a gap between 0 and $M$ (red numbers), and therefore at least one of the blue sums must fill this gap. So in this case the maximum would be even smaller.

Is my proof a valid one? I am asking because I feel a bit uncomfortable that I don't proof the $\Leftarrow$ and $\Rightarrow$ separately.
I feel that this "fact" has a more elegant proof. Could you provide your own?
 A: Firstly, the proof I provided in the post is not correct.

The conventions and notations are the same as in the post. Also let's denote the set of $N$-tuples which coordinates are distinct powers of two as $B(N)$ so
$$
B(N) := \{a\in \mathbb{N} | a_i = 2^{i-1} \}
$$

It is obvious that
$$
|\mathcal{P}(\overline{[1,N]})| = 2^N
$$
So the maximum possible number in $S(a):a\in G(N)$ is $2^N -1$. This is because there are at most $2^N$ numbers in $S(a)$ for any $a\in \mathbb{N}^N$. But for good $N$-tuples these numbers are consequent from 0 to $M$. So the maximum possible $M$ will be for $a\in G(N)$ for which $S(a)$ contains exactly $2^N$ elements (i.e. there are no duplicates).
So we prove
$$
\big[ a\in G(N)\, \wedge\, \max S(a) = 2^N - 1\big] \iff a\in B(N)
$$
($\Leftarrow$). Obviously, the tuple $a = (1, 2, 4, ..., 2^{N-1})$ doesn't duplicate elements in $S(a)$. And
$$\forall n\in \mathbb{N}, n<2^{N} \quad\exists s\in S(a) : s = n$$
So this $N$-tuple possesses the desired feature.
($\Rightarrow$). Suppose that there exists an $N$-tuple $b\notin B(N)$ such that $$\max_{s_i\in S(b)}s_i = 2^{N} - 1$$
Without loss of generality we assume that
$$b_N \neq 2^{N-1}\tag{1}\label{one}$$
Indeed, otherwise we throw away $b_N$ and consider $b' = (b_1, b_2, b_3, ..., b_{N-1})$ such that its maximum is $2^{N-1}-1$ and the problem stands the same.
Let's consider the tuple $b'=(b_1, b_2, ..., b_{N-1})$. We know that $|S(b')| = 2^{N-1}$. What is the maximum of $S(b')$.
Case 1. $\max S(b') = 2^{N-1} -1$. Then we obtain $b_N = 2^{N-1}$. Contradiction with \ref{one}.
Case 2. $\max S(b') < 2^{N-1} - 1$. Then $b_N > 2^{N-1}$. But then gaps are guaranteed (e.g. the number $2^{N-1} - 1$ would not appear in $S(b)$).$\tag{*}\label{aster}$
Case 3. $\max S(b') > 2^{N-1} - 1$. Then $b_N < 2^{N-1}$. We notice that $S(b)$ divides into two disjoint subsets
$$
S(b) = S(b') \cup S(b)\setminus S(b')\tag{**}\label{twoaster}
$$
In $S(b')$ are the sums which are obtained only from $b'$ and in $S(b)\setminus S(b')$ are the sums which are of the form $b_{N} + X, X\in S(b')$ so, obviously,
$$
\overline{[0, b_{N} - 1]} \subset S(b')\tag{2}\label{two}
$$
But then
$$
\overline{[b_{N}, 2b_{N} -1]} \subset S(b)\setminus S(b')\tag{3}\label{three}
$$
Those numbers may fill the entire set $S(b)$ but then $2b_{N} - 1 = 2^{N} - 1$ so $b_{N} = 2^{N-1}$. Contradiction with \ref{one}. Therefore there are more numbers in $S(b)$. The next one is $2b_{N}$ and it is, obviously, lies in $S(b')$. And then $3b_{N}$ is also in $S(b)$ and lies in $S(b)\setminus S(b')$. But then all the numbers between $2b_{N}$ and $3b_{N}$ are also in $S(b)$:
$$
\overline{[2b_N, 3b_N-1]}\subset S(b')\tag{4}\label{four}
$$
But then
$$
\overline{[3b_N, 4b_N-1]}\subset S(b)\setminus S(b')\tag{5}\label{five}
$$
Again it may be the entire $S(b)$. Then $b_N = 2^{N-2}$. Acting in the same manner one can prove that
$$
b_N = 2^K, K < N-1\tag{6}\label{six}
$$
I feel that there should be a shorter proof using \ref{six} but I couldn't find it.
We see that the last element $b_{N}$ divides the entire range into equal intervals of length of $2^{K}$.
Let's look at the first such interval $I_1 := [0, b_{N}-1]$ together with the tuple $b'' = (b_1', b_2', ..., b_{N-2}')$ i.e. $b'$ without its last element. Obviously, $S(b'')\subset S(b')$ and $S(b')$ itself can be represented as a union of two disjoint subsets (cf.\ref{twoaster}):
$$
S(b') = S(b'') \cup S(b')\setminus S(b'')
$$
But more important, $I_1\subset S(b')$ is also divided by $b_{N-1}$ into two subsets
$$
\overline{[0, b_{N}-1]} = A \cup B, A\subset S(b''), B\subset S(b')\setminus S(b'')
$$
The first we can say is that $B\neq \emptyset$. Indeed if it was so it would mean that the tuple $S(b'')$ fills the entire interval including the number $b_{N}-1$ but then adding $b_{N-1} < b_{N}$ to it would give the number between $b_{N}-1$ and $2b_{N}$ which is in contradiction with \ref{three}. So we know that
$$
\max_{s_i''\in S(b''),s_i''\in I_1} s_i'' < b_{N} -1
$$
More over, it must be greater then or equal to $b_{N}/2 - 1$. Otherwise, $S(b')$ wouldn't fill the entire interval (cf.\ref{aster}):
$$
b_{N}/2 - 1 \leq \max_{s_i''\in S(b''),s_i''\in I_1} s_i'' < b_{N}-1
$$
Using the same reasoning as in the first part of this proof one can show that
$$
b_{N-1} = 2^{K}, K\leq N-3
$$
And eventually
$$
b_{i} = 2^{K_i}, b_{N}  < 2^{N-1}, i < j \Rightarrow b_{i} < b_{j}\tag{7}\label{seven}
$$
But this makes no sense.
