Update: I've now found two solutions that require only $25$ weighings:
$$
01011011100010111101000001100111110011010100011010\;,\\
00101110001101001010111110001000110010011111011101\;.
$$
These are the only solutions known (in the context of this post) with $k=n\,/\,2$, for any $n$. I found them using Tad's recipe of repeatedly maximizing the determinant of $AA^\top$ by hill climbing and testing the candidate with the highest determinant among several runs. The determinants are
$$
87546852131623099566361867104\;,\\
93001686149176479553585114299\;.
$$
I only had to test half a dozen candidates to find these two solutions, which seems consistent with $f(50)\simeq 22.4$. I'll now be testing these two with $24$ weighings; that will take the better part of a day and will require a bit more than a trillion difference vectors to be checked in each case.
I believe the bounty should go to Tad, who developed the mathematical approach. I tried to improve on it but couldn't, so instead I coded it a whole lot faster :-). Here's the code. It uses a special bit encoding for efficient arithmetic with vectors over $\mathbb F_3$. On my MacBook, it takes half a minute to test a candidate with $31$ weighings and two hours for a candidate with $26$ weighings.
The best solution it's found so far is $01010111100010011111110000111011000101101100100001$, with $26$ weighings required. I'm now looking for solutions with $25$ weighings, but that requires up to six hours of testing per candidate. I expect the minimum to be near $23$ (see below). I haven't implemented Tad's determinant maximization approach yet; the above solution was just the third candidate I tried with uniform sampling among the vectors (uniform sampling among the rotational equivalence classes would be worse since it gives higher weight to periodic sequences).
Here's a short summary of the mathematical basis of the code, mostly following Tad's ideas and notation; $n$ is the number of coins, and $k$ is the number of weighings. We can treat the two weights of the coins as $0$ and $1$. Then every possible weight configuration is characterized by a binary weight vector, and the differences between weight vectors are ternary vectors with entries $-1$, $0$ and $+1$. A solution is admissible if the $k\times n$ matrix $A$, with entries $A_{ij}\in\{0,1\}$ according as coin $j$ is weighed in weighing $i$, has different products with all possible weight vectors, or equivalently, if its product with all possible difference vectors is non-zero.
We can regard the difference vectors as vectors over $\mathbb F_3$. Only the difference vectors in the kernel of $A$ over $\mathbb F_3$ can be in the kernel of $A$ over $\mathbb Z$. The kernel over $\mathbb F_3$ will generally contain $3^{n-k}$ difference vectors, and these we have to check over $\mathbb Z$ (actually only half of them, due to invariance under negation). The key to doing this efficiently is to bit-encode $\mathbb F_3$ such that the time-limiting steps (addition, componentwise multiplication and summation of vectors) can be performed compactly using integer operations and table lookups instead of iterating over $n$ vector components.
Here are some results that expand on Tad's results for values well below $50$. $n_\text{c}$ is the number of equivalence classes under rotation; the numbers in the header row are $n-k$, and the numbers in the bodies of the two tables are the numbers and fractions, respectively, of rotational equivalence classes that yield a solution.
\begin{array}{c|c|cc}
n_\text{c}&n&0&1&2&3&4&5&6&7&8&9\\\hline
2&1&1\\
3&2&1\\
4&3&2\\
6&4&2&1\\
8&5&6&3\\
14&6&6&4\\
20&7&18&12\\
36&8&20&17&8\\
60&9&50&45&25&6\\
108&10&74&71&56&17\\
188&11&186&178&130&83&4\\
352&12&216&214&200&135&42\\
632&13&630&614&520&469&261&6\\
1182&14&916&910&878&787&535&101\\
2192&15&2002&1988&1924&1831&1427&648&22\\
4116&16&3040&3037&3000&2926&2605&1686&330&2\\
7712&17&7710&7699&7464&7330&6915&5361&2131&34\\
14602&18&10806&10801&10760&10649&10310&9014&5547&875\\
27596&19&27594&27582&27270&27110&26565&24773&18964&7307&152\\
52488&20&40642&40639&40592&40474&40024&38487&33414&20254&3102&2\\
99880&21&94658&94640&94440&94281&93518&91759&85546&65010&24368&508\\
\end{array}
\begin{array}{c|c|cc}
n_\text{c}&n&0&1&2&3&4&5&6&7&8&9\\\hline
2&1&.5000\\
3&2&.3333\\
4&3&.5000\\
6&4&.3333&.1667\\
8&5&.7500&.3750\\
14&6&.4286&.2857\\
20&7&.9000&.6000\\
36&8&.5556&.4722&.2222\\
60&9&.8333&.7500&.4167&.1000\\
108&10&.6852&.6574&.5185&.1574\\
188&11&.9894&.9468&.6915&.4415&.0213\\
352&12&.6136&.6080&.5682&.3835&.1193\\
632&13&.9968&.9715&.8228&.7421&.4130&.0095\\
1182&14&.7750&.7699&.7428&.6658&.4526&.0854\\
2192&15&.9133&.9069&.8777&.8353&.6510&.2956&.0100\\
4116&16&.7386&.7379&.7289&.7109&.6329&.4096&.0802&.0005\\
7712&17&.9997&.9983&.9678&.9505&.8967&.6952&.2763&.0044\\
14602&18&.7400&.7397&.7369&.7293&.7061&.6173&.3799&.0599\\
27596&19&.9999&.9995&.9882&.9824&.9626&.8977&.6872&.2648&.0055\\
52488&20&.7743&.7743&.7734&.7711&.7625&.7333&.6366&.3859&.0591&.0000\\
99880&21&.9477&.9475&.9455&.9439&.9363&.9187&.8565&.6509&.2440&.0051\\
\end{array}
Here are the minimal $k$ values up to $n=26$ (denoted by $f(n)$ as in Tad's table), together with the numbers and fractions of solution classes at minimal $k$:
\begin{array}{c|c|c|c}
n&f(n)&\#&\#\,/\,n_\text{c}\\\hline
1&1&1&.5000\\
2&2&1&.3333\\
3&3&2&.5000\\
4&3&1&.1667\\
5&4&3&.3750\\
6&5&4&.2857\\
7&6&12&.6000\\
8&6&8&.2222\\
9&6&6&.1000\\
10&7&17&.1574\\
11&7&4&.0213\\
12&8&42&.1193\\
13&8&6&.0095\\
14&9&101&.0854\\
15&9&22&.0100\\
16&9&2&.0005\\
17&10&34&.0044\\
18&11&875&.0599\\
19&11&152&.0055\\
20&11&2&.0000\\
21&12&508&.0051\\
22&12&8&.0000\\
23&13&2340&.0064\\
24&13&36&.0001\\
25&14&10688&.0080\\
26&14&216&.0001
\end{array}
The solution class counts of $2$ for $n=16$ and $n=20$ that both come at the end of an unusual triple of identical values of $f(n)$ are salient; here are the solution vectors:
$$
0000101110111011\;,\\
0000110111011101\;,\\
00000101111011110111\;,\\
00000111011110111101\;.\\
$$
Note that in both cases the two solutions are related by inversion, they contain slightly longer runs of $0$s than of $1$s, and the runs of $1$s are separate by single zeros.
It's tempting to speculate that the bound $f(n)\gt n/2$ that holds up to $n=26$ will continue to hold for higher $n$. However, I don't believe that this is the case. There's already a slight indication of this in the data; the solutions are gradually encroaching on the bound. (The trends in the absolute numbers are relevant here, not in the fractions, since we only need a single solution.) The fact that the third uniformly randomly guessed candidate turned out to be a solution for $k=26$, $n=50$ also points in this direction, since it suggests that the density of solutions at that point is much higher than it is for $k=n/2-1$ at lower $n$ values (where it's almost negligible).
Moreover, statistical considerations suggest that the asymptotic behaviour of $k$ might be $n\,/\log n$. There are various handwaving arguments that could be used to support this, based on entropy or collision probabilities; the details may be hard to get right, but the bigger picture is the same in each case: The $k$ measurements each have an effective space of order some power $n^\alpha$ to spread across ($\sqrt n$ if we consider them to be binomially distributed), so our discerning ability grows with $n^{\alpha k}$, while the number of possibilities we need to be able to discern grows with $2^n$; equating the logarithms yields $\alpha k\log n=n\log2$ and thus $k\propto n\,/\log n$.
Such $n\,/\log n$ behaviour seems consistent with the data up to $n=26$, so we can get a rough estimate e.g. from $f(26)=14$:
$$f(50)\simeq\frac{50\log26}{26\log50}\,f(26)\simeq1.6\,f(26)\simeq22.4\;.$$
That leaves ample space for improvement. :-)
Best Answer
Nice idea to take different amount of coins from different bags. After first measurement, You will have at most 5 suspected pairs as You mentioned.
Hint: if You take $i$ coins from the $i$ - th pairs and weigh them together, You will be able to tell which pair is defective. That means only TWO weighings
Edit: it does not need to be any specific number from the pairs. as long as you take different amount from different pair