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 1-10 of 10 results.

A047780 Number of inequivalent ways to color faces of a cube using at most n colors.

Original entry on oeis.org

0, 1, 10, 57, 240, 800, 2226, 5390, 11712, 23355, 43450, 76351, 127920, 205842, 319970, 482700, 709376, 1018725, 1433322, 1980085, 2690800, 3602676, 4758930, 6209402, 8011200, 10229375, 12937626, 16219035, 20166832, 24885190, 30490050
Offset: 0

Views

Author

Keywords

Comments

Here inequivalent means under the action of the rotation group of the cube, of order 24, which in its action on the faces has cycle index (x1^6 + 3*x1^2*x2^2 + 6*x1^2*x4 + 6*x2^3 + 8*x3^2)/24.
a(n) is also the number of inequivalent colorings of the vertices of a regular octahedron using at most n colors. - José H. Nieto S., Jan 19 2012
From Robert A. Russell, Oct 08 2020: (Start)
Each chiral pair is counted as two when enumerating oriented arrangements. The Schläfli symbols for the regular octahedron and cube are {3,4} and {4,3} respectively. They are mutually dual.
There are 24 elements in the rotation group of the regular octahedron/cube. They divide into five conjugacy classes. The first formula is obtained by averaging the cube face (octahedron vertex) cycle indices after replacing x_i^j with n^j according to the Pólya enumeration theorem.
Conjugacy Class Count Even Cycle Indices
Identity 1 x_1^6
Vertex rotation 8 x_3^2
Edge rotation 6 x_2^3
Small face rotation 6 x_1^2x_4^1
Large face rotation 3 x_1^2x_2^2 (End)

References

  • L. Comtet, Advanced Combinatorics, Reidel, 1974, p. 254 (corrected).
  • N. G. De Bruijn, Polya's theory of counting, in E. F. Beckenbach, ed., Applied Combinatorial Mathematics, Wiley, 1964, pp. 144-184 (see p. 147).
  • M. Gardner, New Mathematical Diversions from Scientific American. Simon and Schuster, NY, 1966, p. 246 (the formula given is incorrect but was corrected in the second printing).
  • J.-P. Delahaye, 'Le miraculeux "lemme de Burnside"','Le coloriage du cube' p. 147 in 'Pour la Science' (French edition of 'Scientific American') No.350 December 2006 Paris.
  • N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Crossrefs

Cf. A198833 (unoriented), A093566(n+1) (chiral), A337898 (achiral).
Other elements: A060530 (edges), A000543 (cube vertices, octahedron faces).
Cf. A006008 (tetrahedron), A000545 (dodecahedron faces, icosahedron vertices), A054472 (icosahedron faces, dodecahedron vertices).
Row 3 of A325004 (orthoplex vertices, orthotope facets) and A337887 (orthotope faces, orthoplex peaks).

Programs

  • Magma
    [(n^6 + 3*n^4 + 12*n^3 + 8*n^2)/24: n in [1..30]]; // Vincenzo Librandi, Apr 27 2012
  • Mathematica
    CoefficientList[Series[x*(1+3*x+8*x^2+16*x^3+2*x^4)/(1-x)^7,{x,0,33}],x] (* Vincenzo Librandi, Apr 27 2012 *)

Formula

a(n) = (n^6 + 3*n^4 + 12*n^3 + 8*n^2)/24 = n+8*C(n, 2)+30*C(n, 3)+68*C(n, 4)+75*C(n, 5)+30*C(n, 6). Each term of the RHS indicates the number of ways to use n colors to color the cube faces (octahedron vertices) with exactly 1, 2, 3, 4, 5, or 6 colors.
G.f.: x*(1+3*x+8*x^2+16*x^3+2*x^4)/(1-x)^7. - Colin Barker, Jan 29 2012
a(n) = A198833(n) + A093566(n+1) = 2*A198833(n) - A337898(n) = 2*A093566(n+1) + A337898(n). - Robert A. Russell, Oct 08 2020

Extensions

Corrected version of A006550 and A006529.
Entry revised by N. J. A. Sloane, Jan 03 2005

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).

A325006 Array read by descending antidiagonals: A(n,k) is the number of chiral pairs of colorings of the facets of a regular n-dimensional orthotope using up to k colors.

Original entry on oeis.org

0, 1, 0, 3, 0, 0, 6, 3, 0, 0, 10, 15, 1, 0, 0, 15, 45, 20, 0, 0, 0, 21, 105, 120, 15, 0, 0, 0, 28, 210, 455, 210, 6, 0, 0, 0, 36, 378, 1330, 1365, 252, 1, 0, 0, 0, 45, 630, 3276, 5985, 3003, 210, 0, 0, 0, 0, 55, 990, 7140, 20475, 20349, 5005, 120, 0, 0, 0, 0, 66, 1485, 14190, 58905, 98280, 54264, 6435, 45, 0, 0, 0, 0
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 up to k colors.

Examples

			Array begins with A(1,1):
0 1 3  6  10   15     21       28        36         45          55 ...
0 0 3 15  45  105    210      378       630        990        1485 ...
0 0 1 20 120  455   1330     3276      7140      14190       26235 ...
0 0 0 15 210 1365   5985    20475     58905     148995      341055 ...
0 0 0  6 252 3003  20349    98280    376992    1221759     3478761 ...
0 0 0  1 210 5005  54264   376740   1947792    8145060    28989675 ...
0 0 0  0 120 6435 116280  1184040   8347680   45379620   202927725 ...
0 0 0  0  45 6435 203490  3108105  30260340  215553195  1217566350 ...
0 0 0  0  10 5005 293930  6906900  94143280  886163135  6358402050 ...
0 0 0  0   1 3003 352716 13123110 254186856 3190187286 29248649430 ...
For a(2,3)=3, each chiral pair consists of two adjacent edges of the square with one of the three colors.
		

Crossrefs

Cf. A325004 (oriented), A325005 (unoriented), A325007 (achiral), A325010 (exactly k colors)
Other n-dimensional polytopes: A007318(k,n+1) (simplex), A325014 (orthoplex)
Rows 1-3 are A161680, A050534, A093566(n+1), A234249(n-1)

Programs

  • Mathematica
    Table[Binomial[Binomial[d-n+1,2],n],{d,1,12},{n,1,d}] // Flatten
  • PARI
    a(n, k) = binomial(binomial(k, 2), n)
    array(rows, cols) = for(x=1, rows, for(y=1, cols, print1(a(x, y), ", ")); print(""))
    /* Print initial 10 rows and 11 columns of array as follows: */
    array(10, 11) \\ Felix Fröhlich, May 30 2019

Formula

A(n,k) = binomial(binomial(k,2),n).
A(n,k) = Sum_{j=1..2*n} A325010(n,j) * binomial(k,j).
A(n,k) = A325004(n,k) - A325005(n,k) = (A325004(n,k) - A325007(n,k)) / 2 = A325005(n,k) - A325007(n,k).
G.f. for row n: Sum{j=1..2*n} A325010(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+x)^binomial(k,2) - 1.

A325005 Array read by descending antidiagonals: A(n,k) is the number of unoriented colorings of the facets of a regular n-dimensional orthotope using up to k colors.

Original entry on oeis.org

1, 3, 1, 6, 6, 1, 10, 21, 10, 1, 15, 55, 56, 15, 1, 21, 120, 220, 126, 21, 1, 28, 231, 680, 715, 252, 28, 1, 36, 406, 1771, 3060, 2002, 462, 36, 1, 45, 666, 4060, 10626, 11628, 5005, 792, 45, 1, 55, 1035, 8436, 31465, 53130, 38760, 11440, 1287, 55, 1
Offset: 1

Views

Author

Robert A. Russell, Mar 23 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 up to k colors.

Examples

			Array begins with A(1,1):
1  3    6    10     15      21       28        36        45         55 ...
1  6   21    55    120     231      406       666      1035       1540 ...
1 10   56   220    680    1771     4060      8436     16215      29260 ...
1 15  126   715   3060   10626    31465     82251    194580     424270 ...
1 21  252  2002  11628   53130   201376    658008   1906884    5006386 ...
1 28  462  5005  38760  230230  1107568   4496388  15890700   50063860 ...
1 36  792 11440 116280  888030  5379616  26978328 115775100  436270780 ...
1 45 1287 24310 319770 3108105 23535820 145008513 752538150 3381098545 ...
For A(1,2) = 3, 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. A325004 (oriented), A325006 (chiral), A325007 (achiral), A325009 (exactly k colors).
Other n-dimensional polytopes: A325000 (simplex), A325013 (orthoplex).
Rows 1-3 are A000217, A002817, A198833.

Programs

  • Mathematica
    Table[Binomial[Binomial[d-n+2,2]+n-1,n],{d,1,11},{n,1,d}] // Flatten

Formula

A(n,k) = binomial(n + binomial(k+1,2) - 1, n).
A(n,k) = Sum_{j=1..2n} A325009(n,j) * binomial(k,j).
A(n,k) = A325004(n,k) - A325006(n,k) = (A325004(n,k) + A325007(n,k)) / 2 = A325006(n,k) + A325007(n,k).
G.f. for row n: Sum_{j=1..2n} A325009(n,j) * x^j / (1-x)^(j+1).
Linear recurrence for row n: T(n,k) = Sum_{j=0..2n} binomial(-2-j,2n-j) * T(n,k-1-j).
G.f. for column k: 1/(1-x)^binomial(k+1,2) - 1.

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).

A324999 Array read by descending antidiagonals: A(n,k) is the number of oriented colorings of the facets (or vertices) of a regular n-dimensional simplex using up to k colors.

Original entry on oeis.org

1, 4, 1, 9, 4, 1, 16, 11, 5, 1, 25, 24, 15, 6, 1, 36, 45, 36, 21, 7, 1, 49, 76, 75, 56, 28, 8, 1, 64, 119, 141, 127, 84, 36, 9, 1, 81, 176, 245, 258, 210, 120, 45, 10, 1, 100, 249, 400, 483, 463, 330, 165, 55, 11, 1, 121, 340, 621, 848, 931, 792, 495, 220, 66, 12, 1
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. Two oriented colorings are the same if one is a rotation of the other; chiral pairs are counted as two.

Examples

			The array begins with A(1,1):
  1  4  9  16  25   36   49    64    81   100   121    144    169    196 ...
  1  4 11  24  45   76  119   176   249   340   451    584    741    924 ...
  1  5 15  36  75  141  245   400   621   925  1331   1860   2535   3381 ...
  1  6 21  56 127  258  483   848  1413  2254  3465   5160   7475  10570 ...
  1  7 28  84 210  463  931  1744  3087  5215  8470  13300  20280  30135 ...
  1  8 36 120 330  792 1717  3440  6471 11560 19778  32616  52104  80952 ...
  1  9 45 165 495 1287 3003  6436 12879 24355 43923  76077 127257 206493 ...
  1 10 55 220 715 2002 5005 11440 24311 48630 92433 168180 294645 499422 ...
  ...
For A(1,2) = 4, the two achiral colorings use just one of the two colors for both vertices; the chiral pair uses two colors. For A(2,2)=4, the triangle may have 0, 1, 2, or 3 edges of one color.
		

Crossrefs

Cf. A325000 (unoriented), A325000(n,k-n) (chiral), A325001 (achiral), A325002 (exactly k colors), A327083 (edges, ridges), A337883 (faces, peaks), A325004 (orthotope facets, orthoplex vertices), A325012 (orthoplex facets, orthotope vertices).
Rows 1-4 are A000290, A006527, A006008, A337895.

Programs

  • Mathematica
    Table[Binomial[d+1,n+1] + Binomial[d+1-n,n+1], {d,1,15}, {n,1,d}] // Flatten

Formula

A(n,k) = binomial(n+k,n+1) + binomial(k,n+1).
A(n,k) = Sum_{j=1..n+1} A325002(n,j) * binomial(k,j).
A(n,k) = A325000(n,k) + A325000(n,k-n) = 2*A325000(n,k) - A325001(n,k) = 2*A325000(n,k-n) + A325001(n,k).
G.f. for row n: (x + x^(n+1)) / (1-x)^(n+2).
Linear recurrence for row n: A(n,k) = Sum_{j=1..n+2} -binomial(j-n-3,j) * A(n,k-j).
G.f. for column k: (1 - 2*(1-x)^k + (1-x^2)^k) / (x*(1-x)^k) - 2*k.

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).

A337411 Array read by descending antidiagonals: T(n,k) is the number of oriented colorings of the edges of a regular n-dimensional orthoplex (cross polytope) using k or fewer colors.

Original entry on oeis.org

1, 2, 1, 3, 6, 1, 4, 24, 218, 1, 5, 70, 2285, 90054, 1, 6, 165, 703760, 1471640157, 573439556, 1, 7, 336, 10194250, 1466049174160, 6332134720430727, 50043770249328, 1, 8, 616, 90775566, 310441584462375, 629648890639384572032, 1839894096099964270283469, 59966884221697869216, 1
Offset: 1

Views

Author

Robert A. Russell, Aug 26 2020

Keywords

Comments

Each chiral pair is counted as two when enumerating oriented arrangements. For n=1, the figure is a line segment with one edge. For n=2, the figure is a square with 4 edges. For n=3, the figure is an octahedron with 12 edges. The number of edges is 2n*(n-1) for n>1.
Also the number of oriented colorings of the regular (n-2)-dimensional orthotopes (hypercubes) in a regular n-dimensional orthotope.

Examples

			Table begins with T(1,1):
1   2     3      4        5        6         7          8           9 ...
1   6    24     70      165      336       616       1044        1665 ...
1 218 22815 703760 10194250 90775566 576941778 2863870080 11769161895 ...
For T(2,2)=6, the arrangements are AAAA, AAAB, AABB, ABAB, ABBB, and BBBB.
		

Crossrefs

Cf. A337412 (unoriented), A337413 (chiral), A337414 (achiral).
Rows 1-4 are A000027, A006528, A060530, A331354.
Cf. A327083 (simplex edges), A337407 (orthotope edges), A325004 (orthoplex vertices).

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
    Table[array[n,d+m-n], {d,8}, {n,m,d+m-1}] // Flatten

Formula

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).
T(n,k) = A337412(n,k) + A337413(n,k) = 2*A337412(n,k) - A337414(n,k) = 2*A337413(n,k) + A337414(n,k).

A337956 Number of oriented colorings of the 8 cubic facets of a tesseract or of the 8 vertices of a hyperoctahedron.

Original entry on oeis.org

1, 15, 126, 730, 3270, 11991, 37450, 102726, 253485, 573265, 1205556, 2384460, 4475926, 8031765, 13858860, 23106196, 37372545, 58837851, 90421570, 135971430, 200486286, 290376955, 413769126, 580852650, 804281725
Offset: 1

Views

Author

Robert A. Russell, Oct 03 2020

Keywords

Comments

Each chiral pair is counted as two when enumerating oriented arrangements. The Schläfli symbols for the tesseract and the hyperoctahedron are {4,3,3} and {3,3,4} respectively. Both figures are regular 4-D polyhedra and they are mutually dual.

Crossrefs

Cf. A337957 (unoriented), A234249(n+1) (chiral), A337958 (achiral).
Other elements: A331354 (hyperoctahedron edges, tesseract faces), A331358 (hyperoctahedron faces, tesseract edges), A337952 (hyperoctahedron facets, tesseract vertices).
Other polychora: A337895 (5-cell), A338948 (24-cell), A338964 (120-cell, 600-cell).
Row 4 of A325004 (orthotope facets, orthoplex vertices).

Programs

  • Mathematica
    Table[Binomial[Binomial[n+1,2]+3,4] + Binomial[Binomial[n,2],4],{n,30}]

Formula

a(n) = binomial(binomial(n+1,2)+3,4) + binomial(binomial(n,2),4).
a(n) = n * (n+1) * (n^6 - n^5 + 7*n^4 + 29*n^3 + 16*n^2 - 4*n + 48) / 192.
a(n) = 1*C(n,1) + 13*C(n,2) + 84*C(n,3) + 312*C(n,4) + 735*C(n,5) + 1020*C(n,6) + 735*C(n,7) + 210*C(n,8), where the coefficient of C(n,k) is the number of oriented colorings using exactly k colors.
a(n) = A337957(n) + A234249(n+1) = 2*A337957(n) - A337958(n) = 2*A234249(n+1) + A337958(n).
From Stefano Spezia, Oct 04 2020: (Start)
G.f.: x*(1 + 6*x + 27*x^2 + 52*x^3 + 102*x^4 + 21*x^5 + x^6)/(1 - x)^9.
a(n) = 9*a(n-1)-36*a(n-2)+84*a(n-3)-126*a(n-4)+126*a(n-5)-84*a(n-6)+36*a(n-7)-9*a(n-8)+a(n-8) for n > 8.
(End)

A337891 Array read by descending antidiagonals: T(n,k) is the number of oriented colorings of the faces of a regular n-dimensional orthoplex (cross polytope) using k or fewer colors.

Original entry on oeis.org

1, 2, 1, 3, 23, 1, 4, 333, 22409620, 1, 5, 2916, 9651199594275, 629648865588086369152, 1, 6, 16725, 96076801068337216, 76983765319971901895960429658208179, 63433230786931550329738915431918588874940416, 1
Offset: 2

Views

Author

Robert A. Russell, Sep 28 2020

Keywords

Comments

Each chiral pair is counted as two when enumerating oriented arrangements. For n=2, the figure is a square with one square face. For n=3, the figure is an octahedron with 8 triangular faces. For higher n, the number of triangular faces is 8*C(n,3).
Also the number of oriented colorings of the peaks of an n-dimensional orthotope (hypercube). A peak is an (n-3)-dimensional orthotope.
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

			Array begins with T(2,1):
1        2             3                 4                     5 ...
1       23           333              2916                 16725 ...
1 22409620 9651199594275 96076801068337216 121265960728368199375 ...
		

Crossrefs

Cf. A337892 (unoriented), A337893 (chiral), A337894 (achiral).
Other elements: A325004 (vertices), A337411 (edges).
Other polytopes: A337883 (simplex), A337887 (orthotope).
Rows 2-4 are A000027, A000543, A331358

Programs

  • Mathematica
    m=2; (* dimension of color element, here a face *)
    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
    Table[array[n,d+m-n], {d,6}, {n,m,d+m-1}] // Flatten

Formula

T(n,k) = A337892(n,k) + A337893(n,k) = 2*A337892(n,k) - A337894(n,k) = 2*A337893(n,k) + A337894(n,k).
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