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.

Showing 1-8 of 8 results.

A325007 Array read by descending antidiagonals: A(n,k) is the number of achiral colorings of the facets of a regular n-dimensional orthotope using up to k colors.

Original entry on oeis.org

1, 2, 1, 3, 6, 1, 4, 18, 10, 1, 5, 40, 55, 15, 1, 6, 75, 200, 126, 21, 1, 7, 126, 560, 700, 252, 28, 1, 8, 196, 1316, 2850, 1996, 462, 36, 1, 9, 288, 2730, 9261, 11376, 5004, 792, 45, 1, 10, 405, 5160, 25480, 50127, 38550, 11440, 1287, 55, 1, 11, 550, 9075, 61776, 181027, 225225, 116160, 24310, 2002, 66, 1
Offset: 1

Views

Author

Robert A. Russell, May 27 2019

Keywords

Comments

Also called hypercube, n-dimensional cube, and measure polytope. For n=1, the figure is a line segment with two vertices. For n=2 the figure is a square with four edges. For n=3 the figure is a cube with six square faces. For n=4, the figure is a tesseract with eight cubic facets. The Schläfli symbol, {4,3,...,3}, of the regular n-dimensional orthotope (n>1) consists of a four followed by n-2 threes. Each of its 2n facets is an (n-1)-dimensional orthotope. An achiral coloring is identical to its reflection.
Also the number of achiral colorings of the vertices of a regular n-dimensional orthoplex using up to k colors.

Examples

			Array begins with A(1,1):
1  2   3     4      5      6       7        8         9        10 ...
1  6  18    40     75    126     196      288       405       550 ...
1 10  55   200    560   1316    2730     5160      9075     15070 ...
1 15 126   700   2850   9261   25480    61776    135675    275275 ...
1 21 252  1996  11376  50127  181027   559728   1529892   3784627 ...
1 28 462  5004  38550 225225 1053304  4119648  13942908  41918800 ...
1 36 792 11440 116160 881595 5263336 25794288 107427420 390891160 ...
For a(2,2)=6, all colorings are achiral: two with just one of the colors, two with one color on just one edge, one with opposite colors the same, and one with opposite colors different.
		

Crossrefs

Cf. A325004 (oriented), A325005 (unoriented), A325006 (chiral), A325011 (exactly k colors).
Other n-dimensional polytopes: A325001 (simplex), A325015 (orthoplex).
Rows 1-2 are A000027, A002411; column 2 is A186783(n+2).

Programs

  • Mathematica
    Table[Binomial[Binomial[d-n+2,2]+n-1,n]-Binomial[Binomial[d-n+1,2],n],{d,1,11},{n,1,d}] // Flatten
  • PARI
    a(n, k) = binomial(binomial(k+1, 2)+n-1, n) - binomial(binomial(k, 2), n)
    array(rows, cols) = for(x=1, rows, for(y=1, cols, print1(a(x, y), ", ")); print(""))
    /* Print initial 6 rows and 8 columns of array as follows: */
    array(6, 8) \\ Felix Fröhlich, May 30 2019

Formula

A(n,k) = binomial(binomial(k+1,2) + n-1, n) - binomial(binomial(k,2),n).
A(n,k) = Sum_{j=1..2*n} A325011(n,j) * binomial(k,j).
A(n,k) = 2*A325005(n,k) - A325004(n,k) = (A325004(n,k) - 2*A325006(n,k)) / 2 = A325005(n,k) + A325006(n,k).
G.f. for row n: Sum{j=1..2*n} A325011(n,j) * x^j / (1-x)^(j+1).
Linear recurrence for row n: T(n,k) = Sum_{j=0..2*n} binomial(-2-j,2*n-j) * T(n,k-1-j).
G.f. for column k: 1/(1-x)^binomial(k+1,2) - (1+x)^binomial(k,2).

A325008 Triangle read by rows: T(n,k) is the number of oriented colorings of the facets of a regular n-dimensional orthotope using exactly k colors. Row n has 2n columns.

Original entry on oeis.org

1, 2, 1, 4, 9, 6, 1, 8, 30, 68, 75, 30, 1, 13, 84, 312, 735, 1020, 735, 210, 1, 19, 192, 1122, 4155, 10242, 16380, 15960, 8505, 1890, 1, 26, 381, 3322, 18285, 67679, 173936, 308056, 363825, 270900, 114345, 20790, 1, 34, 687, 8484, 66765, 352359, 1305612, 3479268, 6668865, 9035460, 8378370, 5031180, 1756755, 270270
Offset: 1

Views

Author

Robert A. Russell, May 27 2019

Keywords

Comments

Also called hypercube, n-dimensional cube, and measure polytope. For n=1, the figure is a line segment with two vertices. For n=2 the figure is a square with four edges. For n=3 the figure is a cube with six square faces. For n=4, the figure is a tesseract with eight cubic facets. The Schläfli symbol, {4,3,...,3}, of the regular n-dimensional orthotope (n>1) consists of a four followed by n-2 threes. Each of its 2n facets is an (n-1)-dimensional orthotope. Two oriented colorings are the same if one is a rotation of the other; chiral pairs are counted as two.
Also the number of oriented colorings of the vertices of a regular n-dimensional orthoplex using exactly k colors.

Examples

			Triangle begins with T(1,1):
 1  2
 1  4   9    6
 1  8  30   68    75    30
 1 13  84  312   735  1020    735    210
 1 19 192 1122  4155 10242  16380  15960   8505   1890
 1 26 381 3322 18285 67679 173936 308056 363825 270900 114345 20790
For T(2,2)=4, there are two squares with just one edge for one color, one square with opposite edges the same color, and one square with opposite edges different colors.
		

Crossrefs

Cf. A325009 (unoriented), A325010 (chiral), A325011 (achiral), A325004 (up to k colors).
Other n-dimensional polytopes: A325002 (simplex), A325016 (orthoplex).

Programs

  • Mathematica
    Table[Sum[Binomial[-j-2,k-j-1] Binomial[n + Binomial[j+2,2]-1, n], {j,0,k-1}] + Sum[Binomial[j-k-1,j] Binomial[Binomial[k-j,2],n],{j,0,k-2}], {n,1,10},{k,1,2n}] // Flatten

Formula

T(n,k) = Sum_{j=0..k-1} binomial(-j-2,k-j-1) * binomial(n + binomial(j+2,2)-1, n) + Sum_{j=0..k-2} binomial(j-k-1,j) * binomial(binomial(k-j,2),n).
T(n,k) = A325009(n,k) + A325010(n,k) = (A325009(n,k) + A325011(n,k)) / 2 = 2*A325010(n,k) + A325011(n,k).

A325009 Triangle read by rows: T(n,k) is the number of unoriented colorings of the facets of a regular n-dimensional orthotope using exactly k colors. Row n has 2n columns.

Original entry on oeis.org

1, 1, 1, 4, 6, 3, 1, 8, 29, 52, 45, 15, 1, 13, 84, 297, 600, 690, 420, 105, 1, 19, 192, 1116, 3933, 8661, 11970, 10080, 4725, 945, 1, 26, 381, 3321, 18080, 63919, 150332, 236978, 247275, 163800, 62370, 10395, 1, 34, 687, 8484, 66645, 346644, 1231857, 3052008, 5316885, 6483330, 5415795, 2952180, 945945, 135135
Offset: 1

Views

Author

Robert A. Russell, May 27 2019

Keywords

Comments

Also called hypercube, n-dimensional cube, and measure polytope. For n=1, the figure is a line segment with two vertices. For n=2 the figure is a square with four edges. For n=3 the figure is a cube with six square faces. For n=4, the figure is a tesseract with eight cubic facets. The Schläfli symbol, {4,3,...,3}, of the regular n-dimensional orthotope (n>1) consists of a four followed by n-2 threes. Each of its 2n facets is an (n-1)-dimensional orthotope. Two unoriented colorings are the same if congruent; chiral pairs are counted as one.
Also the number of unoriented colorings of the vertices of a regular n-dimensional orthoplex using exactly k colors.

Examples

			The triangle begins with T(1,1):
1  1
1  4   6    3
1  8  29   52    45    15
1 13  84  297   600   690    420    105
1 19 192 1116  3933  8661  11970  10080   4725    945
1 26 381 3321 18080 63919 150332 236978 247275 163800 62370 10395
For T(2,2)=4, there are two squares with just one edge for one color, one square with opposite edges the same color, and one square with opposite edges different colors.
		

Crossrefs

Cf. A325008 (oriented), A325010 (chiral), A325011 (achiral), A325005 (up to k colors).
Other n-dimensional polytopes: A007318(n,k-1) (simplex), A325017 (orthoplex).

Programs

  • Mathematica
    Table[Sum[Binomial[-j-2,k-j-1]Binomial[n+Binomial[j+2,2]-1,n],{j,0,k-1}],{n,1,10},{k,1,2n}] // Flatten

Formula

T(n,k) = Sum{j=0..k-1} binomial(-j-2, k-j-1) * binomial(n+binomial(j+2, 2)-1, n).
T(n,k) = A325009(n,k) + A325010(n,k) = 2*A325009(n,k) - A325011(n,k) = 2*A325010(n,k) + A325011(n,k).

A325019 Triangle read by rows: T(n,k) is the number of achiral colorings of the facets of a regular n-dimensional orthoplex using exactly k colors. Row n has 2^n columns.

Original entry on oeis.org

1, 0, 1, 4, 3, 0, 1, 19, 141, 394, 450, 180, 0, 0, 1, 306, 33207, 921908, 10359075, 59584470, 197644440, 400752240, 505197000, 386694000, 164656800, 29937600, 0, 0, 0, 0
Offset: 1

Views

Author

Robert A. Russell, Jun 09 2019

Keywords

Comments

Also called cross polytope and hyperoctahedron. For n=1, the figure is a line segment with two vertices. For n=2 the figure is a square with four edges. For n=3 the figure is an octahedron with eight triangular faces. For n=4, the figure is a 16-cell with sixteen tetrahedral facets. The Schläfli symbol, {3,...,3,4}, of the regular n-dimensional orthoplex (n>1) consists of n-2 threes followed by a four. Each of its 2^n facets is an (n-1)-dimensional simplex. An achiral coloring is identical to its reflection. The last 2^(n-2) columns of row n are zero; there are no achiral colorings with that many colors.
Also the number of achiral colorings of the vertices of a regular n-dimensional orthotope (cube) using exactly k colors.

Examples

			Triangle begins with T(1,1):
1  0
1  4   3   0
1 19 141 394 450 180 0 0
For T(2,3)=3, each square has one of the three colors on two opposite edges.
		

Crossrefs

Cf. A325016 (oriented), A325017 (unoriented), A325018 (chiral), A325015 (up to k colors).
Other n-dimensional polytopes: A325003 (simplex), A325011 (orthotope).

Programs

  • Mathematica
    a48[n_] := a48[n] = DivisorSum[NestWhile[#/2&,n,EvenQ],MoebiusMu[#]2^(n/#)&]/(2n); (* A000048 *)
    a37[n_] := a37[n] = DivisorSum[n,MoebiusMu[n/#]2^#&]/n; (* A001037 *)
    CI0[{n_Integer}] := CI0[{n}] = CI[Transpose[If[EvenQ[n], p2 = IntegerExponent[n, 2]; sub = Divisors[n/2^p2]; {2^(p2+1) sub, a48 /@ (2^p2 sub) }, sub = Divisors[n]; {sub, a37 /@ sub}]]] 2^(n-1); (* even permutation *)
    CI1[{n_Integer}] := CI1[{{n}}] = CI[sub = Divisors[n]; Transpose[If[EvenQ[n], {sub, a37 /@ sub}, {2 sub, (a37 /@ sub)/2}]]] 2^(n-1); (* odd permutation *)
    compress[x : {{, } ...}] := (s = Sort[x]; For[i = Length[s], i > 1, i -= 1, If[s[[i, 1]] == s[[i - 1, 1]], s[[i - 1, 2]] += s[[i, 2]]; s = Delete[s, i], Null]]; s)
    cix[{a_, b_}, {c_, d_}] := {LCM[a, c], (a b c d)/LCM[a, c]};
    Unprotect[Times]; Times[CI[a_List], CI[b_List]] :=  (* combine *) CI[compress[Flatten[Outer[cix, a, b, 1], 1]]]; Protect[Times];
    CI0[p_List] := CI0[p] = Expand[CI0[Drop[p, -1]] CI0[{Last[p]}] + CI1[Drop[p, -1]] CI1[{Last[p]}]]
    CI1[p_List] := CI1[p] = Expand[CI0[Drop[p, -1]] CI1[{Last[p]}] + CI1[Drop[p, -1]] CI0[{Last[p]}]]
    pc[p_List] := Module[{ci,mb},mb = DeleteDuplicates[p]; ci = Count[p, #] & /@ mb; n!/(Times @@ (ci!) Times @@ (mb^ci))] (* partition count *)
    row[n_Integer] := row[n] = Factor[(Total[(CI1[#] pc[#]) & /@ IntegerPartitions[n]])/(n! 2^(n - 1))] /. CI[l_List] :> j^(Total[l][[2]])
    array[n_, k_] := row[n] /. j -> k (* A325012 *)
    Table[LinearSolve[Table[Binomial[i,j],{i,1,2^n},{j,1,2^n}],Table[array[n,k],{k,1,2^n}]],{n,1,6}] // Flatten

Formula

A325015(n,k) = Sum_{j=1..2^n} T(n,j) * binomial(k,j).
T(n,k) = 2*A325017(n,k) - A325016(n,k) = A325016(n,k) - 2*A325018(n,k) = A325017(n,k) - A325018(n,k).

A325003 Triangle read by rows: T(n,k) is the number of achiral colorings of the facets (or vertices) of a regular n-dimensional simplex using exactly k colors.

Original entry on oeis.org

1, 0, 1, 2, 0, 1, 3, 3, 0, 1, 4, 6, 4, 0, 1, 5, 10, 10, 5, 0, 1, 6, 15, 20, 15, 6, 0, 1, 7, 21, 35, 35, 21, 7, 0, 1, 8, 28, 56, 70, 56, 28, 8, 0, 1, 9, 36, 84, 126, 126, 84, 36, 9, 0, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 0, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 0
Offset: 1

Views

Author

Robert A. Russell, Mar 23 2019

Keywords

Comments

For n=1, the figure is a line segment with two vertices. For n=2, the figure is a triangle with three edges. For n=3, the figure is a tetrahedron with four triangular faces. The Schläfli symbol, {3,...,3}, of the regular n-dimensional simplex consists of n-1 threes. Each of its n+1 facets is a regular (n-1)-dimensional simplex. An achiral coloring is the same as its reflection. For k <= n all the colorings are achiral.
The final zero in each row indicates no achiral colorings when each facet has a different color.

Examples

			Triangle begins with T(1,1):
1  0
1  2   0
1  3   3   0
1  4   6   4    0
1  5  10  10    5    0
1  6  15  20   15    6    0
1  7  21  35   35   21    7    0
1  8  28  56   70   56   28    8    0
1  9  36  84  126  126   84   36    9    0
1 10  45 120  210  252  210  120   45   10    0
1 11  55 165  330  462  462  330  165   55   11    0
1 12  66 220  495  792  924  792  495  220   66   12   0
1 13  78 286  715 1287 1716 1716 1287  715  286   78  13   0
1 14  91 364 1001 2002 3003 3432 3003 2002 1001  364  91  14  0
1 15 105 455 1365 3003 5005 6435 6435 5005 3003 1365 455 105 15 0
For T(3,2)=3, the tetrahedron may have one, two, or three faces of one color.
		

Crossrefs

Cf. A325002 (oriented), A007318(n,k-1) (unoriented), A325001 (up to k colors).
Other n-dimensional polytopes: A325011 (orthotope), A325019 (orthoplex).
Cf. A198321.

Programs

  • Mathematica
    Table[Binomial[n, k-1] - Boole[k==n+1], {n,1,15}, {k,1,n+1}] \\ Flatten

Formula

T(n,k) = binomial(n,k-1) - [k==n+1] = A007318(n,k-1) - [k==n+1].
T(n,k) = A325002(n,k) - 2*[k==n+1] = 2*A007318(n,k-1) - A325002(n,k).
G.f. for row n: x * (1+x)^n - x^(n+1).
G.f. for column k>1: x^(k-1)/(1-x)^k - x^(k-1).

A325010 Triangle read by rows: T(n,k) is the number of chiral pairs of colorings of the facets of a regular n-dimensional orthotope using exactly k colors. Row n has 2n columns.

Original entry on oeis.org

0, 1, 0, 0, 3, 3, 0, 0, 1, 16, 30, 15, 0, 0, 0, 15, 135, 330, 315, 105, 0, 0, 0, 6, 222, 1581, 4410, 5880, 3780, 945, 0, 0, 0, 1, 205, 3760, 23604, 71078, 116550, 107100, 51975, 10395, 0, 0, 0, 0, 120, 5715, 73755, 427260, 1351980, 2552130, 2962575, 2079000, 810810, 135135
Offset: 1

Views

Author

Robert A. Russell, May 27 2019

Keywords

Comments

Also called hypercube, n-dimensional cube, and measure polytope. For n=1, the figure is a line segment with two vertices. For n=2 the figure is a square with four edges. For n=3 the figure is a cube with six square faces. For n=4, the figure is a tesseract with eight cubic facets. The Schläfli symbol, {4,3,...,3}, of the regular n-dimensional orthotope (n>1) consists of a four followed by n-2 threes. Each of its 2n facets is an (n-1)-dimensional orthotope. The chiral colorings of its facets come in pairs, each the reflection of the other.
Also the number of chiral pairs of colorings of the vertices of a regular n-dimensional orthoplex using exactly k colors.

Examples

			The triangle begins with T(1,1):
 0 1
 0 0 3  3
 0 0 1 16  30   15
 0 0 0 15 135  330   315    105
 0 0 0  6 222 1581  4410   5880    3780     945
 0 0 0  1 205 3760 23604  71078  116550  107100   51975   10395
 0 0 0  0 120 5715 73755 427260 1351980 2552130 2962575 2079000 810810 135135
For T(2,3)=3, the three squares have the two edges with the same color adjacent.
		

Crossrefs

Cf. A325008 (oriented), A325009 (unoriented), A325011 (achiral), A325006 (up to k colors).
Other n-dimensional polytopes: A325018 (orthoplex).

Programs

  • Mathematica
    Table[Sum[Binomial[j-k-1,j]Binomial[Binomial[k-j,2],n],{j,0,k-2}],{n,1,10},{k,1,2n}] // Flatten

Formula

T(n,k) = Sum{j=0..k-2} binomial(j-k-1,j) * binomial(binomial(k-j,2),n).
T(n,k) = A325008(n,k) - A325009(n,k) = (A325008(n,k) - A325011(n,k)) / 2 = A325009(n,k) - A325011(n,k).

A338145 Triangle read by rows: T(n,k) is the number of achiral colorings of the edges of a regular n-D orthotope (or ridges of a regular n-D orthoplex) using exactly k colors. Row n has n*2^(n-1) columns.

Original entry on oeis.org

1, 1, 4, 3, 0, 1, 68, 1200, 7268, 20025, 27750, 18900, 5040, 0, 0, 0, 0, 1, 93022, 293878020, 90807857080, 7503022894800, 258528829444320, 4681671089961600, 50981530073846400, 363246007692204000
Offset: 1

Views

Author

Robert A. Russell, Oct 12 2020

Keywords

Comments

An achiral coloring is identical to its reflection. A ridge is an (n-2)-face of an n-D polytope. For n=1, the figure is a line segment with one edge. For n=2, the figure is a square with 4 edges (vertices). For n=3, the figure is a cube (octahedron) with 12 edges. The number of edges (ridges) is n*2^(n-1). The Schläfli symbols for the n-D orthotope (hypercube) and the n-D orthoplex (hyperoctahedron, cross polytope) are {4,...,3,3} and {3,3,...,4} respectively, with n-2 3's in each case. The figures are mutually dual.
The algorithm used in the Mathematica program below assigns each permutation of the axes to a partition of n and then considers separate conjugacy classes for axis reversals. It uses the formulas in Balasubramanian's paper. If the value of m is increased, one can enumerate colorings of higher-dimensional elements beginning with T(m,1).

Examples

			Triangle begins with T(1,1):
  1
  1  4    3    0
  1 68 1200 7268 20025 27750 18900 5040 0 0 0 0
  ...
For T(2,2)=4, the achiral colorings are AAAB, AABB, ABAB, and ABBB. For T(2,3)=3, the achiral colorings are ABAC, ABCB, and ACBC.
		

Crossrefs

Cf. A338142 (oriented), A338143 (unoriented), A338144 (chiral), A337410 (k or fewer colors), A325019 (orthotope vertices, orthoplex facets).
Cf. A327090 (simplex), A338149 (orthoplex edges, orthotope ridges).

Programs

  • Mathematica
    m=1; (* dimension of color element, here an edge *)
    Fi1[p1_] := Module[{g, h}, Coefficient[Product[g = GCD[k1, p1]; h = GCD[2 k1, p1]; (1 + 2 x^(k1/g))^(r1[[k1]] g) If[Divisible[k1, h], 1, (1+2x^(2 k1/h))^(r2[[k1]] h/2)], {k1, Flatten[Position[cs, n1_ /; n1 > 0]]}], x, n - m]];
    FiSum[] := (Do[Fi2[k2] = Fi1[k2], {k2, Divisors[per]}]; DivisorSum[per, DivisorSum[d1 = #, MoebiusMu[d1/#] Fi2[#] &]/# &]);
    CCPol[r_List] := (r1 = r; r2 = cs - r1; If[EvenQ[Sum[If[EvenQ[j3], r1[[j3]], r2[[j3]]], {j3, n}]], 0, (per = LCM @@ Table[If[cs[[j2]] == r1[[j2]], If[0 == cs[[j2]], 1, j2], 2j2], {j2, n}]; Times @@ Binomial[cs, r1] 2^(n-Total[cs]) b^FiSum[])]);
    PartPol[p_List] := (cs = Count[p, #]&/@ Range[n]; Total[CCPol[#]&/@ Tuples[Range[0, cs]]]);
    pc[p_List] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #]&/@ mb; n!/(Times@@(ci!) Times@@(mb^ci))] (*partition count*)
    row[n_Integer] := row[n] = Factor[(Total[(PartPol[#] pc[#])&/@ IntegerPartitions[n]])/(n! 2^(n-1))]
    array[n_, k_] := row[n] /. b -> k
    Table[LinearSolve[Table[Binomial[i,j],{i,2^(n-m)Binomial[n,m]},{j,2^(n-m)Binomial[n,m]}], Table[array[n,k],{k,2^(n-m)Binomial[n,m]}]], {n,m,m+4}] // Flatten

Formula

A337410(n,k) = Sum_{j=1..n*2^(n-1)} T(n,j) * binomial(k,j).
T(n,k) = 2*A338143(n,k) - A338142(n,k) = A338142(n,k) - 2*A338144(n,k) = A338143(n,k) - A338144(n,k).
T(2,k) = A338149(2,k) = A325019(2,k) = A325011(2,k); T(3,k) = A338149(3,k).

A338149 Triangle read by rows: T(n,k) is the number of achiral colorings of the edges of a regular n-D orthoplex (or ridges of a regular n-D orthotope) using exactly k colors. Row 1 has 1 column; row n>1 has 2*n*(n-1) columns.

Original entry on oeis.org

1, 1, 4, 3, 0, 1, 68, 1200, 7268, 20025, 27750, 18900, 5040, 0, 0, 0, 0, 1, 8198, 9055962, 1467050480, 74035775370, 1679679306420, 20864180531565, 159341117375160, 804216787965360, 2808560520334800, 6981656802951600
Offset: 1

Views

Author

Robert A. Russell, Oct 12 2020

Keywords

Comments

An achiral coloring is identical to its reflection. A ridge is an (n-2)-face of an n-D polytope. For n=1, the figure is a line segment with one edge. For n=2, the figure is a square with 4 edges (vertices). For n=3, the figure is an octahedron (cube) with 12 edges. For n>1, the number of edges (ridges) is 2*n*(n-1). The Schläfli symbols for the n-D orthotope (hypercube) and the n-D orthoplex (hyperoctahedron, cross polytope) are {4,3,...,3,3} and {3,3,...,3,4} respectively, with n-2 3's in each case. The figures are mutually dual.
The algorithm used in the Mathematica program below assigns each permutation of the axes to a partition of n and then considers separate conjugacy classes for axis reversals. It uses the formulas in Balasubramanian's paper. If the value of m is increased, one can enumerate colorings of higher-dimensional elements beginning with T(m,1).

Examples

			Triangle begins with T(1,1):
  1
  1    4       3          0
  1   68    1200       7268       20025         27750 18900 5040 0 0 0 0
  1 8198 9055962 1467050480 74035775370 1679679306420 ...
  ...
For T(2,2)=4, the achiral colorings are AAAB, AABB, ABAB, and ABBB. For T(2,3)=3, the 3 achiral colorings are ABAC, ABCB, and ACBC.
		

Crossrefs

Cf. A338146 (oriented), A338147 (unoriented), A338148 (chiral), A337414 (k or fewer colors), A325011 (orthoplex vertices, orthotope facets).
Cf. A327090 (simplex), A338145 (orthotope edges, orthoplex ridges).

Programs

  • Mathematica
    m=1; (* dimension of color element, here an edge *)
    Fi1[p1_] := Module[{g, h}, Coefficient[Product[g = GCD[k1, p1]; h = GCD[2 k1, p1]; (1 + 2 x^(k1/g))^(r1[[k1]] g) If[Divisible[k1, h], 1, (1+2x^(2 k1/h))^(r2[[k1]] h/2)], {k1, Flatten[Position[cs, n1_ /; n1 > 0]]}], x, m+1]];
    FiSum[] := (Do[Fi2[k2] = Fi1[k2], {k2, Divisors[per]}]; DivisorSum[per, DivisorSum[d1 = #, MoebiusMu[d1/#] Fi2[#] &]/# &]);
    CCPol[r_List] := (r1 = r; r2 = cs - r1; If[EvenQ[Sum[If[EvenQ[j3], r1[[j3]], r2[[j3]]], {j3, n}]], 0, (per = LCM @@ Table[If[cs[[j2]] == r1[[j2]], If[0 == cs[[j2]], 1, j2], 2j2], {j2, n}]; Times @@ Binomial[cs, r1] 2^(n-Total[cs]) b^FiSum[])]);
    PartPol[p_List] := (cs = Count[p, #]&/@ Range[n]; Total[CCPol[#]&/@ Tuples[Range[0, cs]]]);
    pc[p_List] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #]&/@ mb; n!/(Times@@(ci!) Times@@(mb^ci))] (*partition count*)
    row[m]=b; row[n_Integer] := row[n] = Factor[(Total[(PartPol[#] pc[#])&/@ IntegerPartitions[n]])/(n! 2^(n-1))]
    array[n_, k_] := row[n] /. b -> k
    Join[{{1}},Table[LinearSolve[Table[Binomial[i,j],{i,2^(m+1)Binomial[n,m+1]},{j,2^(m+1)Binomial[n,m+1]}], Table[array[n,k],{k,2^(m+1)Binomial[n,m+1]}]], {n,m+1,m+4}]] // Flatten

Formula

For n>1, A337414(n,k) = Sum_{j=1..2*n*(n-1)} T(n,j) * binomial(k,j).
T(n,k) = 2*A338147(n,k) - A338146(n,k) = A338146(n,k) - 2*A338148(n,k) = A338147(n,k) - A338148(n,k).
T(2,k) = A338145(2,k) = A325019(2,k) = A325011(2,k); T(3,k) = A338145(3,k).
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