Mia DeStefano - cp's OEIS frontend

User: Mia DeStefano

Mia DeStefano has authored 3 sequences.

A347637 Table read by ascending antidiagonals. T(n, k) 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 (k+1, k) pebbling game. T(n, k) for n >= 5 and k >= 1.

7, 13, 15, 9, 21, 21, 15, 17, 35, 27, 11, 25, 25, 37, 33, 17, 21, 41, 33, 59, 39, 13, 29, 31, 45, 41, 53

Offset: 5

Kayla Barker, Mia DeStefano, Eugene Fiorini, Michael Gohn, Joe Miller, Jacob Roeder, Wing Hong Tony Wong, Sep 09 2021

A (k+1, k) pebbling move involves removing k + 1 pebbles from a vertex in a simple graph and placing k pebbles on an adjacent vertex.
A two-player impartial (k+1, k) pebbling game involves two players alternating (k+1, k) pebbling moves. The first player unable to make a move loses.
T(3, k) = A016921(k) for k >= 0. The proof will appear in a paper that is currently in preparation.
It is conjectured that T(4, k) for odd k>=3 is infinite, so we start with n = 5.
T(5, k) = A346197(k) for k >= 1.
T(n, 1) = A340631(n) for n >= 3.
T(n, 2) = A346401(n) for n >= 3.

			The data is organized in a table beginning with row n = 5 and column k = 1. The data is read by ascending antidiagonals. The formula binomial(n + k - 5, 2) + k converts the indices from table form to sequence form.
The table T(n, k) begins:
  [n/k]  1   2   3   4   5   6  ...
  ---------------------------------
  [ 5]   7, 15, 21, 27, 33, 39, ...
  [ 6]  13, 21, 35, 37, 59, 53, ...
  [ 7]   9, 17, 25, 33, 41, 51, ...
  [ 8]  15, 25, 41, 45, 61, ...
  [ 9]  11, 21, 31, 41, 51, ...
  [10]  17, 29, 45, 53, 71, ...
  [11]  13, 25, 37, 49, 61, ...
  [12]  19, 33, 51, ...
  [13]  15, 29, 43, ...
  [14]  21, 37, ...
  [15]  17, 33, ...
  [16]  23, 41, ...

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

Kayla Barker, Mia DeStefano, Eugene Fiorini, Michael Gohn, Joe Miller, Jacob Roeder, and Tony W. H. Wong, Generalized Impartial Two-player Pebbling Games on K_3 and C_4, J. Int. Seq. (2024) Vol. 27, Issue 5, Art. No. 24.5.8. See p. 4.
Eugene Fiorini, Max Lind, Andrew Woldar, and Tony W. H. Wong, Characterizing Winning Positions in the Impartial Two-Player Pebbling Game on Complete Graphs, J. Int. Seq., Vol. 24 (2021), Article 21.6.4.

Cf. A340631, A346197, A346401.

(* m represents number of vertices in the complete graph. Each pebbling move removes k+1 pebbles from a vertex and adds k pebbles to an adjacent vertex. *)
Do[(* Given m and a, list all possible assignments with a pebbles. *)
alltuples[m_, a_] := IntegerPartitions[a + m, {m}] - 1;
(* Given an assignment, list all resultant assignments after one pebbling move; only works for m>=3. *)
pebblemoves[config_] :=
  Block[{m, temp}, m = Length[config];
   temp = Table[config, {i, m (m - 1)}] +
     Permutations[Join[{-(k + 1), k}, Table[0, {i, m - 2}]]];
   temp = Select[temp, Min[#] >= 0 &];
   temp = ReverseSort[DeleteDuplicates[ReverseSort /@ temp]]];
(* Given m and a, list all assignments that are P-games. *)
Plist = {};
plist[m_, a_] :=
  Block[{index, tuples},
   While[Length[Plist] < m, index = Length[Plist];
    AppendTo[Plist, {{Join[{1}, Table[0, {i, index}]]}}]];
   Do[AppendTo[Plist[[m]], {}]; tuples = alltuples[m, i];
    Do[If[
      Not[IntersectingQ[pebblemoves[tuples[[j]]],
        If[i > 2, Plist[[m, i - 1]], {}]]],
      AppendTo[Plist[[m, i]], tuples[[j]]]], {j, Length[tuples]}], {i,
      Length[Plist[[m]]] + 1, a}]; Plist[[m, a]]];
(* Given m, print out the minimum a such that there are no P-games with a pebbles *)
Do[a = 1; While[plist[m, a] != {}, a++];
  Print["k=", k, " m=", m, " a=", a], {m, 5, 10}], {k, 1, 6}]

A346197 a(n) is the minimum number of pebbles such that any assignment of those pebbles on K_5 is a next-player winning game in the two-player impartial (n+1,n) pebbling game.

Original entry on oeis.org

7, 15, 21, 27, 33, 39, 47, 53, 59, 67, 73, 79, 87, 93, 99, 107, 113, 119, 127, 133, 139

Offset: 1

Kayla Barker, Mia DeStefano, Eugene Fiorini, Michael Gohn, Joe Miller, Jacob Roeder, Wing Hong Tony Wong, Jul 09 2021

For n>0, an (n+1,n) pebbling move involves removing n+1 pebbles from a vertex in a simple graph and placing n pebbles on an adjacent vertex.

A two-player impartial (n+1,n) pebbling game involves two players alternating (n+1,n) pebbling moves. The first player unable to make a move loses.

			For n=1, a(1)=7 is the least number of pebbles for which every (2,1) game on K_5 is a next-player winning game regardless of assignment.

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

Kayla Barker, Mia DeStefano, Eugene Fiorini, Michael Gohn, Joe Miller, Jacob Roeder, and Tony W. H. Wong, Generalized Impartial Two-player Pebbling Games on K_3 and C_4, J. Int. Seq. (2024) Vol. 27, Issue 5, Art. No. 24.5.8. See p. 4.
Eugene Fiorini, Max Lind, Andrew Woldar, and Tony W. H. Wong, Characterizing Winning Positions in the Impartial Two-Player Pebbling Game on Complete Graphs, Journal of Integer Sequences, (2021) Vol. 24, Issue 6, Art. No. 21.6.4.

Cf. A340631, A084964.

Do[remove = k + 1; add = k;
(*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[" k=", k, " m=", m], {n, 5, 5}],
{k, 1, 21}]

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.

Original entry on oeis.org

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

Offset: 3

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

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.

			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.

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

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.

Cf. A340631, A084964, A346197.

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}]

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

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

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User: Mia DeStefano

A347637 Table read by ascending antidiagonals. T(n, k) 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 (k+1, k) pebbling game. T(n, k) for n >= 5 and k >= 1.

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A346197 a(n) is the minimum number of pebbles such that any assignment of those pebbles on K_5 is a next-player winning game in the two-player impartial (n+1,n) pebbling game.

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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.

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