A subset S of {1,2,…,n} is said to be packed if whenever i,j∈S the number ⌊(i+j)/2⌋ is also in S. Determine how many subsets of {1,2,…,25} are packed.

Details and assumptions
i and j need not be distinct. If i=j is in the set, then clearly so is ⌊(i+j)2⌋

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Yup it's cheating

To determine the number of packed subsets of {1, 2, ..., 25}, we can use a recursive approach.

First, let's consider the base case: for an empty set, there is only one packed subset, which is the empty set itself.

Now, for each number n in {1, 2, ..., 25}, we have two possibilities:

1. n is not included in the subset:
In this case, the packed subset will be the same as the packed subset of {1, 2, ..., n-1}. So, we need to count the number of packed subsets of {1, 2, ..., n-1}.

2. n is included in the subset:
In this case, we need to find the packed subsets of {1, 2, ..., n-1} that also include n. For any subset S that includes n, we can remove n from S and find the packed subset of {1, 2, ..., floor(n/2)}. So, we need to count the number of packed subsets of {1, 2, ..., floor(n/2)}.

Using this recursive approach, we can build a dynamic programming solution to count the number of packed subsets.

Here is a Python code implementation:

```python
def count_packed_subsets(n):
if n == 0:
return 1 # Base case: empty set is packed

count = count_packed_subsets(n - 1) # Case 1: n is not included

for i in range(1, n//2 + 1):
count += count_packed_subsets(i - 1) # Case 2: n is included in subset

return count

n = 25
num_packed_subsets = count_packed_subsets(n)
print("Number of packed subsets of {1, 2, ..., 25}:", num_packed_subsets)
```

By running this code, you will get the number of packed subsets of {1, 2, ..., 25}.