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.

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A325012 Array read by descending antidiagonals: A(n,k) is the number of oriented colorings of the facets of a regular n-dimensional orthoplex using up to k colors.

Original entry on oeis.org

1, 4, 1, 9, 6, 1, 16, 24, 23, 1, 25, 70, 333, 496, 1, 36, 165, 2916, 230076, 2275974, 1, 49, 336, 16725, 22456756, 965227578201, 800648638402240, 1, 64, 616, 70911, 795467350, 9607713956430560, 149031415906337877339236058, 1054942853799126580390222487977120, 1
Offset: 1

Views

Author

Robert A. Russell, May 27 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. 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 orthotope (cube) using up to k colors.

Examples

			Array begins with A(1,1):
1   4      9       16        25          36           49            64 ...
1   6     24       70       165         336          616          1044 ...
1  23    333     2916     16725       70911       241913        701968 ...
1 496 230076 22456756 795467350 14697611496 173107727191 1466088119056 ...
For A(1,2) = 4, the two achiral colorings use just one of the two colors for both vertices; the chiral pair uses one color for each vertex.
		

Crossrefs

Cf. A325013 (unoriented), A325014 (chiral), A325015 (achiral), A325016 (exactly k colors).
Other n-dimensional polytopes: A324999 (simplex), A325004 (orthotope).
Rows 1-3 are A000290, A006528, A000543; column 2 is A237748.

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 perm. *)
    CI1[{n_Integer}] := CI1[{n}] = CI[sub = Divisors[n]; Transpose[If[EvenQ[n], {sub, a37 /@ sub}, {2 sub, a48 /@ sub}]]] 2^(n-1); (* odd perm. *)
    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[(CI0[#] pc[#]) & /@ IntegerPartitions[n]])/(n! 2^(n - 1))] /. CI[l_List] :> j^(Total[l][[2]])
    array[n_, k_] := row[n] /. j -> k
    Table[array[n, d-n+1], {d, 1, 10}, {n, 1, d}] // Flatten

Formula

The algorithm used in the Mathematica program below assigns each permutation of the axes to a partition of n. It then determines the number of permutations for each partition and the cycle index for each partition.
A(n,k) = A325013(n,k) + A325014(n,k) = 2*A325013(n,k) - A325015(n,k) = 2*A325014(n,k) + A325015(n,k).
A(n,k) = Sum_{j=1..2^n} A325016(n,j) * binomial(k,j).

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).

A325017 Triangle read by rows: T(n,k) is the number of unoriented 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, 1, 1, 4, 6, 3, 1, 20, 204, 1056, 2850, 4080, 2940, 840, 1, 400, 130899, 11230666, 347919225, 5158324560, 43174480650, 225086553300, 775894225050, 1831178115900, 3008073915000, 3439243962000, 2685727044000, 1366701336000, 408648240000, 54486432000
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. 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 orthotope (cube) using exactly k colors.

Examples

			Triangle begins with T(1,1):
1  1
1  4   6    3
1 20 204 1056 2850 4080 2940 840
For T(2,2)=4, two squares have three edges the same color, one has opposite edges the same color, and one has opposite edges different colors.
		

Crossrefs

Cf. A325016 (oriented), A325018 (chiral), A325019 (achiral), A325013 (up to k colors).
Other n-dimensional polytopes: A007318(n,k-1) (simplex), A325009 (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 perm. *)
    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 perm. *)
    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[((CI0[#] + CI1[#]) pc[#]) & /@ IntegerPartitions[n]])/(n! 2^n)] /. CI[l_List] :> j^(Total[l][[2]])
    array[n_, k_] := row[n] /. j -> k (* A325013 *)
    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

A325013(n,k) = Sum_{j=1..2^n} T(n,j) * binomial(k,j).
T(n,k) = A325016(n,k) - A325018(n,k) = (A325016(n,k) + A325019(n,k)) / 2 = A325018(n,k) + A325019(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).

A325002 Triangle read by rows: T(n,k) is the number of oriented colorings of the facets (or vertices) of a regular n-dimensional simplex using exactly k colors.

Original entry on oeis.org

1, 2, 1, 2, 2, 1, 3, 3, 2, 1, 4, 6, 4, 2, 1, 5, 10, 10, 5, 2, 1, 6, 15, 20, 15, 6, 2, 1, 7, 21, 35, 35, 21, 7, 2, 1, 8, 28, 56, 70, 56, 28, 8, 2, 1, 9, 36, 84, 126, 126, 84, 36, 9, 2, 1, 10, 45, 120, 210, 252, 210, 120, 45, 10, 2, 1, 11, 55, 165, 330, 462, 462, 330, 165, 55, 11, 2
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 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. Two oriented colorings are the same if one is a rotation of the other; chiral pairs are counted as two. The only chiral pair occurs when k=n+1; for k <= n all the colorings are achiral.

Examples

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

Crossrefs

Cf. A007318(n,k-1) (unoriented), A325003 (achiral), A325001 (up to k colors).
Other n-dimensional polytopes: A325008 (orthotope), A325016 (orthoplex).

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) = 2*A007318(n,k-1) - A325003(n,k) = [k==n+1] + A325003(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).

A325018 Triangle read by rows: T(n,k) is the number of chiral pairs of 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

0, 1, 0, 0, 3, 3, 0, 1, 63, 662, 2400, 3900, 2940, 840, 0, 94, 97692, 10308758, 337560150, 5098740090, 42976836210, 224685801060, 775389028050, 1830791421900, 3007909258200, 3439214024400, 2685727044000, 1366701336000, 408648240000, 54486432000
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. 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 orthotope (cube) using exactly k colors.

Examples

			Triangle begins with T(1,1):
0 1
0 0  3   3
0 1 63 662 2400 3900 2940 840
For T(2,3)=3, each square has one of the three colors on two adjacent edges.
		

Crossrefs

Cf. A325016 (oriented), A325017 (unoriented), A325019 (achiral), A325014 (up to k colors).
Other n-dimensional polytopes: A325010 (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 perm. *)
    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 perm. *)
    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[((CI0[#] - CI1[#]) pc[#]) & /@ IntegerPartitions[n]])/(n! 2^n)] /. CI[l_List] :> j^(Total[l][[2]])
    array[n_, k_] := row[n] /. j -> k (* A325014 *)
    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

A325014(n,k) = Sum_{j=1..2^n} T(n,j) * binomial(k,j).
T(n,k) = A325016(n,k) - A325017(n,k) = (A325016(n,k) - A325019(n,k)) / 2 = A325017(n,k) - A325019(n,k).

A338142 Triangle read by rows: T(n,k) is the number of oriented 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, 9, 6, 1, 216, 22164, 613804, 6901425, 39713430, 131754420, 267165360, 336798000, 257796000, 109771200, 19958400, 1, 22409618, 9651132365418, 96038196404417832, 120785673234798359850
Offset: 1

Views

Author

Robert A. Russell, Oct 12 2020

Keywords

Comments

Each chiral pair is counted as two when enumerating oriented arrangements. 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     9      6
  1 216 22164 613804 6901425 39713430 131754420 267165360 336798000
  ...
		

Crossrefs

Cf. A338143 (unoriented), A338144 (chiral), A338145 (achiral), A337407 (k or fewer colors), A325016 (orthotope vertices, orthoplex facets).
Cf. A327087 (simplex), A338146 (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}]], (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[]), 0]);
    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

A337407(n,k) = Sum_{j=1..n*2^(n-1)} T(n,j) * binomial(k,j).
T(n,k) = A338143(n,k) + A338144(n,k) = 2*A338143(n,k) - A338145(n,k) = 2*A338144(n,k) + A338145(n,k).
T(2,k) = A338146(2,k) = A325016(2,k) = A325008(2,k); T(3,k) = A338146(3,k).

A338146 Triangle read by rows: T(n,k) is the number of oriented 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, 9, 6, 1, 216, 22164, 613804, 6901425, 39713430, 131754420, 267165360, 336798000, 257796000, 109771200, 19958400, 1, 90052, 1471369998, 1460163153852, 303126054092610, 22838390261305920, 831533453035309605
Offset: 1

Views

Author

Robert A. Russell, Oct 12 2020

Keywords

Comments

Each chiral pair is counted as two when enumerating oriented arrangements. 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     9      6
  1 216 22164 613804 6901425 39713430 131754420 267165360 336798000
  ...
For T(2,3)=9, the 3 achiral colorings are ABAC, ABCB, and ACBC. The three chiral pairs are AABC-AACB, ABBC-ACBB, and ABCC-ACCB.
		

Crossrefs

Cf. A338147 (unoriented), A338148 (chiral), A338149 (achiral), A337411 (k or fewer colors), A325008 (orthoplex vertices, orthotope facets).
Cf. A327087 (simplex), A338142 (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}]], (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[]), 0]);
    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, A337411(n,k) = Sum_{j=1..2*n*(n-1)} T(n,j) * binomial(k,j).
T(n,k) = A338147(n,k) + A338148(n,k) = 2*A338147(n,k) - A338149(n,k) = 2*A338148(n,k) + A338149(n,k).
T(2,k) = A338142(2,k) = A325016(2,k) = A325008(2,k); T(3,k) = A338142(3,k).
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