A346401 a(n) is the minimum number of pebbles such that any assignment of those pebbles on a complete graph with n vertices is a next-player winning game in the two-player impartial (3, 2) pebbling game.

13, 21, 15, 21, 17, 25, 21, 29, 25, 33, 29, 37, 33, 41, 37, 45, 41, 49, 45, 53, 49, 57

Offset: 3

Author

Kayla Barker, Mia DeStefano, Eugene Fiorini, Michael Gohn, Joe Miller, Jacob Roeder, Wing Hong Tony Wong, Jul 15 2021

Keywords

• nonn
• more

Comments

A (3,2) pebbling move involves removing 3 pebbles from a vertex in a simple graph and placing 2 pebbles on an adjacent vertex.

A two-player impartial (3,2) pebbling game involves two players alternating (3,2) pebbling moves. The first player unable to make a move loses.

Examples

			For n=6, a(6)=21 is the least number of pebbles for which every (3,2) game on K_6 is a next-player winning game regardless of assignment.
For n=7, a(7)=17 is the least number of pebbles for which every (3,2) game on K_7 is a next-player winning game regardless of assignment.
		

References

• E. R. Berlekamp, J. H. Conway, and R. K. Guy, Winning Ways for Your Mathematical Plays, Vol. 1, CRC Press, 2001.

Links

• E. Fiorini, M. Lind, A. Woldar, and T. W. H. Wong,Characterizing Winning Positions in the Impartial Two-Player Pebbling Game on Complete Graphs, Journal of Integer Sequences, 24(6), 2021.

Crossrefs

Cf. A340631, A084964, A346197.

Programs

Mathematica

remove = 3; add = 2;
(*Given n and m,list all possible assignments.*)
alltuples[n_, m_] := IntegerPartitions[m + n, {n}] - 1;
(*Given an assignment,list all resultant assignments after one pebbling move; only work for n>=3.*)
pebblemoves[config_] := Block[{n, temp},
    n = Length[config];
    temp = Table[config, {i, n (n - 1)}] +
        Permutations[Join[{-remove, add}, Table[0, {i, n - 2}]]];
    temp = Select[temp, Min[#] >= 0 &];
    temp = ReverseSort[DeleteDuplicates[ReverseSort /@ temp]]];
(*Given n and m,list all assignments that are P-games.*)
Plist = {};
plist[n_, m_] :=  Block[{index, tuples},
    While[Length[Plist] < n, index = Length[Plist];
        AppendTo[Plist, {{Join[{1}, Table[0, {i, index}]]}}]];
    Do[AppendTo[Plist[[n]], {}]; tuples = alltuples[n, i];
        Do[If[Not[IntersectingQ[pebblemoves[tuples[[j]]],
                If[i > (remove - add), Plist[[n, i - (remove - add)]], {}]]],
            AppendTo[Plist[[n, i]], tuples[[j]]]], {j, Length[tuples]}],
    {i, Length[Plist[[n]]] + 1, m}]; Plist[[n, m]]];
Do[m = 1; While[plist[n, m] != {}, m++]; Print[" n=", n, " m=", m], {n, 3, 24}]

Formula

a(n) = 2n+3 when n >= 7 is odd (conjectured).

a(n) = 2n+9 when n >= 6 is even (conjectured).