A387704 Size of the maximal subset S of {1,2,...,n} such that for all a, b, c in S not necessarily distinct, a+b+c is unique up to permutation.

0, 1, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 6, 7, 7, 7, 7

Offset

0,3

Comments

It is straightforward to prove a(n) <= a(n+1) <= a(n)+1, and a(n) < a(3n+1).

Links

Sharvil Kesarwani, Table of n, a(n) for n = 0..150

Thomas Bloom Problem 241, Erdős Problems.

Example

For n=5, the maximal subset is {1,2,5}, which generates sums 3,4,5,6,7,8,9,11,12,15 uniquely. Adding 3 would create the duplicate sums 1+1+3=1+2+2, and adding 4 would create the duplicate sum 1+1+4=2+2+2. Hence, a(5)=3.

Mathematica

FindMaxSubset[t_] := Module[{best={}, curr={}, used=ConstantArray[False, 3n+1], search},
search[x_] := Module[{newSums}, If[Length[curr]>Length[best], best=curr]; If[Length[curr] > Length[best], best=curr]; If[Length[curr]+(n-x+1)<=Length[best], Return[]]; Do[newSums=Flatten[{3k, k+2curr, k+Total /@ Subsets[curr, {2}], k+2curr}]; newSums = Flatten[{3k, Table[2k+c, {c, curr}], Table[k+curr[[i]]+curr[[j]], {i, Length@curr}, {j, i, Length@curr}]}]; If[FreeQ[used[[newSums]], True] && DuplicateFreeQ[newSums], used[[newSums]]=True; AppendTo[curr, k]; search[k + 1]; curr=Delete[curr, -1]; used[[newSums]] = False; ], {k, x, n}]]; search[1]; best]

Crossrefs

Cf. A143824 (for 2-sums), A385931 (for distinct 3-sums).

Sequence in context: A385654 A186188 A227177 * A132944 A210568 A262887

Adjacent sequences: A387701 A387702 A387703 * A387705 A387706 A387707

Keyword

nonn,hard

Author

Sharvil Kesarwani, Dec 29 2025

Status

approved