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 1-7 of 7 results.

A093566 a(n) = n*(n-1)*(n-2)*(n-3)*(n^2-3*n-2)/48.

Original entry on oeis.org

0, 0, 0, 0, 1, 20, 120, 455, 1330, 3276, 7140, 14190, 26235, 45760, 76076, 121485, 187460, 280840, 410040, 585276, 818805, 1125180, 1521520, 2027795, 2667126, 3466100, 4455100, 5668650, 7145775, 8930376, 11071620, 13624345, 16649480, 20214480
Offset: 0

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Author

Keywords

Comments

a(n+1) is the number of chiral pairs of colorings of the faces of a cube (vertices of a regular octahedron) using n or fewer colors. - Robert A. Russell, Sep 28 2020

Examples

			For a(3+1) = 1, each of the three colors is applied to a pair of adjacent faces of the cube (vertices of the octahedron). - _Robert A. Russell_, Sep 28 2020
		

Crossrefs

From Robert A. Russell, Sep 28 2020: (Start)
Cf. A047780 (oriented), A198833 (unoriented), A337898 (achiral) colorings.
a(n+1) = A325006(3,n) (chiral pairs of colorings of orthotope facets or orthoplex vertices).
a(n+1) = A337889(3,n) (chiral pairs of colorings of orthotope faces or orthoplex peaks).
Other polyhedra: A000332 (tetrahedron), A337896 (cube/octahedron).
(End)

Programs

  • Mathematica
    Table[ Binomial[ Binomial[n-1, 2], 3], {n,0,32}]
    LinearRecurrence[{7,-21,35,-35,21,-7,1},{0,0,0,0,1,20,120},40] (* Harvey P. Dale, Feb 18 2016 *)
  • PARI
    a(n)=n*(n-1)*(n-2)*(n-3)*(n^2-3*n-2)/48 \\ Charles R Greathouse IV, Jun 11 2015
  • Sage
    [(binomial(binomial(n,2),3)) for n in range(-1, 33)] # Zerinvary Lajos, Nov 30 2009
    

Formula

a(n) = binomial(binomial(n-1, 2), 3).
G.f.: -x^4*(1+13*x+x^2)/(x-1)^7. - R. J. Mathar, Dec 08 2010
a(n+1) = 1*C(n,3) + 16*C(n,4) + 30*C(n,5) + 15*C(n,6), where the coefficient of C(n,k) is the number of chiral pairs of colorings using exactly k colors. - Robert A. Russell, Sep 28 2020
a(n) = A000217(n-1)*A239352(n-2)/6. - R. J. Mathar, Mar 25 2022

Extensions

Edited (with a new definition) by N. J. A. Sloane, Jul 02 2008

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).

A337885 Array read by descending antidiagonals: T(n,k) is the number of chiral pairs of colorings of the triangular faces of a regular n-dimensional simplex using k or fewer colors.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 0, 6, 0, 0, 1, 405, 1368, 0, 0, 5, 7904, 4775706, 6711288, 0, 0, 15, 76880, 1522540416, 9923557498416, 1785683627824, 0, 0, 35, 486522, 132342705750, 234239763858347776, 12979826761630383196344, 53302696800142157920, 0
Offset: 2

Views

Author

Robert A. Russell, Sep 28 2020

Keywords

Comments

Each member of a chiral pair is a reflection, but not a rotation, of the other. An n-simplex has n+1 vertices. For n=2, the figure is a triangle with one triangular face. For n=3, the figure is a tetrahedron with 4 triangular faces. For higher n, the number of triangular faces is C(n+1,3).
Also the number of chiral pairs of colorings of the peaks of a regular n-dimensional simplex. A peak of an n-simplex is an (n-3)-dimensional simplex.

Examples

			Table begins with T(2,1):
 0    0       0          0            0             0               0 ...
 0    0       0          1            5            15              35 ...
 0    6     405       7904        76880        486522         2300305 ...
 0 1368 4775706 1522540416 132342705750 5076500214744 110809322249220 ...
For T(3,4)=1, the chiral pair is ABCD-ABDC.
		

Crossrefs

Cf. A337883 (oriented), A337884 (unoriented), A337886 (achiral), A051168 (binary Lyndon words).
Other elements: A325000(n,k-n) (vertices), A327085 (edges).
Other polytopes: A337889 (orthotope), A337893 (orthoplex).
Rows 2-4 are A000004, A000332, A331352.

Programs

  • Mathematica
    m=2; (* dimension of color element, here a triangular face *)
    lw[n_,k_]:=lw[n, k]=DivisorSum[GCD[n,k],MoebiusMu[#]Binomial[n/#,k/#]&]/n (*A051168*)
    cxx[{a_, b_},{c_, d_}]:={LCM[a, c], GCD[a, c] b d}
    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)
    combine[a : {{, } ...}, b : {{, } ...}] := Outer[cxx, a, b, 1]
    CX[p_List, 0] := {{1, 1}} (* cycle index for partition p, m vertices *)
    CX[{n_Integer}, m_] := If[2m>n, CX[{n}, n-m], CX[{n},m] = Table[{n/k, lw[n/k, m/k]}, {k, Reverse[Divisors[GCD[n, m]]]}]]
    CX[p_List, m_Integer] := CX[p, m] = Module[{v = Total[p], q, r}, If[2 m > v, CX[p, v - m], q = Drop[p, -1]; r = Last[p]; compress[Flatten[Join[{{CX[q, m]}}, Table[combine[CX[q, m - j], CX[{r}, j]], {j, Min[m, r]}]], 2]]]]
    pc[p_] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #] &/@ mb; Total[p]!/(Times @@ (ci!) Times @@ (mb^ci))] (* partition count *)
    row[n_Integer] := row[n] = Factor[Total[If[EvenQ[Total[1-Mod[#, 2]]],1,-1] pc[#] j^Total[CX[#, m+1]][[2]] & /@ IntegerPartitions[n+1]]/(n+1)!]
    array[n_, k_] := row[n] /. j -> 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 vertices to a partition of n+1. It then determines the number of permutations for each partition and the cycle index for each partition using a formula for binary Lyndon words. If the value of m is increased, one can enumerate colorings of higher-dimensional elements beginning with T(m,1).
T(n,k) = A337883(n,k) - A337884(n,k) = (A337883(n,k) - A337886(n,k)) / 2 = A337884(n,k) - A337886(n,k).

A337887 Array read by descending antidiagonals: T(n,k) is the number of oriented 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, 57, 90054, 1, 5, 240, 1471640157, 629648865588086369152, 1, 6, 800, 1466049174160, 76983765319971901895960429658208179, 76686070519895153193719509580895099970955878067526648007224125292544, 1
Offset: 2

Views

Author

Robert A. Russell, Sep 28 2020

Keywords

Comments

Each chiral pair is counted as two when enumerating oriented arrangements. 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 oriented colorings of peaks of an n-dimensional orthoplex. A peak is an (n-3)-dimensional simplex.

Examples

			Array begins with T(2,1):
 1     2          3             4               5                 6 ...
 1    10         57           240             800              2226 ...
 1 90054 1471640157 1466049174160 310441584462375 24679078461920106 ...
		

Crossrefs

Cf. A337888 (unoriented), A337889 (chiral), A337890 (achiral).
Other elements: A325012 (vertices), A337407 (edges).
Other polytopes: A337883 (simplex), A337891 (orthoplex).
Rows 2-4 are A000027, A047780, A331354.

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}]], (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[array[n,d+m-n], {d,6}, {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) = A337888(n,k) + A337889(n,k) = 2*A337888(n,k) - A337890(n,k) = 2*A337889(n,k) + A337890(n,k).

A337888 Array read by descending antidiagonals: T(n,k) is the number of unoriented 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, 56, 49127, 1, 5, 220, 740360358, 314824532572147370464, 1, 6, 680, 733776248840, 38491882660671134164965704408524083, 38343035259947576596859948806931124970404417593861154473053467181056, 1
Offset: 2

Views

Author

Robert A. Russell, Sep 28 2020

Keywords

Comments

Each chiral pair is counted as one when enumerating unoriented arrangements. 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 unoriented colorings of peaks of an n-dimensional orthoplex. A peak is an (n-3)-dimensional simplex.

Examples

			Array begins with T(2,1):
 1     2         3            4               5                 6 ...
 1    10        56          220             680              1771 ...
 1 49127 740360358 733776248840 155261523065875 12340612271439081 ...
		

Crossrefs

Cf. A337887 (oriented), A337889 (chiral), A337890 (achiral).
Other elements: A325013 (vertices), A337408 (edges).
Other polytopes: A337884 (simplex), A337892 (orthoplex).
Rows 2-4 are A000027, A198833, A331355.

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; 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)]
    array[n_, k_] := row[n] /. b -> k
    Table[array[n,d+m-n], {d,6}, {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) = A337887(n,k) - A337889(n,k) = (A337887(n,k) + A337890(n,k)) / 2 = A337889(n,k) + A337890(n,k).

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).

A337893 Array read by descending antidiagonals: T(n,k) is the number of chiral pairs of colorings of the faces of a regular n-dimensional orthoplex (cross polytope) using k or fewer colors.

Original entry on oeis.org

0, 0, 0, 0, 1, 0, 0, 66, 11158298, 0, 0, 920, 4825452718593, 314824333015938998688, 0, 0, 6350, 48038354542204960, 38491882659300767730994725249684096, 31716615393292685397985382790580028572676096, 0
Offset: 2

Views

Author

Robert A. Russell, Sep 28 2020

Keywords

Comments

Each member of a chiral pair is a reflection, but not a rotation, of the other. 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 chiral pairs of colorings of the peaks of an n-dimensional orthotope (hypercube). A peak is an (n-3)-dimensional orthotope.

Examples

			Table begins with T(2,1):
 0        0             0                 0                    0 ...
 0        1            66               920                 6350 ...
 0 11158298 4825452718593 48038354542204960 60632976384183154375 ...
		

Crossrefs

Cf. A337891 (oriented), A337892 (unoriented), A337894 (achiral).
Other elements: A325006 (vertices), A337413 (edges).
Other polytopes: A337885 (simplex), A337889 (orthotope).
Rows 2-4 are A000004, A337896, A331360.

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; per = LCM @@ Table[If[cs[[j2]] == r1[[j2]], If[0 == cs[[j2]],1,j2], 2j2], {j2,n}]; If[EvenQ[Sum[If[EvenQ[j3], r1[[j3]], r2[[j3]]], {j3,n}]],1,-1]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)]
    array[n_, k_] := row[n] /. b -> k
    Table[array[n,d+m-n], {d,6}, {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) = A337891(n,k) - A337892(n,k) = (A337891(n,k) - A337894(n,k)) / 2 = A337892(n,k) - A337894(n,k).
Showing 1-7 of 7 results.