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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A331353 Number of achiral colorings of the edges (or triangular faces) of a regular 4-dimensional simplex with n available colors.

Original entry on oeis.org

1, 28, 387, 2784, 13125, 46836, 137543, 349952, 797769, 1667500, 3248971, 5973408, 10459917, 17571204, 28479375, 44742656, 68393873, 102041532, 148984339, 213340000, 300189141, 415735188, 567481047, 764423424
Offset: 1

Views

Author

Robert A. Russell, Jan 14 2020

Keywords

Comments

A 4-dimensional simplex has 5 vertices and 10 edges. Its Schläfli symbol is {3,3,3}. An achiral coloring is identical to its reflection,
There are 60 elements in the automorphism group of the 4-dimensional simplex that are not in its rotation group. Each is an odd permutation of the vertices and can be associated with a partition of 5 based on the conjugacy group of the permutation. The first formula is obtained by averaging their cycle indices after replacing x_i^j with n^j according to the Pólya enumeration theorem.
Partition Count Odd Cycle Indices
41 30 x_2^1x_4^2
32 20 x_1^1x_3^1x_6^1
2111 10 x_1^4x_2^3

Crossrefs

Cf. A331350 (oriented), A063843 (unoriented), A331352 (chiral).
Other polychora: A331361 (8-cell), A331357 (16-cell), A338955 (24-cell), A338967 (120-cell, 600-cell).
Row 4 of A327086 (simplex edges and ridges) and A337886 (simplex faces and peaks).

Programs

  • Mathematica
    Table[(5 n^3 + n^7)/6, {n, 1, 25}]
  • PARI
    Vec(x*(1 + 20*x + 191*x^2 + 416*x^3 + 191*x^4 + 20*x^5 + x^6) / (1 - x)^8 + O(x^25)) \\ Colin Barker, Jan 15 2020

Formula

a(n) = (5*n^3 + n^7) / 6.
a(n) = C(n,1) + 26*C(n,2) + 306*C(n,3) + 1400*C(n,4) + 2800*C(n,5) + 2520*C(n,6) + 840*C(n,7), where the coefficient of C(n,k) is the number of colorings using exactly k colors.
a(n) = 2*A063843(n) - A331350(n) = A331350(n) - 2*A331352(n) = A063843(n) - A331352(n).
From Colin Barker, Jan 15 2020: (Start)
G.f.: x*(1 + 20*x + 191*x^2 + 416*x^3 + 191*x^4 + 20*x^5 + x^6) / (1 - x)^8.
a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 70*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8) for n>8.
(End)

A331354 Number of oriented colorings of the edges of a regular 4-dimensional orthoplex with n available colors.

Original entry on oeis.org

1, 90054, 1471640157, 1466049174160, 310441584462375, 24679078461920106, 997818989210621704, 24595659246351652992, 415450226822646218895, 5208333343963621522750, 51300691059764724112161, 414046079318115654521904
Offset: 1

Views

Author

Robert A. Russell, Jan 14 2020

Keywords

Comments

A regular 4-dimensional orthoplex (also hyperoctahedron or cross polytope) has 8 vertices and 24 edges. Its Schläfli symbol is {3,3,4}. 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 square faces of a tesseract {4,3,3} with n available colors.
There are 192 elements in the rotation group of the 4-dimensional orthoplex. Each is associated with a partition of 4 based on the conjugacy group of the permutation of the axes. The first formula is obtained by averaging their cycle indices after replacing x_i^j with n^j according to the Pólya enumeration theorem.
Partition Count Even Cycle Indices
4 6 8x_8^3
31 8 4x_3^8 + 4x_6^4
22 3 4x_1^4x_2^10 + 4x_4^6
211 6 4x_1^2x_2^11 + 2x_1^4x_4^5 + 2x_2^2x_4^5
1111 1 6x_1^4x_2^10 + x_1^24 + x_2^12

Crossrefs

Cf. A331355 (unoriented), A331356 (chiral), A331357 (achiral).
Other polychora: A331350 (5-cell), A331358 (8-cell), A338952 (24-cell), A338964 (120-cell, 600-cell).
Row 4 of A337411 (orthoplex edges, orthotope ridges) and A337887 (orthotope faces, orthoplex peaks).

Programs

  • Mathematica
    Table[(48n^3 + 32n^4 + 12n^6 + 12n^7 + 32n^8 + 12n^9 + n^12 + 24n^13 + 18n^14 + n^24)/192, {n, 1, 25}]

Formula

a(n) = (48*n^3 + 32*n^4 + 12*n^6 + 12*n^7 + 32*n^8 + 12*n^9 + n^12 + 24*n^13 + 18*n^14 + n^24) / 192.
a(n) = C(n,1) + 90052*C(n,2) + 1471369998*C(n,3) + 1460163153852*C(n,4) + 303126054092610*C(n,5) + 22838390261305920*C(n,6) + 831533453035309605*(n,7) + 17286839341903413240*C(n,8) + 227976665667323280750*C(n,9) + 2046002146009161624900*C(n,10) + 13118524448411114548200*C(n,11) + 62195874413179579657200*C(n,12) + 223421486565003375448800*C(n,13) + 618462331903782130564800*C(n,14) + 1333693289177381452320000*C(n,15) + 2253251792722109699520000*C(n,16) + 2984347082566196867520000*C(n,17) + 3083974243985846090880000*C(n,18) + 2458713052058007064320000*C(n,19) + 1482204734016157831680000*C(n,20) + 653167360418390737920000*C(n,21) + 198468086839148206080000*C(n,22) + 37162274062147153920000*C(n,23) + 3231502092360622080000*C(n,24), where the coefficient of C(n,k) is the number of colorings using exactly k colors.
a(n) = A331355(n) + A331356(n) = 2*A331355(n) - A331357(n) = 2*A331356(n) + A331357(n).

A331355 Number of unoriented colorings of the edges of a regular 4-dimensional orthoplex with n available colors.

Original entry on oeis.org

1, 49127, 740360358, 733776248840, 155261523065875, 12340612271439081, 498926608780739307, 12298018390569089088, 207726683413584244680, 2604177120221402303875, 25650403577338260144611, 207023317470352041578712
Offset: 1

Views

Author

Robert A. Russell, Jan 14 2020

Keywords

Comments

A regular 4-dimensional orthoplex (also hyperoctahedron or cross polytope) has 8 vertices and 24 edges. Its Schläfli symbol is {3,3,4}. Two unoriented colorings are the same if congruent; chiral pairs are counted as one. Also the number of unoriented colorings of the square faces of a tesseract {4,3,3} with n available colors.

Crossrefs

Cf. A331354 (oriented), A331356 (chiral), A331357 (achiral).
Other polychora: A063843 (5-cell), A331359 (8-cell), A338953 (24-cell), A338965 (120-cell, 600-cell).
Row 4 of A337412 (orthoplex edges, orthotope ridges) and A337888 (orthotope faces, orthoplex peaks).

Programs

  • Mathematica
    Table[(48 n^3 + 64 n^4 + 44 n^6 + 84 n^7 + 56 n^8 + 12 n^9 + 5 n^12 +
        36 n^13 + 18 n^14 + 12 n^15 + 4 n^18 + n^24)/384, {n, 1, 25}]

Formula

a(n) = (48*n^3 + 64*n^4 + 44*n^6 + 84*n^7 + 56*n^8 + 12*n^9 + 5*n^12 +
36*n^13 + 18*n^14 + 12*n^15 + 4*n^18 + n^24) / 384.
a(n) = C(n,1) + 49125*C(n, 2) + 740212980*C(n, 3) + 730815102166*C(n, 4) + 151600044933990*C(n, 5) + 11420034970306170*C(n, 6) + 415777158607920585*C(n, 7) + 8643499341510394200*C(n, 8) + 113988734942055623055*C(n, 9) + 1023002477284840979850*C(n, 10) + 6559265715033958749900*C(n, 11) + 31097943476763200314200*C(n, 12) + 111710751446923209781200*C(n, 13) + 309231173588248964052000*C(n, 14) + 666846649590586048584000*C(n, 15) + 1126625898539640346848000*C(n, 16) + 1492173541849975272288000*C(n, 17) + 1541987122059614438208000*C(n, 18) + 1229356526029003532160000*C(n, 19) + 741102367008078915840000*C(n, 20) + 326583680209195368960000*C(n, 21) + 99234043419574103040000*C(n, 22) + 18581137031073576960000*C(n, 23) + 1615751046180311040000*C(n, 24), where the coefficient of C(n,k) is the number of colorings using exactly k colors.
a(n) = A331354(n) - A331356(n) = (A331354(n) + A331357(n)) / 2 = A331356(n) + A331357(n).

A331356 Number of chiral pairs of colorings of the edges of a regular 4-dimensional orthoplex with n available colors.

Original entry on oeis.org

0, 40927, 731279799, 732272925320, 155180061396500, 12338466190481025, 498892380429882397, 12297640855782563904, 207723543409061974215, 2604156223742219218875, 25650287482426463967550, 207022761847763612943192
Offset: 1

Views

Author

Robert A. Russell, Jan 14 2020

Keywords

Comments

A regular 4-dimensional orthoplex (also hyperoctahedron or cross polytope) has 8 vertices and 24 edges. Its Schläfli symbol is {3,3,4}. The chiral colorings of its edges come in pairs, each the reflection of the other. Also the number of chiral pairs of colorings of the square faces of a tesseract {4,3,3} with n available colors.

Crossrefs

Cf. A331354 (oriented), A331355 (unoriented), A331357 (achiral).
Other polychora: A331352 (5-cell), A331360 (8-cell), A338954 (24-cell), A338966 (120-cell, 600-cell).
Row 4 of A337413 (orthoplex edges, orthotope ridges) and A337889 (orthotope faces, orthoplex peaks).

Programs

  • Mathematica
    Table[(48n^3 - 20n^6 - 60n^7 + 8n^8 + 12n^9 - 3n^12 + 12n^13 + 18n^14 - 12n^15 - 4n^18 + n^24)/384, {n, 1, 25}]

Formula

a(n) = (48*n^3 - 20*n^6 - 60*n^7 + 8*n^8 + 12*n^9 - 3*n^12 + 12*n^13 + 18*n^14 - 12*n^15 - 4*n^18 + n^24) / 384.
a(n) = 40927*C(n,2) + 731157018*C(n,3) + 729348051686*C(n,4) + 151526009158620*C(n,5) + 11418355290999750*C(n,6) + 415756294427389020*C(n,7) + 8643340000393019040*C(n,8) + 113987930725267657695*C(n,9) + 1022999668724320645050*C(n,10) + 6559258733377155798300*C(n,11) + 31097930936416379343000*C(n,12) + 111710735118080165667600*C(n,13) + 309231158315533166512800*C(n,14) + 666846639586795403736000*C(n,15) + 1126625894182469352672000*C(n,16) + 1492173540716221595232000*C(n,17) + 1541987121926231652672000*C(n,18) + 1229356526029003532160000*C(n,19) + 741102367008078915840000*C(n,20) + 326583680209195368960000*C(n,21) + 99234043419574103040000*C(n,22) + 18581137031073576960000*C(n,23) + 1615751046180311040000*C(n,24), where the coefficient of C(n,k) is the number of colorings using exactly k colors.
a(n) = A331354(n) - A331355(n) = (A331354(n) - A331357(n)) / 2 = A331355(n) - A331357(n).

A331361 Number of achiral colorings of the edges of a tesseract with n available colors.

Original entry on oeis.org

1, 93024, 294157089, 91983927296, 7960001890625, 304914963625056, 6652124939544609, 96100248309858304, 1013293206632601441, 8334166666733500000, 56066328722011832961, 319495406392484665344
Offset: 1

Views

Author

Robert A. Russell, Jan 14 2020

Keywords

Comments

A tesseract is a regular 4-dimensional orthotope or hypercube with 16 vertices and 32 edges. Its Schläfli symbol is {4,3,3}. An achiral coloring is identical to its reflection. Also the number of achiral colorings of the triangular faces of a regular 4-dimensional orthoplex {3,3,4} with n available colors.
There are 192 elements in the automorphism group of the tesseract that are not in its rotation group. Each is associated with a partition of 4 based on the conjugacy group of the permutation of the axes. The first formula is obtained by averaging their cycle indices after replacing x_i^j with n^j according to the Pólya enumeration theorem.
Partition Count Odd Cycle Indices
4 6 8x_4^8
31 8 4x_1^2x_3^2x_6^4 + 4x_2^1x_6^5
22 3 8x_4^8
211 6 2x_1^8x_2^12 + 2x_2^16 + 4x_4^8
1111 1 4x_1^8x_2^12 + 4x_2^16

Crossrefs

Cf. A331358 (oriented), A331359 (unoriented), A331360 (chiral).
Cf. A331353 (simplex), A331357 (orthoplex), A338955 (24-cell), A338967 (120-cell, 600-cell).

Programs

  • Mathematica
    Table[(2n^6 + 8n^8 + n^16 + n^20)/12, {n, 1, 25}]

Formula

a(n) = (2*n^6 + 8*n^8 + n^16 + n^20) / 12.
a(n) = C(n,1) + 93022*C(n,2) + 293878020*C(n,3) + 90807857080*C(n,4) + 7503022894800*C(n,5) + 258528829444320*C(n,6) + 4681671089961600*C(n,7) + 50981530073846400*C(n,8) + 363246007692204000*C(n,9) + 1789536284820648000*C(n,10) + 6323058513173001600*C(n,11) + 16406578807069651200*C(n,12) + 31689737477798400000*C(n,13) + 45786987328642560000*C(n,14) + 49291621471572480000*C(n,15) + 38970361271761920000*C(n,16) + 21972146261345280000*C(n,17) + 8363100653107200000*C(n,18) + 1926047423139840000*C(n,19) + 202741834014720000*C(n,20), where the coefficient of C(n,k) is the number of colorings using exactly k colors.
a(n) = 2*A331359(n) - A331358(n) = A331358(n) - 2*A331360(n) = A331359(n) - A331360(n).

A337414 Array read by descending antidiagonals: T(n,k) is the number of achiral 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, 18, 70, 1, 5, 40, 1407, 8200, 1, 6, 75, 12480, 9080559, 12804908, 1, 7, 126, 69050, 1503323520, 4906480368591, 304899216832, 1, 8, 196, 281946, 81461669375, 48226825456539776, 187380251418565888983, 103685962258536432, 1
Offset: 1

Views

Author

Robert A. Russell, Aug 26 2020

Keywords

Comments

An achiral arrangement is identical to its reflection. 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 achiral 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       10 ...
1  6   18    40    75    126    196     288     405      550 ...
1 70 1407 12480 69050 281946 931490 2632512 6598935 15041950 ...
For T(2,2)=6, the arrangements are AAAA, AAAB, AABB, ABAB, ABBB, and BBBB.
		

Crossrefs

Cf. A337411 (oriented), A337412 (unoriented), A337413 (chiral).
Rows 1-4 are A000027, A002411, A331351, A331357.
Cf. A327086 (simplex edges), A337410 (orthotope edges), A325007 (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}]],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
    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) = 2*A337412(n,k) - A337411(n,k) = A337411(n,k) - 2*A337413(n,k) = A337412(n,k) - A337413(n,k).

A338955 Number of achiral colorings of the 96 edges (or triangular faces) of the 4-D 24-cell using subsets of a set of n colors.

Original entry on oeis.org

1, 24124751133507584, 883287060208158070437496209, 27692675763559261523047959805034496, 18070082615414169898334284655914306640625, 1018202231744161700740376040914469837333037056
Offset: 1

Views

Author

Robert A. Russell, Nov 17 2020

Keywords

Comments

An achiral coloring is identical to its reflection. The Schläfli symbol of the 24-cell is {3,4,3}. It has 24 octahedral facets. It is self-dual. There are 576 elements in the automorphism group of the 24-cell that are not in its rotation group. They divide into 10 conjugacy classes. The first formula is obtained by averaging the edge (or face) cycle indices after replacing x_i^j with n^j according to the Pólya enumeration theorem.
Count Odd Cycle Indices Count Odd Cycle Indices
12 x_1^24x_2^36 96 x_1^2x_2^2x_3^2x_6^14
12 x_1^8x_2^44 96 x_3^8x_6^12
12+12 x_3^48 96 x_2^3x_6^15
72+72 x_4^24 96 x_6^16

Crossrefs

Cf. A338952 (oriented), A338953 (unoriented), A338954 (chiral), A338959 (exactly n colors), A338951 (vertices, facets), A331353 (5-cell), A331361 (8-cell edges, 16-cell faces), A331357 (16-cell edges, 8-cell faces), A338967 (120-cell, 600-cell).

Programs

  • Mathematica
    Table[(8n^16+8n^18+16n^20+12n^24+2n^48+n^52+n^60)/48,{n,15}]

Formula

a(n) = (8*n^16 + 8*n^18 + 16*n^20 + 12*n^24 + 2*n^48 + n^52 + n^60) / 48.
a(n) = Sum_{j=1..Min(n,60)} A338959(n) * binomial(n,j).
a(n) = 2*A338953(n) - A338952(n) = A338952(n) - 2*A338954(n) = A338953(n) - A338954(n).

A337955 Number of achiral colorings of the 16 tetrahedral facets of a hyperoctahedron or of the 16 vertices of a tesseract.

Original entry on oeis.org

1, 308, 34128, 1056576, 15303750, 136236276, 865711763, 4296782848, 17656466751, 62510672500, 196174554026, 557301826368, 1456216515468, 3543525156276, 8109415963125, 17592637669376, 36414622551373
Offset: 1

Views

Author

Robert A. Russell, Oct 03 2020

Keywords

Comments

An achiral coloring is identical to its reflection. 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. There are 192 elements in the automorphism group of the tesseract that are not in its rotation group. Each involves a permutation of the axes that can be associated with a partition of 4 based on the conjugacy class of the permutation. This table shows the hyperoctahedron facet (tesseract vertex) cycle indices for each member of such a class. The first formula is obtained by averaging these cycle indices after replacing x_i^j with n^j according to the Pólya enumeration theorem.
Partition Count Odd Cycle Indices
4 6 8x_1^2x_2^1x_4^3
31 8 8x_2^2x_6^2
22 3 8x_4^4
211 6 2x_1^8x_2^4 + 2x_2^8 + 4x_4^4
1111 1 8x_2^8

Crossrefs

Cf. A337952 (oriented), A128767 (unoriented), A337954 (chiral).
Other elements: A331361 (tesseract edges, hyperoctahedron faces), A331357 (tesseract faces, hyperoctahedron edges), A337958 (tesseract facets, hyperoctahedron vertices).
Other polychora: A132366(n-1) (4-simplex facets/vertices), A338951 (24-cell), A338967 (120-cell, 600-cell).
Row 4 of A325015 (orthoplex facets, orthotope vertices).

Programs

  • Mathematica
    Table[(3n^12+5n^8+12n^6+28n^4)/48,{n,30}]

Formula

a(n) = n^4 * (3*n^8 + 5*n^4 + 12*n^2 + 28) / 48.
a(n) = 1*C(n,1) + 306*C(n,2) + 33207*C(n,3) + 921908*C(n,4) + 10359075*C(n,5) + 59584470*C(n,6) + 197644440*C(n,7) + 400752240*C(n,8) + 505197000*C(n,9) + 386694000*C(n,10) + 164656800*C(n,11) + 29937600*C(n,12), where the coefficient of C(n,k) is the number of achiral colorings using exactly k colors.
a(n) = 2*A128767(n) - A337952(n) = A337952(n) - 2*A337954(n) = A128767(n) - A337954(n).

A337958 Number of achiral 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, 700, 2850, 9261, 25480, 61776, 135675, 275275, 523446, 943020, 1623076, 2686425, 4298400, 6677056, 10104885, 14942151, 21641950, 30767100, 43008966, 59208325, 80378376, 107730000, 142699375, 186978051, 242545590
Offset: 1

Views

Author

Robert A. Russell, Oct 03 2020

Keywords

Comments

An achiral coloring is identical to its reflection. 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. A337956 (oriented), A337956 (unoriented), A234249(n+1) (chiral).
Other elements: A331357 (hyperoctahedron edges, tesseract faces), A331361 (hyperoctahedron faces, tesseract edges), A337955 (hyperoctahedron facets, tesseract vertices).
Other polychora: A132366(n-1) (5-cell), A338951 (24-cell), A338967 (120-cell, 600-cell).
Row 4 of A325007 (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^2 * (n+1)^2 * (n+3) * (n^2 -2n +4) / 48.
a(n) = 1*C(n,1) + 13*C(n,2) + 84*C(n,3) + 282*C(n,4) + 465*C(n,5) + 360*C(n,6) + 105*C(n,7), where the coefficient of C(n,k) is the number of achiral colorings using exactly k colors.
a(n) = 2*A337957(n) - A337956(n) = A337956(n) - 2 * A234249(n+1) = A337957(n) - A234249(n+1).
From Stefano Spezia, Oct 04 2020: (Start)
G.f.: x*(1 + 7*x + 34*x^2 + 56*x^3 + 8*x^4 - x^5)/(1 - x)^8.
a(n) = 8*a(n-1) - 28*a(n-2) + 56*a(n-3) - 70*a(n-4) + 56*a(n-5) - 28*a(n-6) + 8*a(n-7) - a(n-8) for n > 8.
(End)

A337890 Array read by descending antidiagonals: T(n,k) is the number of achiral colorings of the square faces of a regular n-dimensional orthotope (hypercube) using k or fewer colors.

Original entry on oeis.org

1, 2, 1, 3, 10, 1, 4, 55, 8200, 1, 5, 200, 9080559, 199556208371776, 1, 6, 560, 1503323520, 1370366433970979158839987, 388032967149969852957120195660938882809069568, 1
Offset: 2

Views

Author

Robert A. Russell, Sep 28 2020

Keywords

Comments

An achiral arrangement is identical to its reflection. Each face is a square bounded by four edges. For n=2, the figure is a square with one face. For n=3, the figure is a cube with 6 faces. For n=4, the figure is a tesseract with 24 faces. The number of faces is 2^(n-2)*C(n,2).
Also the number of chiral pairs of colorings of peaks of an n-dimensional orthoplex. A peak is an (n-3)-dimensional simplex.
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             6              7 ...
1   10      55        200         560          1316           2730 ...
1 8200 9080559 1503323520 81461669375 2146080958056 34228350856910 ...
		

Crossrefs

Cf. A337887 (oriented), A337888 (unoriented), A337889 (chiral).
Other elements: A325015 (vertices), A337410 (edges).
Other polytopes: A337886 (simplex), A337894 (orthoplex).
Rows 2-4 are A000027, A337897, A331357.

Programs

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

Formula

T(n,k) = 2*A337888(n,k) - A337887(n,k) = A337887(n,k) - 2*A337889(n,k) = A337888(n,k) - A337889(n,k).
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