Yu Hin Au - cp's OEIS frontend

User: Yu Hin Au

Yu Hin Au has authored 4 sequences.

A336529 a(n) = (n^3+5n+3)/3 + 2floor(n/2) + a(n-2), with a(0)=1 and a(1)=3.

1, 3, 10, 20, 43, 75, 132, 208, 325, 475, 686, 948, 1295, 1715, 2248, 2880, 3657, 4563, 5650, 6900, 8371, 10043, 11980, 14160, 16653, 19435, 22582, 26068, 29975, 34275, 39056, 44288, 50065, 56355, 63258, 70740, 78907, 87723, 97300, 107600, 118741, 130683, 143550, 157300

Offset: 0

Yu Hin Au, Jul 24 2020

Let S be a fixed matching of size 2 in a complete n-uniform hypergraph G with >= 4n vertices. Given T,T' (each also a matching of size 2), define the equivalence relation where T ~ T' if and only if there exists an automorphism of G that maps every hyperedge in T to a hyperedge in T' while mapping every hyperedge in S to a hyperedge in S. Then the number of equivalence classes is a(n).

a(n) is the number of equivalence classes of 2 X 2 matrices with nonnegative integer entries where each row and column sum to at most n, such that two matrices are related if one can be obtained from the other by permuting rows and columns.

			To see a(2)=10, let S = {{1,2},{3,4}}. Then a representative from each of the 10 equivalence classes are
  1. {{1,2}, {3,4}}
  2. {{1,3}, {2,4}}
  3. {{1,5}, {3,4}}
  4. {{1,3}, {4,5}}
  5. {{1,2}, {5,6}}
  6. {{1,3}, {5,6}}
  7. {{1,5}, {2,6}}
  8. {{1,5}, {3,6}}
  9. {{1,5}, {6,7}}
  10. {{5,6}, {7,8}}
Likewise, in the 2 X 2 matrix interpretation, a representative from each of the a(2)=10 equivalence classes are
  [2 0 ; 0 2]
  [1 1 ; 1 1]
  [2 0 ; 0 1]
  [1 1 ; 1 0]
  [2 0 ; 0 0]
  [1 1 ; 0 0]
  [1 0 ; 1 0]
  [1 0 ; 0 1]
  [1 0 ; 0 0]
  [0 0 ; 0 0]

Michael De Vlieger, Table of n, a(n) for n = 0..10000
Yu Hin Au, Nathan Lindzey, and Levent Tunçel, Matchings, hypergraphs, association schemes, and semidefinite optimization, arXiv:2008.08628 [math.CO], 2020.
Index entries for linear recurrences with constant coefficients, signature (3,-1,-5,5,1,-3,1).

Cf. A316587.

Nest[Append[#1, (#2^3 + 5 #2 + 3)/3 + 2*Floor[#2/2] + #1[[-2]] ] & @@ {#, Length@ #} &, {1, 3}, 42] (* Michael De Vlieger, Nov 04 2020 *)
LinearRecurrence[{3,-1,-5,5,1,-3,1},{1,3,10,20,43,75,132},60] (* Harvey P. Dale, May 28 2021 *)

Vec((1 + 2*x^2 - 2*x^3 + 3*x^4) / ((1 - x)^5*(1 + x)^2) + O(x^40)) \\ Colin Barker, Nov 05 2020

a(n) = 3*a(n-1) - a(n-2) - 5*a(n-3) + 5*a(n-4) + a(n-5) - 3*a(n-6) + a(n-7). - Wesley Ivan Hurt, Nov 04 2020

G.f.: (1 + 2*x^2 - 2*x^3 + 3*x^4) / ((1 - x)^5*(1 + x)^2). - Colin Barker, Nov 05 2020

A322543 Number of triadic partitions of the unit square into (2n+1) subrectangles.

Original entry on oeis.org

1, 2, 12, 96, 879, 8712, 90972, 985728, 10979577, 124937892, 1446119664, 16972881120, 201526230555, 2416309004872, 29215072931136, 355800894005760, 4360705642282569, 53744080256387478, 665667989498682936, 8281518339078928800, 103441301833577854041, 1296713265300164761632

Offset: 0

Yu Hin Au, Dec 14 2018

nonn

A kind of two-dimensional ternary Catalan number. This sequence enumerates the decompositions of the unit square into 2n+1 rectangles obtained by the following algorithm.
(a) Start with the unit square.
(b) Perform the following operation n times:
    (1) Choose a rectangle in the current decomposition.
    (2) Trisect this rectangle into three rectangles horizontally or vertically.
Note that different sequences of trisections can produce the same decomposition.

Alois P. Heinz, Table of n, a(n) for n = 0..889
Yu Hin (Gary) Au, Fatemeh Bagherzadeh, Murray R. Bremner, Enumeration and Asymptotic Formulas for Rectangular Partitions of the Hypercube, arXiv:1903.00813 [math.CO], Mar 03 2019.

Cf. A000108 (Catalan numbers), A005408, A236339 (decompositions of unit square using bisections).

a:= n-> coeff(series(RootOf(G^9-2*G^3+G-x, G), x, 2*n+2), x, 2*n+1):
seq(a(n), n=0..25);  # Alois P. Heinz, Dec 14 2018

a[n_] := SeriesCoefficient[InverseSeries[x - 2 x^3 + x^9 + O[x]^(2n+2), x], {x, 0, 2n+1}];
Table[a[n], {n, 0, 21}] (* Jean-François Alcover, Aug 13 2019, from PARI *)

a(n)={polcoef(serreverse(x - 2*x^3 + x^9 + O(x^(2*n+2))), 2*n+1)} \\ Andrew Howroyd, Dec 14 2018

Recurrence relation: a(n) = C(2n+1) with C(1) = 1 and C(n) = 2 Sum_{i1,i2,i3} C(i1)C(i2)C(i3) - Sum_{i1,i2,i3,i4,i5,i6,i7,i8,i9} C(i1)C(i2)C(i3)C(i4)C(i5)C(i6)C(i7)C(i8)C(i9). The first sum is over all 3-compositions of n into positive integers (i1+i2+i3=n), and the second sum is over all 9-compositions of n into positive integers (i1+i2+...+i9=n).

a(n) = [x^(2n+1)] G(x), where G(x) satisfies: G(x)^9 - 2*G(x)^3 + G(x) - x = 0.

A316587 a(n) = [x^(2n)y^n] Product_{i>=1} 1/((1-x^(2i-1)y^i)(1-x^(2i-1)y^(i-1))(1-x^(2i)y^i)^2).

Original entry on oeis.org

1, 3, 10, 27, 69, 161, 361, 767, 1578, 3134, 6064, 11432, 21105, 38175, 67863, 118658, 204455, 347439, 583063, 966952, 1586231, 2575474, 4141832, 6600731, 10430455, 16349788, 25434178, 39280676, 60250276, 91810915, 139034070, 209294256, 313269591, 466343647

Offset: 0

Yu Hin Au, Aug 31 2018

nonn

Let S be a fixed matching of size n in a complete graph G with >= 4n vertices. Given T,T' (also matchings of size n), define the equivalence relation where T ~ T' if and only if there exists an automorphism of G that maps edges in T to edges in T' while mapping edges in S to edges in S. Then the number of equivalence classes is a(n).

a(n) is the number of partitions of 2n with 4 kinds of parts (types 1,2,3,4) where (i) all parts of types 1,2 are odd and all parts of types 3,4 are even; and (ii) the number of type 1 and type 2 parts are equal.

			To see a(2)=10, let S = {{1,2},{3,4}}. Then a representative from each of the 10 equivalence classes are
  1. {{1,2}, {3,4}}
  2. {{1,3}, {2,4}}
  3. {{1,5}, {3,4}}
  4. {{1,3}, {4,5}}
  5. {{1,2}, {5,6}}
  6. {{1,3}, {5,6}}
  7. {{1,5}, {2,6}}
  8. {{1,5}, {3,6}}
  9. {{1,5}, {6,7}}
  10. {{5,6}, {7,8}}

Yu Hin Au, Nathan Lindzey, and Levent Tunçel, Matchings, hypergraphs, association schemes, and semidefinite optimization, arXiv:2008.08628 [math.CO], 2020.

If the equivalence relation is defined as T~T' if and only if there exists an automorphism of G mapping union of S,T to union of S,T' (i.e., the map does not necessarily fix edges in S), then we obtain A305168.

A305168 Number of non-isomorphic graphs on 4n vertices whose edges are the union of two n-edge matchings.

Original entry on oeis.org

1, 3, 9, 23, 54, 118, 246, 489, 940, 1751, 3177, 5630, 9776, 16659, 27922, 46092, 75039, 120615, 191611, 301086, 468342, 721638, 1102113, 1669226, 2508429, 3741741, 5542532, 8155720, 11925654, 17334077, 25051940, 36009468, 51491111, 73263043, 103744575

Offset: 0

Yu Hin Au, Aug 17 2018

nonn

a(n) is also the number of partitions of 2n with two kinds of parts where all parts of the second kind are even. E.g., the a(2) = 9 such partitions are (2', 2'), (4'), (2,2'), (4), (1,1,2'), (3,1), (2,2), (2,1,1), (1,1,1,1). A bijection is to take each component in the graph whose edges are the union of two n-edge matchings, map each path of length p to a part p and each cycle (which must be even) of length p to a part p'.

			To see a(2)=9, observe that all graphs that are the union of two matchings of size n=2 are isomorphic to the union of S = {{1,2},{3,4}} and one of T=
  1. {{1,2}, {3,4}} --> (2',2')
  2. {{1,3}, {2,4}} --> (4')
  3. {{1,5}, {3,4}} --> (2,2')
  4. {{1,3}, {4,5}} --> (4)
  5. {{1,2}, {5,6}} --> (1,1,2')
  6. {{1,3}, {5,6}} --> (3,1)
  7. {{1,5}, {3,6}} --> (2,2)
  8. {{1,5}, {6,7}} --> (2,1,1)
  9. {{5,6}, {7,8}} --> (1,1,1,1)
Note that the partitions correspond to the bijection mentioned in the comments above.

Bisection (even part) of A002513.

Cf. A000041.

b:= proc(n) option remember; `if`(n=0, 1, add(b(n-j)*add(d*
      (2-irem(d, 2)), d=numtheory[divisors](j)), j=1..n)/n)
    end:
a:= n-> b(2*n):
seq(a(n), n=0..40);  # Alois P. Heinz, Aug 18 2018

a[n_] := Sum[PartitionsP[2k] PartitionsP[n-k], {k, 0, n}];
a /@ Range[0, 40] (* Jean-François Alcover, Nov 27 2020 *)

a(n) = sum(i=0, n, numbpart(2*i)*numbpart(n-i)); \\ Michel Marcus, Aug 18 2018

a(n) = [x^2n] (Product_{i>=1} 1/(1-x^i))*(Product_{j>=1} 1/(1-x^(2j))).
a(n) = Sum_{i=0..n} b(2i)*b(n-i) where b(n) is the number of partitions of n (A000041).
a(n) = A002513(2n). - Alois P. Heinz, Aug 18 2018

cp's OEIS Frontend

User: Yu Hin Au

A336529 a(n) = (n^3+5*n+3)/3 + 2*floor(n/2) + a(n-2), with a(0)=1 and a(1)=3.

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A322543 Number of triadic partitions of the unit square into (2n+1) subrectangles.

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A316587 a(n) = [x^(2n)y^n] Product_{i>=1} 1/((1-x^(2i-1)y^i)(1-x^(2i-1)y^(i-1))(1-x^(2i)y^i)^2).

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A305168 Number of non-isomorphic graphs on 4n vertices whose edges are the union of two n-edge matchings.

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A336529 a(n) = (n^3+5n+3)/3 + 2floor(n/2) + a(n-2), with a(0)=1 and a(1)=3.