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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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.

A208544 T(n,k) = Number of n-bead necklaces of k colors allowing reversal, with no adjacent beads having the same color.

Original entry on oeis.org

1, 2, 0, 3, 1, 0, 4, 3, 0, 0, 5, 6, 1, 1, 0, 6, 10, 4, 6, 0, 0, 7, 15, 10, 21, 3, 1, 0, 8, 21, 20, 55, 24, 13, 0, 0, 9, 28, 35, 120, 102, 92, 9, 1, 0, 10, 36, 56, 231, 312, 430, 156, 30, 0, 0, 11, 45, 84, 406, 777, 1505, 1170, 498, 29, 1, 0, 12, 55, 120, 666, 1680, 4291, 5580, 4435
Offset: 1

Views

Author

R. H. Hardin, Feb 27 2012

Keywords

Comments

Table starts
.1.2..3...4....5.....6......7......8.......9......10......11.......12.......13
.0.1..3...6...10....15.....21.....28......36......45......55.......66.......78
.0.0..1...4...10....20.....35.....56......84.....120.....165......220......286
.0.1..6..21...55...120....231....406.....666....1035....1540.....2211.....3081
.0.0..3..24..102...312....777...1680....3276....5904....9999....16104....24882
.0.1.13..92..430..1505...4291..10528...23052...46185...86185...151756...254618
.0.0..9.156.1170..5580..19995..58824..149796..341640..714285..1391940..2559414
.0.1.30.498.4435.25395.107331.365260.1058058.2707245.6278140.13442286.26942565

Examples

			All solutions for n=7, k=3:
..1....1....1....1....1....1....1....1....1
..2....2....2....2....2....2....2....2....2
..3....3....1....1....3....1....3....1....3
..1....1....2....2....1....2....2....3....2
..2....3....3....3....3....1....3....1....3
..3....1....1....2....2....2....2....2....1
..2....3....3....3....3....3....3....3....3
		

Crossrefs

Main diagonal is A208538.
Columns 3..7 are A208539, A208540, A208541, A208542, A208543.
Row 2 is A000217(n-1).
Row 3 is A000292(n-2).
Row 4 is A002817(n-1).
Row 5 is A164938(n-1).
Row 6 is A027670(n-1).

Programs

  • Mathematica
    T[n_, k_] := If[n == 1, k, (DivisorSum[n, EulerPhi[n/#]*(k-1)^#&]/n + If[ OddQ[n], 1-k, k*(k-1)^(n/2)/2])/2]; Table[T[n-k+1, k], {n, 1, 12}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Oct 30 2017, after Andrew Howroyd *)
  • PARI
    T(n, k) = if(n==1, k, (sumdiv(n, d, eulerphi(n/d)*(k-1)^d)/n + if(n%2, 1-k, k*(k-1)^(n/2)/2))/2);
    for(n=1, 10, for(k=1, 10, print1(T(n, k), ", ")); print) \\ Andrew Howroyd, Oct 14 2017

Formula

T(2n+1,k) = A208535(2n+1,k)/2 for n > 0, T(2n,k) = (A208535(2n,k) + (k*(k-1)^n)/2)/2. - Andrew Howroyd, Mar 12 2017
Empirical for row n:
n=1: a(k) = k
n=2: a(k) = (1/2)*k^2 - (1/2)*k
n=3: a(k) = (1/6)*k^3 - (1/2)*k^2 + (1/3)*k
n=4: a(k) = (1/8)*k^4 - (1/4)*k^3 + (3/8)*k^2 - (1/4)*k
n=5: a(k) = (1/10)*k^5 - (1/2)*k^4 + k^3 - k^2 + (2/5)*k
n=6: a(k) = (1/12)*k^6 - (1/2)*k^5 + (3/2)*k^4 - (7/3)*k^3 + (23/12)*k^2 - (2/3)*k
n=7: a(k) = (1/14)*k^7 - (1/2)*k^6 + (3/2)*k^5 - (5/2)*k^4 + (5/2)*k^3 - (3/2)*k^2 + (3/7)*k

A232713 Doubly pentagonal numbers: a(n) = n*(3*n-2)*(3*n-1)*(3*n+1)/8.

Original entry on oeis.org

0, 1, 35, 210, 715, 1820, 3876, 7315, 12650, 20475, 31465, 46376, 66045, 91390, 123410, 163185, 211876, 270725, 341055, 424270, 521855, 635376, 766480, 916895, 1088430, 1282975, 1502501, 1749060, 2024785, 2331890, 2672670, 3049501, 3464840, 3921225, 4421275
Offset: 0

Views

Author

Bruno Berselli, Nov 28 2013

Keywords

Crossrefs

Cf. similar sequences: A000583 for A000290(A000290(n)); A002817 for A000217(A000217(n)); A063249 for A000384(A000384(n)).

Programs

  • Magma
    [n*(3*n-2)*(3*n-1)*(3*n+1)/8: n in [0..40]];
    
  • Mathematica
    Table[n (3 n - 2) (3 n - 1) (3 n + 1)/8, {n, 0, 40}]
  • PARI
    a(n)=n*(3*n-2)*(3*n-1)*(3*n+1)/8 \\ Charles R Greathouse IV, Oct 07 2015

Formula

G.f.: x*(1 + 30*x + 45*x^2 + 5*x^3) / (1 - x)^5.
a(n) = A000326(A000326(n)) = A000332(3n+1).
From Amiram Eldar, Aug 25 2022: (Start)
Sum_{n>=1} 1/a(n) = 4 + 2*Pi/sqrt(3) - 6*log(3).
Sum_{n>=1} (-1)^(n+1)/a(n) = 32*log(2)/3 - 4*Pi/(3*sqrt(3)) - 4. (End)

A325013 Array read by descending antidiagonals: A(n,k) is the number of unoriented colorings of the facets of a regular n-dimensional orthoplex using up to k colors.

Original entry on oeis.org

1, 3, 1, 6, 6, 1, 10, 21, 22, 1, 15, 55, 267, 402, 1, 21, 120, 1996, 132102, 1228158, 1, 28, 231, 10375, 11756666, 484086357207, 400507806843728, 1, 36, 406, 41406, 405385550, 4805323147589984, 74515759884862073604656433, 527471432057653004017274030725792, 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 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 up to k colors.

Examples

			Array begins with A(1,1):
1   3      6       10        15         21          28           36 ...
1   6     21       55       120        231         406          666 ...
1  22    267     1996     10375      41406      135877       384112 ...
1 402 132102 11756666 405385550 7416923886 86986719477 735192450952 ...
For A(2,2)=6, two squares have all edges the same color, two have three edges the same color, one has opposite edges the same color, and one has opposite edges different colors.
		

Crossrefs

Cf. A325012 (oriented), A325014 (chiral), A325015 (achiral), A325017 (exactly k colors).
Other n-dimensional polytopes: A325000 (simplex), A325005 (orthotope).
Rows 1-4 are A000217, A002817, A128766, A128767; column 2 is A000616.

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[#] + CI1[#]) pc[#]) & /@ IntegerPartitions[n]])/(n! 2^n)] /. 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) = A325012(n,k) - A325014(n,k) = (A325012(n,k) + A325015(n,k)) / 2 = A325014(n,k) + A325015(n,k).
A(n,k) = Sum_{j=1..2^n} A325017(n,j) * binomial(k,j).

A283033 Number of inequivalent 5 X 5 matrices with entries in {1,2,3,...,n} up to rotations and reflections.

Original entry on oeis.org

0, 1, 4211744, 105918450471, 140738033618944, 37252918396015625, 3553786240466361696, 167633579843887699759, 4722366500530551259136, 89737248564744874067889, 1250000000501250002500000, 13543382431328404683826391, 119245270812803151147085824
Offset: 0

Views

Author

David Nacin, Feb 27 2017

Keywords

Comments

Cycle index of dihedral group D4 acting on the 25 entries is (2*s(4)^6*s(1) + s(2)^{12}*s(1) + 4*s(2)^10*s(1)^5 + s(1)^25)/8.

Examples

			For n=2 we get a(2)=4211744 inequivalent 5 X 5 binary matrices up to rotations and reflections.
		

Crossrefs

Row n=5 of A343097.
Cf. A217338 (4 X 4 version), A217331 (3 X 3 version), A002817 (2 X 2 version).

Programs

  • GAP
    List([0..20], n -> n^7*(n^18+4*n^8+n^6+2)/8); # G. C. Greubel, Dec 07 2018
  • Magma
    [n^7*(n^18+4*n^8+n^6+2)/8: n in [0..20]]; // G. C. Greubel, Dec 07 2018
    
  • Maple
    [n^7*(n^18+4*n^8+n^6+2)/8$n=0..16]; # Muniru A Asiru, Dec 07 2018
  • Mathematica
    Table[n^7 (n^18 + 4 n^8 + n^6 + 2)/8, {n, 0, 16}]
  • PARI
    a(n) = n^7*(n^18 + 4*n^8 + n^6 + 2)/8; \\ Indranil Ghosh, Feb 27 2017
    
  • Python
    def A283033(n): return n**7*(n**18 + 4*n**8 + n**6 + 2)/8 # Indranil Ghosh, Feb 27 2017
    
  • Sage
    [n^7*(n^18+4*n^8+n^6+2)/8 for n in range(20)] # G. C. Greubel, Dec 07 2018
    

Formula

a(n) = n^7*(n^18 + 4*n^8 + n^6 + 2)/8.
From Chai Wah Wu, Dec 07 2018: (Start)
a(n) = 26*a(n-1) - 325*a(n-2) + 2600*a(n-3) - 14950*a(n-4) + 65780*a(n-5) - 230230*a(n-6) + 657800*a(n-7) - 1562275*a(n-8) + 3124550*a(n-9) - 5311735*a(n-10) + 7726160*a(n-11) - 9657700*a(n-12) + 10400600*a(n-13) - 9657700*a(n-14) + 7726160*a(n-15) - 5311735*a(n-16) + 3124550*a(n-17) - 1562275*a(n-18) + 657800*a(n-19) - 230230*a(n-20) + 65780*a(n-21) - 14950*a(n-22) + 2600*a(n-23) - 325*a(n-24) + 26*a(n-25) - a(n-26) for n > 25.
G.f.: x*(x^24 + 4211718*x^23 + 105808945452*x^22 + 137985522720898*x^21 + 33628142067806706*x^20 + 2630674898090394666*x^19 + 86978000386844370748*x^18 + 1424113432167998385342*x^17 + 12744486540004851097263*x^16 + 66464282669989885009756*x^15 + 210673587611186802329496*x^14 + 416826570643036689533748*x^13 + 522455888740564118388412*x^12 + 416826570643036689533748*x^11 + 210673587611186802329496*x^10 + 66464282669989885009756*x^9 + 12744486540004851097263*x^8 + 1424113432167998385342*x^7 + 86978000386844370748*x^6 + 2630674898090394666*x^5 + 33628142067806706*x^4 + 137985522720898*x^3 + 105808945452*x^2 + 4211718*x + 1)/(x - 1)^26. (End)

A337412 Array read by descending antidiagonals: T(n,k) is the number of unoriented 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, 21, 144, 1, 5, 55, 12111, 49127, 1, 6, 120, 358120, 740360358, 293122232, 1, 7, 231, 5131650, 733776248840, 3168520600399659, 25174334733080, 1, 8, 406, 45528756, 155261523065875, 314848558732420555904, 920040738175691418086226, 30035285091978202824, 1
Offset: 1

Views

Author

Robert A. Russell, Aug 26 2020

Keywords

Comments

Each chiral pair is counted as one when enumerating unoriented 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 unoriented 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    21     55     120      231       406        666       1035 ...
1 144 12111 358120 5131650 45528756 288936634 1433251296 5887880415 ...
For T(2,2)=6, the arrangements are AAAA, AAAB, AABB, ABAB, ABBB, and BBBB.
		

Crossrefs

Cf. A337411 (oriented), A337413 (chiral), A337414 (achiral).
Rows 1-4 are A000027, A002817, A199406, A331355.
Cf. A327084 (simplex edges), A337408 (orthotope edges), A325005 (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; 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)]
    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) = A337411(n,k) - A337413(n,k) = (A337411(n,k) + A337414(n,k)) / 2 = A337413(n,k) + A337414(n,k).

A058389 Number of 3 X 3 matrices with nonnegative integer entries and all row sums equal to n, up to row and column permutation.

Original entry on oeis.org

1, 3, 14, 44, 129, 316, 714, 1452, 2775, 4963, 8478, 13838, 21827, 33306, 49504, 71754, 101871, 141807, 194128, 261570, 347633, 456026, 591384, 758596, 963657, 1212861, 1513806, 1874440, 2304225, 2813030, 3412466, 4114608, 4933519
Offset: 0

Views

Author

Vladeta Jovovic, Nov 24 2000

Keywords

Crossrefs

Programs

  • Mathematica
    a[n_] := (m = Mod[n, 6]; (n^3 + 9*n^2 + 39*n + 120)*n^3 + Which[m == 0, 12*(23*n^2 + 32*n + 24), m == 1 || m == 5, 249*n^2 + 303*n + 143, m == 2 || m == 4, 4*(69*n^2 + 96*n + 56), m == 3, 3*(83*n^2 + 101*n + 69)])/288; Table[a[n], {n, 0, 32}] (* Jean-François Alcover, Oct 12 2011, after Vladeta Jovovic *)
  • PARI
    \\ See A318951 for RowSumMats
    a(n)=RowSumMats(3, 3, n); \\ Andrew Howroyd, Sep 05 2018

Formula

a(n) = (1/6)*(C(C(n + 2, 2) + 2, 3) + 3/2*floor((n + 2)/2)*(C(n + 2, 2) - floor((n + 2)/2)) + 3*C(floor((n + 2)/2) + 2, 3) + 2*floor(C(n + 2, 2)/3) + 2*C(C(n + 2, 2) - 3*floor(C(n + 2, 2)/3) + 2, 3)).
Empirical G.f.: -(x^8 + 3*x^7 + 14*x^6 + 12*x^5 + 15*x^4 + 9*x^3 + 5*x^2 + 1) / ((x-1)^7*(x+1)^3*(x^2+x+1)). - Colin Barker, Dec 27 2012

Extensions

More terms from Marc LeBrun, Dec 11 2000

A060446 Number of ways to color vertices of a pentagon using <= n colors, allowing rotations and reflections.

Original entry on oeis.org

0, 1, 8, 39, 136, 377, 888, 1855, 3536, 6273, 10504, 16775, 25752, 38233, 55160, 77631, 106912, 144449, 191880, 251047, 324008, 413049, 520696, 649727, 803184, 984385, 1196936, 1444743, 1732024, 2063321, 2443512, 2877823
Offset: 0

Views

Author

N. J. A. Sloane, Apr 07 2001

Keywords

Comments

a(n) is also the number of 5-cycles in the (n+4)-path complement graph, - Eric W. Weisstein, Apr 11 2018

Crossrefs

Cf. A054620.
Cf. A000292 (3-cycle count of \bar P_{n+4}), A002817 (4-cycle count of \bar P_{n+4}), A302695 (6-cycle count of \bar P_{n+5}).

Programs

  • Mathematica
    Table[n (n^2 + 1) (n^2 + 4)/10, {n, 0, 20}] (* Eric W. Weisstein, Apr 11 2018 *)
    LinearRecurrence[{6, -15, 20, -15, 6, -1}, {1, 8, 39, 136, 377, 888}, {0, 20}] (* Eric W. Weisstein, Apr 11 2018 *)
    CoefficientList[Series[x (1 + 2 x + 6 x^2 + 2 x^3 + x^4)/(-1 + x)^6, {x, 0, 20}], x] (* Eric W. Weisstein, Apr 11 2018 *)
  • PARI
    for (n=0, 1000, write("b060446.txt", n, " ", (n^5 + 5*n^3 + 4*n)/10); ) \\ Harry J. Smith, Jul 05 2009

Formula

a(n) = (n^5+5*n^3+4*n)/10.
G.f.: x*(1+2*x+6*x^2+2*x^3+x^4)/(1-x)^6. - Colin Barker, Jan 29 2012
a(n) = 6*a(n-1) - 15*a(n-2) + 20*a(n-3) - 15*a(n-4) + 6*a(n-5) - a(n-6). - Eric W. Weisstein, Apr 11 2018

A337408 Array read by descending antidiagonals: T(n,k) is the number of unoriented colorings of the edges of a regular n-dimensional orthotope (hypercube) using k or fewer colors.

Original entry on oeis.org

1, 2, 1, 3, 6, 1, 4, 21, 144, 1, 5, 55, 12111, 11251322, 1, 6, 120, 358120, 4825746875682, 314824456456819827136, 1, 7, 231, 5131650, 48038446526132256, 38491882660019692002988737797054040, 136221825854745676520058554256163406987047485113810944, 1
Offset: 1

Views

Author

Robert A. Russell, Aug 26 2020

Keywords

Comments

Each chiral pair is counted as one when enumerating unoriented 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 a cube with 12 edges. The number of edges is n*2^(n-1).
Also the number of unoriented colorings of the regular (n-2)-dimensional simplexes in a regular n-dimensional orthoplex.

Examples

			Table begins with T(1,1):
1   2     3      4       5        6         7          8          9 ...
1   6    21     55     120      231       406        666       1035 ...
1 144 12111 358120 5131650 45528756 288936634 1433251296 5887880415 ...
For T(2,2)=6, the arrangements are AAAA, AAAB, AABB, ABAB, ABBB, and BBBB.
		

Crossrefs

Cf. A337407 (oriented), A337409 (chiral), A337410 (achiral).
Rows 1-4 are A000027, A002817, A199406, A331359.
Cf. A327084 (simplex edges), A337412 (orthoplex edges), A325013 (orthotope 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, 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,7}, {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) = A337407(n,k) - A337409(n,k) = (A337407(n,k) - A337410(n,k)) / 2 = A337409(n,k) + A337410(n,k).

A063249 Doubly hexagonal numbers.

Original entry on oeis.org

0, 1, 66, 435, 1540, 4005, 8646, 16471, 28680, 46665, 72010, 106491, 152076, 210925, 285390, 378015, 491536, 628881, 793170, 987715, 1216020, 1481781, 1788886, 2141415, 2543640, 3000025, 3515226, 4094091, 4741660, 5463165
Offset: 0

Views

Author

Henry Bottomley, Jul 11 2001

Keywords

Examples

			a(2)=66 since 6 is the 2nd hexagonal number and 66 is the 6th hexagonal number.
		

Crossrefs

Cf. A002817 for doubly triangular numbers, A000583 for doubly square numbers and A232713 for doubly pentagonal numbers.

Programs

  • Mathematica
    LinearRecurrence[{5,-10,10,-5,1},{0,1,66,435,1540},30] (* Harvey P. Dale, Dec 02 2016 *)
  • PARI
    concat(0, Vec(-x*(15*x^3+115*x^2+61*x+1)/(x-1)^5 + O(x^100))) \\ Colin Barker, Sep 14 2014

Formula

a(n) = n*(2*n-1)(4*n^2-2*n-1) = A000384(A000384(n)).
G.f.: -x*(15*x^3+115*x^2+61*x+1) / (x-1)^5. - Colin Barker, Sep 14 2014
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