The 2n vertices of a graph G corresponds to all subsets of a set of size n,…

Concept: Fix a subset S of the n-element universe with |S| = k. A subset T is a neighbour of S exactly when T shares precisely two elements with S: choose those 2 shared elements from S in kC2 ways, then freely include or exclude each of the remaining n − k elements of the universe, giving 2n−k possibilities. Let r(k) = kC2 · 2n−k denote this raw count for any subset size k ≥ 2. This equals the true vertex degree d(k) = r(k) for k ≥ 3. One exception: when k = 2, choosing ‘both elements of S’ is the only way (2C2 = 1), so one of the 2n−2 choices — the one adding none of the outside elements — reproduces T = S itself, which is not a valid neighbour (no self-loops); so d(2) = r(2) − 1 = 2n−2 − 1, strictly less than r(2). The graph's maximum degree is the largest value of d(k) over all subset sizes k — not the value at any single k.

Application:

1. For k ≥ 2, compare the raw values r(k) = kC2 · 2n−k consecutively via the ratio r(k+1)/r(k) = [(k+1)C2 / kC2] · (1/2) = [(k+1)/(k−1)] · (1/2).

2. This ratio exceeds 1 when k < 3, equals 1 at k = 3 (so r(4) = r(3)), and is below 1 when k > 3 — so r(k) rises up to k = 3, ties at k = 4, then falls; in particular r(2) < r(3).

3. Since d(k) = r(k) exactly for every k ≥ 3, and d(2) = r(2) − 1 < r(2) < r(3) = d(3), the true degree sequence d(k) also peaks at k = 3 (tied with k = 4) over the entire range k = 2, …, n — the k = 2 correction only lowers d(2) further below d(3), it never changes which k is maximal.

4. Hence the maximum degree is d(3) = r(3) = 3C2 · 2n−3 = 3 · 2n−3.

Cross-check: Verify directly for n = 6 (the same case used in the reported approach), tabulating d(k) at every subset size:

d(6) = 6C2 · 20 = 15 is exactly the count obtained by fixing S as the full 6-element set and listing its 2-element subsets — the same direct-enumeration approach used in the report. That count is a valid vertex degree, but only for that one vertex; it is not the graph's maximum. The true maximum, 24, occurs at k = 3 or k = 4, matching 3 · 2n−3 = 3 · 23 = 24 for n = 6.

Conclusion: the maximum degree of G, valid for every n ≥ 6, is 3 · 2n−3.