cp's OEIS Frontend

This is a front-end for the Online Encyclopedia of Integer Sequences, made by Christian Perfect. The idea is to provide OEIS entries in non-ancient HTML, and then to think about how they're presented visually. The source code is on GitHub.

User: Wing Hong Tony Wong

Wing Hong Tony Wong's wiki page.

Wing Hong Tony Wong has authored 38 sequences. Here are the ten most recent ones:

A375354 T(m,n) is the number of suitably connected m X n Legendrian mosaics read by rows, with 1<=n<=m.

Original entry on oeis.org

1, 1, 2, 1, 4, 20, 1, 8, 104, 1504, 1, 16, 544, 22208, 948032, 1, 32, 2848, 329216, 40930304, 5204262912, 1, 64, 14912, 4883968, 1772261888, 666548862976, 254112496082944, 1, 128, 78080, 72464384, 76795762688, 85575149027328, 97392800416399360, 111879597850371293184
Offset: 1

Keywords

Comments

An m X n Legendrian mosaic is an m X n array of the 10 tiles given in Figure 5 of Pezzimenti and Pandey. These tiles represent part of a Legendrian curve in the front projection. The condition of being suitably connected means that the connection points of each tile coincide with those of the contiguous tiles.
The Mathematica program below is based on the algorithm given in Theorem 1 of Oh, Hong, Lee, and Lee, adapted to the Legendrian setting: since Legendrian mosaic tiles omit the crossing tile T_9 used in general knot mosaics, the bottom-right submatrix of O_(k+1) is 3*O_k rather than 4*O_k. See Theorem 6 of Kipe et al.
T(m,2) = A375353(m,2) = A000079(m-1) for all m >= 2 since neither classical nor Legendrian link mosaics with only 2 columns or rows can use T_9 tiles.

Examples

			Triangle begins:
  1;
  1,  2;
  1,  4,   20;
  1,  8,  104,   1504;
  1, 16,  544,  22208,   948032;
  1, 32, 2848, 329216, 40930304, 5204262912;
  ...
T(2,2) = 2 since the only suitably connected 2 X 2 Legendrian mosaics are the empty mosaic and the mosaic depicting the Legendrian unknot with maximal Thurston-Bennequin invariant.
For all n >= 1, we have T(n,1) = 1 since the only suitably connected Legendrian mosaic with one column is empty.
		

Crossrefs

The main diagonal T(n,n) is A374947.

Programs

  • Mathematica
    x[0] = o[0] = {{1}};
    x[n_] := ArrayFlatten[{{x[n - 1], o[n - 1]}, {o[n - 1], x[n - 1]}}];
    o[n_] := ArrayFlatten[{{o[n - 1], x[n - 1]}, {x[n - 1], 3*o[n - 1]}}];
    legendrian[m_, n_] := If[m > 1 && n > 1, 2*Total[MatrixPower[x[m - 2] + o[m - 2], n - 2], 2], 1];
    Flatten[ParallelTable[legendrian[m, n], {m, 1, 11}, {n, 1, m}]] (* Luc Ta, Aug 13 2024 *)

Formula

T(m,3) = A082761(m-1) for all m >= 1. - Luc Ta, Aug 20 2024

A374946 a(n) is the number of suitably connected Legendrian n-mosaics that form a Legendrian knot.

Original entry on oeis.org

0, 1, 17, 793, 275557, 831699598
Offset: 1

Keywords

Comments

A Legendrian n-mosaic is an n X n array of the 10 tiles given in Figure 5 of Pezzimenti and Pandey. These tiles represent part of a Legendrian curve in the front projection. A Legendrian n-mosaic is suitably connected if the connection points of each tile coincide with those of the contiguous tiles.

Examples

			For n = 2 there is exactly a(2) = 1 Legendrian 2-mosaic forming the front projection of a Legendrian knot, namely the Legendrian unknot with maximal Thurston-Bennequin invariant.
		

A374944 a(n) is the maximum over the minimum crossing numbers of all Legendrian knots that can be realized on a Legendrian n-mosaic.

Original entry on oeis.org

0, 0, 0, 0, 3, 8
Offset: 1

Keywords

Comments

A Legendrian n-mosaic is an n X n array of the 10 tiles given in Figure 5 of Pezzimenti and Pandey. These tiles represent part of a Legendrian curve in the front projection.

Examples

			For n = 5, the only Legendrian knots that can be realized on a Legendrian 5-mosaic are positive and negative Legendrian trefoils, which have a minimal crossing number of 3, and Legendrian unknots, which have a minimal crossing number of 0. Therefore, a(5) = 3.
		

A374943 a(n) is the number of distinct Legendrian unknots with nonnegative rotation numbers that can be realized on a Legendrian n-mosaic.

Original entry on oeis.org

0, 1, 4, 9, 21, 55
Offset: 1

Keywords

Comments

A Legendrian n-mosaic is an n X n array of the 10 tiles given in Figure 5 of Pezzimenti and Pandey. These tiles represent part of a Legendrian curve in the front projection.
By Theorem 1.5 of Eliashberg and Fraser, two Legendrian unknots are equivalent if and only if they share the same Thurston-Bennequin invariant and rotation number.

Examples

			For n = 3 there are exactly a(3) = 4 distinct Legendrian unknots that can be realized on a Legendrian 3-mosaic, namely those whose Thurston-Bennequin invariants are -1, -2, -3, and -3 and whose rotation numbers are 0, 1, 0, and 2, respectively.
		

A374945 a(n) is the number of knots having a Legendrian representative realizable on a Legendrian n-mosaic.

Original entry on oeis.org

0, 1, 1, 1, 2, 11
Offset: 1

Keywords

Comments

A Legendrian n-mosaic is an n X n array of the 10 tiles given in Figure 5 of Pezzimenti and Pandey. These tiles represent part of a Legendrian curve in the front projection.
Two knots have the same smooth knot type if and only if they are related by an ambient isotopy.

Examples

			For n = 2, there is only a(2) = 1 smooth knot family with Legendrian representatives realizable on a Legendrian 2-mosaic, namely unknots.
For n = 5, every Legendrian 5-mosaic depicts either an unknot or a trefoil. Since unknots and trefoils are not ambient-isotopic, we have a(5) = 2.
		

A374942 T(|tb|,r) is the mosaic number of the Legendrian unknot, read by rows of the mountain range organized by Thurston-Bennequin number and rotation number, where 1-|tb|<=r<=|tb|-1.

Original entry on oeis.org

2, 3, 3, 3, 3, 3, 5, 4, 4, 5, 6, 4, 4, 4, 6, 6, 5, 4, 4, 5, 6, 6, 6, 5, 4, 5, 6, 6, 7, 6, 5, 5, 5, 5, 6, 7, 7, 6, 6, 5, 5, 5, 6, 6, 7
Offset: 1

Keywords

Comments

A Legendrian n-mosaic is an n X n array of the 10 tiles given in Figure 5 of Pezzimenti and Pandey. These tiles represent part of a Legendrian curve in the front projection.
The mosaic number of a Legendrian knot L is the smallest integer n such that L is realizable on a Legendrian n-mosaic.
Note that the Thurston-Bennequin number of a Legendrian unknot is always negative, so we take the absolute value in this sequence.
For more entries (but with incomplete rows), see Figure C.1 of Kipe et al. - Luc Ta, Oct 27 2024

Examples

			T(1,0)=2 because the mosaic number of the Legendrian unknot with tb=-1 and r=0 is 2. T(3,-2)=3 because the mosaic number of the Legendrian unknot with tb=-3 and r=-2 is 3.
		

A374939 a(n) is the number of distinct Legendrian knots, up to smooth knot type and classical invariants, with nonnegative rotation numbers that can be realized on a Legendrian n-mosaic.

Original entry on oeis.org

0, 1, 4, 9, 40, 328
Offset: 1

Keywords

Comments

A Legendrian n-mosaic is an n X n array of the 10 tiles given in Figure 5 of Pezzimenti and Pandey. These tiles represent part of a Legendrian curve in the front projection.
The classical invariants of Legendrian knots are the Thurston-Bennequin invariant and the rotation number.

Examples

			For n = 3 there are exactly a(3) = 4 distinct Legendrian knots with nonnegative rotation numbers that can be realized on a Legendrian 3-mosaic, namely the four Legendrian unknots whose Thurston-Bennequin invariants are -1, -2, -3, and -3 and whose rotation numbers are 0, 1, 0, and 2, respectively.
		

A374947 a(n) is the number of suitably connected Legendrian n-Mosaics.

Original entry on oeis.org

1, 2, 20, 1504, 948032, 5204262912, 254112496082944, 111879597850371293184, 448381477417976615986528256, 16469260582635747355818375736459264, 5571666891811926168753521842383673521864704, 17424018517043252553551626372130243982114254609186816
Offset: 1

Keywords

Comments

A Legendrian n-mosaic is an n X n array of the 10 tiles given in Figure 5 of Pezzimenti and Pandey. These tiles represent part of a Legendrian curve in the front projection.
A Legendrian n-mosaic is suitably connected iff the connection points of each tile coincide with those of all contiguous tiles. Note that the n-mosaic consisting of all blank tiles is vacuously suitably connected even though it does not represent a link.
This is the main diagonal of A375354. It appears to grow at a quadratic exponential rate, and the ratios a(n)/A261400(n) seem to converge to 0 at a quadratic exponential rate.
For more information, see Sections 4 and 5 of Kipe et al. In particular, see Figures 20 and 21 for explicit best-fit models. - Luc Ta, Oct 27 2024

Examples

			For n = 2 there are exactly a(2) = 2 suitably connected Legendrian 2-mosaics, namely the empty mosaic and the Legendrian unknot with maximal Thurston-Bennequin invariant.
		

Programs

  • Mathematica
    x[0] = o[0] = {{1}};
    x[n_] := ArrayFlatten[{{x[n - 1], o[n - 1]}, {o[n - 1], x[n - 1]}}];
    o[n_] := ArrayFlatten[{{o[n - 1], x[n - 1]}, {x[n - 1], 3*o[n - 1]}}];
    legendrianSquare[n_] := If[n > 1, 2*Total[MatrixPower[x[n - 2] + o[n - 2], n - 2], 2], 1];
    Flatten[ParallelTable[legendrianSquare[n], {n, 1, 11}]] (* This program is adapted from Theorem 1 of Oh, Hong, Lee, and Lee (see Links, cf. A375354). - Luc Ta, Aug 20 2024 *)
  • Rust
    // See Margaret Kipe link

Extensions

a(7)-a(11) from Luc Ta, Aug 20 2024
a(12) from Alois P. Heinz, Aug 20 2024

A364237 a(n) is the number of non-equivalent permutations of {1,2,...,2n-1} such that no subset of consecutive terms from the permutation sums to 0 modulo 2n, where two permutations are equivalent if one can be obtained from the other by multiplying every entry with an integer relatively prime to 2n and/or reversing the permutation.

Original entry on oeis.org

1, 1, 2, 4, 42, 504, 7492, 172480, 8639632
Offset: 1

Keywords

Comments

If we consider all permutations of {1,2,...,2n-1} such that no subset of consecutive terms from the permutation sums to 0 modulo 2n, then the number of such permutations is given by the number of constructive orderings mentioned in A141599. For example, given the permutation 14325 that satisfies the given conditions, observe that the partial sums modulo 6, namely 1=1, 1+4=5, 1+4+3=2, 1+4+3+2=4, and 1+4+3+2+5=3, are distinct.

Examples

			When n=3, there are four permutations of {1,2,3,4,5} such that no subset of consecutive terms from the permutation sums to 0 modulo 6, namely 14325, 25314, 41352, and 52341. Note that 14325 and 52341 are equivalent by reversing the permutations. Furthermore multiplication by 5 on every entry also yields the same equivalence. Additionally, 25314 and 41352 are analogously equivalent. Hence a(3)=2.
When n=4, 6142573 and 3752416 are equivalent by reversing the permutations but not by multiplying any integer relatively prime to 8, whereas 6142573 and 2346751 are equivalent by multiplication of 3 on every entry.
		

Crossrefs

Cf. A141599.

Programs

  • SageMath
    n = 3 #the index for the sequence a(n)
    orbits = {} #dictionary of permutations that are consecutive zero-sum-free
    seen = [] #list of seen permutations that are consecutive zero-sum-free
    a = 0 #the value of a(n)
    for labeling in Permutations(range(1,2*n)):
        if labeling not in seen:
            sums = [labeling[0]]
            for i in range(1,2*n-1):
                nextsum = (labeling[i] + sums[i-1]) % (2*n)
                if any([nextsum == 0, nextsum in sums]):
                    break
                sums.append(nextsum)
            if len(sums) == (2*n)-1:
                a += 1
                orbits[a] = []
                for m in [x for x in range(1,2*n) if gcd(x,2*n) == 1]:
                    equiv = [(m*labeling[i]) % (2*n) for i in range(2*n-1)]
                    if equiv not in orbits[a]:
                        orbits[a].append(equiv)
                    seen.append(equiv)
                    equiv = [equiv[2*n-2-i] for i in range(2*n-1)]
                    if equiv not in orbits[a]:
                        orbits[a].append(equiv)
                    seen.append(equiv)
    print(f"a({n}) = {a}\n")
    print("Equivalencies:")
    for i in range(1,a+1):
        print(f"{i}.")
        for x in orbits[i]:
            print(x)
        print('\n')

Extensions

a(8)-a(9) from Sean A. Irvine, Aug 15 2023

A361315 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;1,1) pebbling game.

Original entry on oeis.org

31, 26, 19, 17, 17, 19, 21, 23, 25, 27, 29, 31, 33, 35, 37, 39, 41
Offset: 4

Author

Gabrielle Demchak, Eugene Fiorini, Michael J. Herrera, Samuel Murray, Rhaldni Sayaman, Brittany Shelton and Wing Hong Tony Wong, Mar 14 2023

Keywords

Comments

A (3;1,1) move in an impartial two-player pebbling game consists of removing three pebbles from a vertex and adding a pebble to each of two distinct adjacent vertices. The winning player is the one who makes the final allowable move. We start at n = 4 because we have shown that a(3) does not exist while a(2) is clearly undefined.

Examples

			For n = 4, a(4) = 31 is the least number of pebbles for which every game 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.

Crossrefs

Programs

  • Mathematica
    (*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 (3;1,1)-pebbling move; only work for n>=3.*)
    pebblemoves[config_] :=  Block[{n, temp}, n = Length[config];   temp = Table[config, {i, n (n - 1) (n - 2)/2}] +     Permutations[Join[{-3, 1, 1}, Table[0, {i, n - 3}]]];   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]]],        Plist[[n, i - 1]]]],      AppendTo[Plist[[n, i]], tuples[[j]]]], {j, Length[tuples]}], {i,      Length[Plist[[n]]] + 1, m}]; Plist[[n, m]]];
    (*Given n, print out the minimum m such that there are no P-games with m pebbles*)Do[m = 1; While[plist[n, m] != {}, m++];
     Print["n=", n, " m=", m], {n, 4, 20}]