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.

A037270 a(n) = n^2*(n^2 + 1)/2.

Original entry on oeis.org

0, 1, 10, 45, 136, 325, 666, 1225, 2080, 3321, 5050, 7381, 10440, 14365, 19306, 25425, 32896, 41905, 52650, 65341, 80200, 97461, 117370, 140185, 166176, 195625, 228826, 266085, 307720, 354061, 405450, 462241, 524800, 593505, 668746, 750925, 840456, 937765
Offset: 0

Views

Author

Aaron Gulliver (gulliver(AT)elec.canterbury.ac.nz)

Keywords

Comments

Sum of first n^2 positive integers.
Start from xanthene and attach amino acids according to the reaction scheme that describes the reaction between the active sites. See the hyperlink below on chemistry. - Robert G. Wilson v, Aug 02 2002; Amarnath Murthy, Aug 01 2002
Sum of the next n multiples of n. - Amarnath Murthy, Aug 01 2002
The sum of the terms in an n X n spiral. These are also triangular numbers. - William A. Tedeschi, Feb 27 2008
Hypotenuse of Pythagorean triangles with smallest side a cube: A000578(n)^2 + A083374(n)^2 = a(n)^2. - Martin Renner, Nov 12 2011
For n>1, triangular numbers that can be represented as a sum of a square and a triangular number. For example, a(2)=10=4+6=9+1. - Ivan N. Ianakiev, Apr 24 2012
A037270 can be constructed in the following manner: Take A000217 and for every n not in A000290 delete the corresponding A000217(n). - Ivan N. Ianakiev, Apr 26 2012
Starting at a(1)=1 simply take 1*1=1, a(2)= 2*(2+3)=10, a(3)= 3*(4+5+6)=45, a(4)=4*(7+8+9+10) and so on. - J. M. Bergot, May 01 2015
Observation: The digital roots of the terms repeat in the sequence 1, 1, 9; e.g., the digital roots of 1, 10, 45, 136, 325, and 666 are 1, 1, 9, 1, 1, and 9. Verified for the first 10000 terms. - Rob Barton, Mar 28 2018
The above observation is easily explained and proved given that the digital root of a positive number equals the number modulo 9, and a(n + 9k) == a(n) (mod 9). - M. F. Hasler, Apr 05 2018
Number of unoriented rows of length 4 using up to n colors. For a(0)=0, there are no rows using no colors. For a(1)=1, there is one row using that one color for all positions. For a(2)=10, there are 4 achiral (AAAA, ABBA, BAAB, BBBB) and 6 chiral pairs (AAAB-BAAA, AABA-ABAA, AABB-BBAA, ABAB-BABA, ABBB-BBBA, BABB-BBAB). - Robert A. Russell, Nov 14 2018
For n > 0, a(2n+1) is the number of non-isomorphic 6C_m-snakes, where m = 2n+1 or m = 2n (for n>=2). A kC_n-snake is a connected graph in which the k>=2 blocks are isomorphic to the cycle C_n and the block-cutpoint graph is a path. - Christian Barrientos, May 15 2019
Number of achiral colorings of the edges of a tetrahedron with n available colors. - Robert A. Russell, Sep 07 2019

References

  • C. Alsina and R. B. Nelson, Charming Proofs: A Journey into Elegant Mathematics, MAA, 2010. See p. 5.
  • C. Barrientos, Graceful labelings of cyclic snakes, Ars Combin., 60(2001), 85-96.
  • Albert H. Beiler, Recreations in the theory of numbers, New York: Dover, (2nd ed.) 1966, p. 106, table 55.
  • T. A. Gulliver, Sequences from Arrays of Integers, Int. Math. Journal, Vol. 1, No. 4, pp. 323-332, 2002.
  • T. A. Gulliver, Sequences from Cubes of Integers, Int. Math. Journal, 4 (2003), 439-445.
  • R. A. Wilson, Cosmic Trigger, epilogue of S.-P. Sirag.

Crossrefs

Cf. A000217, A236770 (see crossrefs).
Row 4 of A277504.
Cf. A000583 (oriented), A083374 (chiral), A000290 (achiral).
Cf. A317617.
Row 3 of A327086 (achiral simplex edge colorings).

Programs

  • GAP
    a:=List([0..30],n->n^2*(n^2+1)/2); # Muniru A Asiru, Mar 28 2018
    
  • Magma
    [n^2*(n^2 + 1)/2: n in [0..30]] // Stefano Spezia, Jan 15 2019
  • Maple
    seq(n^2*(n^2+1)/2,n=0..30); # Muniru A Asiru, Mar 28 2018
  • Mathematica
    Table[ n^2*((n^2 + 1)/2), {n, 0, 30} ]
    Table[(1/8) Round[N[Sinh[2 ArcSinh[n]]^2, 100]], {n, 0, 30}] (* Artur Jasinski, Feb 10 2010 *)
    LinearRecurrence[{5,-10,10,-5,1},{0,1,10,45,136},30] (* Harvey P. Dale, Aug 03 2014 *)
  • PARI
    a(n)=binomial(n^2+1,2) \\ Charles R Greathouse IV, Apr 25 2012
    
  • Python
    for n in range(0,30): print(n**2*(n**2+1)/2, end=', ') # Stefano Spezia, Jan 10 2019
    

Formula

a(n) = a(n-1) + n^3 + (n-1)^3.
a(n) = A000537(n)+A000537(n-1), i.e., square of sum of first n integers plus square of sum of first n-1 integers. - Henry Bottomley, Oct 15 2001
a(n) = Sum_{k=0..n^2} k. - William A. Tedeschi, Feb 27 2008
a(n) = (1/8)*sinh(2*arcsinh(n)). - Artur Jasinski, Feb 10 2010
G.f.: x*(1+x)*(1+4*x+x^2)/(1-x)^5. - Colin Barker, Mar 22 2012
a(n) = a(n-1) + A005898(n-1). - Ivan N. Ianakiev, May 13 2012
a(n) = 2 * A000217(n-1) * A000217(n) + A000290(n). - Ivan N. Ianakiev, May 26 2012
a(n) = A000217(n^2). - J. M. Bergot, Jun 07 2012
a(n) = 5*a(n-1) -10*a(n-2) +10*a(n-3) -5*a(n-4) +a(n-5) n>4, a(0)=0, a(1)=1, a(2)=10, a(3)=45, a(4)=136. - Yosu Yurramendi, Sep 02 2013
For n>0, a(n) = A000217(n)^2 + A000217(n-1)^2. - Richard R. Forberg, Dec 25 2013
a(n) = T(T(n)) + T(T(n-1)) + T(T(n)-1) + T(T(n-1)-1), where T(n) = A000217(n). - Charlie Marion, Sep 10 2016
a(n) = t(n-3)*t(n)+t(n-1)*t(n+2), with t(n)=A000217(n). - J. M. Bergot, Apr 07 2018
From Robert A. Russell, Nov 14 2018: (Start)
a(n) = (A000583(n) + A000290(n)) / 2 = (n^4 + n^2) / 2.
a(n) = A000583(n) - A083374(n) = A083374(n) + A000290(n).
G.f.: (Sum_{j=1..4} S2(4,j)*j!*x^j/(1-x)^(j+1) + Sum_{j=1..2} S2(2,j)*j!*x^j/(1-x)^(j+1)) / 2, where S2 is the Stirling subset number A008277.
G.f.: Sum_{k=1..4} A145882(4,k) * x^k / (1-x)^5.
E.g.f.: (Sum_{k=1..4} S2(4,k)*x^k + Sum_{k=1..2} S2(2,k)*x^k) * exp(x) / 2, where S2 is the Stirling subset number A008277.
For n>4, a(n) = Sum_{j=1..5} -binomial(j-6,j) * a(n-j). (End)
a(n) = n*A006003(n). - Kritsada Moomuang, Dec 16 2018
For n > 0, a(n) = Sum_{k=1..n} A317617(n,k). - Stefano Spezia, Jan 10 2019
Sum_{n>=1} 1/a(n) = 1 + Pi^2/3 - Pi*coth(Pi) = 1.13652003875929052467672874379... - Vaclav Kotesovec, Jan 21 2019
Sum_{n>=1} (-1)^(n+1)/a(n) = Pi*csch(Pi) + Pi^2/6 - 1. - Amiram Eldar, Nov 02 2021

A327084 Array read by descending antidiagonals: A(n,k) is the number of unoriented colorings of the edges of a regular n-dimensional simplex using up to k colors.

Original entry on oeis.org

1, 2, 1, 3, 4, 1, 4, 10, 11, 1, 5, 20, 66, 34, 1, 6, 35, 276, 792, 156, 1, 7, 56, 900, 10688, 25506, 1044, 1, 8, 84, 2451, 90005, 1601952, 2302938, 12346, 1, 9, 120, 5831, 533358, 43571400, 892341888, 591901884, 274668
Offset: 1

Views

Author

Robert A. Russell, Aug 19 2019

Keywords

Comments

An n-dimensional simplex has n+1 vertices and (n+1)*n/2 edges. For n=1, the figure is a line segment with one edge. For n=2, the figure is a triangle with three edges. For n=3, the figure is a tetrahedron with six edges. The Schläfli symbol, {3,...,3}, of the regular n-dimensional simplex consists of n-1 threes. Two unoriented colorings are the same if congruent; chiral pairs are counted as one.
A(n,k) is also the number of unoriented colorings of (n-2)-dimensional regular simplices in an n-dimensional simplex using up to k colors. Thus, A(2,k) is also the number of unoriented colorings of the vertices (0-dimensional simplices) of an equilateral triangle.

Examples

			Array begins with A(1,1):
  1  2   3     4     5      6       7       8        9       10 ...
  1  4  10    20    35     56      84     120      165      220 ...
  1 11  66   276   900   2451    5831   12496    24651    45475 ...
  1 34 792 10688 90005 533358 2437848 9156288 29522961 84293770 ...
  ...
For A(2,3) = 10, the nine achiral colorings are AAA, AAB, AAC, ABB, ACC, BBB, BBC, BCC, and CCC. The chiral pair is ABC-ACB.
		

Crossrefs

Cf. A327083 (oriented), A327085 (chiral), A327086 (achiral), A327088 (exactly k colors), A325000 (vertices, facets), A337884 (faces, peaks), A337408 (orthotope edges, orthoplex ridges), A337412 (orthoplex edges, orthotope ridges).
Rows 1-4 are A000027, A000292, A063842(n-1), A063843.
Cf. A063841 (k-multigraphs on n nodes).

Programs

  • Mathematica
    CycleX[{2}] = {{1,1}}; (* cycle index for permutation with given cycle structure *)
    CycleX[{n_Integer}] := CycleX[n] = If[EvenQ[n], {{n/2,1}, {n,(n-2)/2}}, {{n,(n-1)/2}}]
    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)
    CycleX[p_List] := CycleX[p] = compress[Join[CycleX[Drop[p, -1]], If[Last[p] > 1, CycleX[{Last[p]}], ## &[]], If[# == Last[p], {#, Last[p]}, {LCM[#, Last[p]], GCD[#, Last[p]]}] & /@ Drop[p, -1]]]
    pc[p_List] := 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[pc[#] j^Total[CycleX[#]][[2]] & /@ IntegerPartitions[n+1]]/(n+1)!]
    array[n_, k_] := row[n] /. j -> k
    Table[array[n,d-n+1], {d,1,10}, {n,1,d}] // Flatten
    (* Using Fripertinger's exponent per Andrew Howroyd code in A063841: *)
    pc[p_] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #] &/@ mb; Total[p]!/(Times @@ (ci!) Times @@ (mb^ci))]
    ex[v_] := Sum[GCD[v[[i]], v[[j]]], {i,2,Length[v]}, {j,i-1}] + Total[Quotient[v,2]]
    array[n_,k_] := Total[pc[#]k^ex[#] &/@ IntegerPartitions[n+1]]/(n+1)!
    Table[array[n,d-n+1], {d,10}, {n,d}] // Flatten
    (* Another program (translated from Andrew Howroyd's PARI code): *)
    permcount[v_] := Module[{m=1, s=0, k=0, t}, For[i=1, i <= Length[v], i++, t = v[[i]]; k = If[i>1 && t == v[[i-1]], k+1, 1]; m *= t*k; s += t]; s!/m];
    edges[v_] := Sum[GCD[v[[i]], v[[j]]], {i, 2, Length[v]}, {j, 1, i-1}] + Total[Quotient[v, 2]];
    T[n_, k_] := Module[{s = 0}, Do[s += permcount[p]*k^edges[p], {p, IntegerPartitions[n+1]}]; s/(n+1)!];
    Table[T[n-k+1, k], {n, 1, 9}, {k, n, 1, -1}] // Flatten (* Jean-François Alcover, Jan 08 2021 *)
  • PARI
    permcount(v) = {my(m=1, s=0, k=0, t); for(i=1, #v, t=v[i]; k=if(i>1&&t==v[i-1], k+1, 1); m*=t*k; s+=t); s!/m}
    edges(v) = {sum(i=2, #v, sum(j=1, i-1, gcd(v[i], v[j]))) + sum(i=1, #v, v[i]\2)}
    T(n, k) = {my(s=0); forpart(p=n+1, s+=permcount(p)*k^edges(p)); s/(n+1)!} \\ Andrew Howroyd, Sep 06 2019
    
  • Python
    from itertools import combinations
    from math import prod, gcd, factorial
    from fractions import Fraction
    from sympy.utilities.iterables import partitions
    def A327084_T(n,k): return int(sum(Fraction(k**(sum(p[r]*p[s]*gcd(r,s) for r,s in combinations(p.keys(),2))+sum((q>>1)*r+(q*r*(r-1)>>1) for q, r in p.items())),prod(q**r*factorial(r) for q, r in p.items())) for p in partitions(n+1))) # Chai Wah Wu, Jul 09 2024

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.
A(n,k) = Sum_{j=1..(n+1)*n/2} A327088(n,j) * binomial(k,j).
A(n,k) = A327083(n,k) - A327085(n,k) = (A327083(n,k) + A327086(n,k)) / 2 = A327085(n,k) + A327086(n,k).
A(n,k) = A063841(n+1,k-1).

A327083 Array read by descending antidiagonals: A(n,k) is the number of oriented colorings of the edges of a regular n-dimensional simplex using up to k colors.

Original entry on oeis.org

1, 2, 1, 3, 4, 1, 4, 11, 12, 1, 5, 24, 87, 40, 1, 6, 45, 416, 1197, 184, 1, 7, 76, 1475, 18592, 42660, 1296, 1, 8, 119, 4236, 166885, 3017600, 4223313, 17072, 1, 9, 176, 10437, 1019880, 85025050, 1748176768, 1139277096, 424992
Offset: 1

Views

Author

Robert A. Russell, Aug 19 2019

Keywords

Comments

An n-dimensional simplex has n+1 vertices and (n+1)*n/2 edges. For n=1, the figure is a line segment with one edge. For n-2, the figure is a triangle with three edges. For n=3, the figure is a tetrahedron with six edges. The Schläfli symbol, {3,...,3}, of the regular n-dimensional simplex consists of n-1 threes. Two oriented colorings are the same if one is a rotation of the other; chiral pairs are counted as two.
A(n,k) is also the number of oriented colorings of (n-2)-dimensional regular simplices in an n-dimensional simplex using up to k colors. Thus, A(2,k) is also the number of oriented colorings of the vertices (0-dimensional simplices) of an equilateral triangle.

Examples

			Array begins with A(1,1):
  1  2    3     4      5       6       7        8        9        10 ...
  1  4   11    24     45      76     119      176      249       340 ...
  1 12   87   416   1475    4236   10437    22912    45981     85900 ...
  1 40 1197 18592 166885 1019880 4738153 17962624 58248153 166920040 ...
  ...
For A(2,3) = 11, the nine achiral colorings are AAA, AAB, AAC, ABB, ACC, BBB, BBC, BCC, and CCC. The chiral pair is ABC-ACB.
		

Crossrefs

Cf. A327084 (unoriented), A327085 (chiral), A327086 (achiral), A327087 (exactly k colors), A324999 (vertices, facets), A337883 (faces, peaks), A337407 (orthotope edges, orthoplex ridges), A337411 (orthoplex edges, orthotope ridges).
Rows 1-4 are A000027, A006527, A046023, A331350.
Column 2 is A218144(n+1).

Programs

  • Mathematica
    CycleX[{2}] = {{1,1}}; (* cycle index for permutation with given cycle structure *)
    CycleX[{n_Integer}] := CycleX[n] = If[EvenQ[n], {{n/2,1}, {n,(n-2)/2}}, {{n,(n-1)/2}}]
    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)
    CycleX[p_List] := CycleX[p] = compress[Join[CycleX[Drop[p,-1]], If[Last[p] > 1, CycleX[{Last[p]}], ## &[]], If[# == Last[p], {#, Last[p]}, {LCM[#, Last[p]], GCD[#, Last[p]]}] & /@ Drop[p,-1]]]
    pc[p_List] := 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]]], pc[#] j^Total[CycleX[#]][[2]], 0] & /@ IntegerPartitions[n+1]]/((n+1)!/2)]
    array[n_, k_] := row[n] /. j -> k
    Table[array[n,d-n+1], {d,1,10}, {n,1,d}] // Flatten
    (* Using Fripertinger's exponent per Andrew Howroyd's code in A063841: *)
    pc[p_] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #] &/@ mb; Total[p]!/(Times @@ (ci!) Times @@ (mb^ci))]
    ex[v_] := Sum[GCD[v[[i]], v[[j]]], {i,2,Length[v]}, {j,i-1}] + Total[Quotient[v,2]]
    array[n_,k_] := Total[If[EvenQ[Total[1-Mod[#,2]]], pc[#]k^ex[#], 0] &/@ IntegerPartitions[n+1]]/((n+1)!/2)
    Table[array[n,d-n+1], {d,10}, {n,d}] // 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.
A(n,k) = Sum_{j=1..(n+1)*n/2} A327087(n,j) * binomial(k,j).
A(n,k) = A327084(n,k) + A327085(n,k) = 2*A327084(n,k) - A327086(n,k) = 2*A327085(n,k) + A327086(n,k).

A327085 Array read by descending antidiagonals: A(n,k) is the number of chiral pairs of colorings of the edges of a regular n-dimensional simplex using up to k colors.

Original entry on oeis.org

0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 4, 21, 6, 0, 0, 10, 140, 405, 28, 0, 0, 20, 575, 7904, 17154, 252, 0, 0, 35, 1785, 76880, 1415648, 1920375, 4726, 0, 0, 56, 4606, 486522, 41453650, 855834880, 547375212, 150324, 0
Offset: 1

Views

Author

Robert A. Russell, Aug 19 2019

Keywords

Comments

An n-dimensional simplex has n+1 vertices and (n+1)*n/2 edges. For n=1, the figure is a line segment with one edge. For n-2, the figure is a triangle with three edges. For n=3, the figure is a tetrahedron with six edges. The Schläfli symbol, {3,...,3}, of the regular n-dimensional simplex consists of n-1 threes. The chiral colorings of its edges come in pairs, each the reflection of the other.
A(n,k) is also the number of chiral pairs of colorings of (n-2)-dimensional regular simplices in an n-dimensional simplex using up to k colors. Thus, A(2,k) is also the number of chiral pairs of colorings of the vertices (0-dimensional simplices) of an equilateral triangle.

Examples

			Array begins with A(1,1):
  0 0   0    0     0      0       0       0        0        0         0 ...
  0 0   1    4    10     20      35      56       84      120       165 ...
  0 1  21  140   575   1785    4606   10416    21330    40425     71995 ...
  0 6 405 7904 76880 486522 2300305 8806336 28725192 82626270 214744629 ...
  ...
For A(2,3) = 1, the chiral pair is ABC-ACB.
		

Crossrefs

Cf. A327083 (oriented), A327084 (unoriented), A327086 (achiral), A327089 (exactly k colors), A325000(n,k-n) (vertices, facets), A337885 (faces, peaks), A337409 (orthotope edges, orthoplex ridges), A337413 (orthoplex edges, orthotope ridges).
Rows 1-4 are A000004, A000292(n-2), A337899, A331352.

Programs

  • Mathematica
    CycleX[{2}] = {{1,1}}; (* cycle index for permutation with given cycle structure *)
    CycleX[{n_Integer}] := CycleX[n] = If[EvenQ[n], {{n/2,1}, {n,(n-2)/2}}, {{n,(n-1)/2}}]
    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)
    CycleX[p_List] := CycleX[p] = compress[Join[CycleX[Drop[p, -1]], If[Last[p] > 1, CycleX[{Last[p]}], ## &[]], If[# == Last[p], {#, Last[p]}, {LCM[#, Last[p]], GCD[#, Last[p]]}] & /@ Drop[p, -1]]]
    pc[p_List] := 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[CycleX[#]][[2]] & /@ IntegerPartitions[n+1]]/(n+1)!]
    array[n_, k_] := row[n] /. j -> k
    Table[array[n,d-n+1], {d,1,10}, {n,1,d}] // Flatten
    (* Using Fripertinger's exponent per Andrew Howroyd's code in A063841: *)
    pc[p_] := Module[{ci, mb}, mb = DeleteDuplicates[p]; ci = Count[p, #] &/@ mb; Total[p]!/(Times @@ (ci!) Times @@ (mb^ci))]
    ex[v_] := Sum[GCD[v[[i]], v[[j]]], {i,2,Length[v]}, {j,i-1}] + Total[Quotient[v,2]]
    array[n_,k_] := Total[If[EvenQ[Total[1-Mod[#,2]]],1,-1] pc[#]k^ex[#] &/@ IntegerPartitions[n+1]]/(n+1)!
    Table[array[n,d-n+1], {d,10}, {n,d}] // 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.
A(n,k) = Sum_{j=1..(n+1)*n/2} A327089(n,j) * binomial(k,j).
A(n,k) = A327083(n,k) - A327084(n,k) = (A327083(n,k) - A327086(n,k)) / 2 = A327084(n,k) - A327086(n,k).

A325001 Array read by descending antidiagonals: A(n,k) is the number of achiral colorings of the facets (or vertices) of a regular n-dimensional simplex using up to k colors.

Original entry on oeis.org

1, 2, 1, 3, 4, 1, 4, 9, 5, 1, 5, 16, 15, 6, 1, 6, 25, 34, 21, 7, 1, 7, 36, 65, 56, 28, 8, 1, 8, 49, 111, 125, 84, 36, 9, 1, 9, 64, 175, 246, 210, 120, 45, 10, 1, 10, 81, 260, 441, 461, 330, 165, 55, 11, 1, 11, 100, 369, 736, 917, 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. An achiral coloring is the same as its reflection.

Examples

			The array begins with A(1,1):
  1  2  3   4   5    6    7     8     9    10    11     12     13 ...
  1  4  9  16  25   36   49    64    81   100   121    144    169 ...
  1  5 15  34  65  111  175   260   369   505   671    870   1105 ...
  1  6 21  56 125  246  441   736  1161  1750  2541   3576   4901 ...
  1  7 28  84 210  461  917  1688  2919  4795  7546  11452  16848 ...
  1  8 36 120 330  792 1715  3424  6399 11320 19118  31032  48672 ...
  1  9 45 165 495 1287 3003  6434 12861 24265 43593  75087 124683 ...
  1 10 55 220 715 2002 5005 11440 24309 48610 92323 167740 293215 ...
  ...
For A(2,2)=4, the triangle may have 0, 1, 2, or 3 edges of one color.
		

Crossrefs

Cf. A324999 (oriented), A325000 (unoriented), A325000(n,k-n) (chiral), A325003 (exactly k colors), A327086 (edges, ridges), A337886 (faces, peaks), A325007 (orthotope facets, orthoplex vertices), A325015 (orthoplex facets, orthotope vertices).
Rows 1-4 are A000027, A000290, A006003, A132366(n-1).
Column 2 is A162880.

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} A325003(n,j) * binomial(k,j).
A(n,k) = 2*A325000(n,k) - A324999(n,k) = A324999(n,k) - 2*A325000(n,k-n) = A325000(n,k) - A325000(n,k-n).
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+1} -binomial(j-n-2,j) * A(n,k-j).
G.f. for column k: (1 - (1-x^2)^k) / (x*(1-x)^k).

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)

A337886 Array read by descending antidiagonals: T(n,k) is the number of achiral colorings of the triangular faces of a regular n-dimensional simplex using k or fewer colors.

Original entry on oeis.org

1, 2, 1, 3, 5, 1, 4, 15, 28, 1, 5, 34, 387, 768, 1, 6, 65, 2784, 202203, 302032, 1, 7, 111, 13125, 11230976, 7109211078, 3098988832, 1, 8, 175, 46836, 254729375, 9393953524224, 50669807706182691, 1831011525739328, 1
Offset: 2

Views

Author

Robert A. Russell, Sep 28 2020

Keywords

Comments

An achiral arrangement is identical to its reflection. 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 achiral 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):
1   2      3        4         5          6           7            8 ...
1   5     15       34        65        111         175          260 ...
1  28    387     2784     13125      46836      137543       349952 ...
1 768 202203 11230976 254729375 3267720576 28271133933 183296831488 ...
For T(3,4)=34, the 34 achiral arrangements are AAAA, AAAB, AAAC, AAAD, AABB, AABC, AABD, AACC, AACD, AADD, ABBB, ABBC, ABBD, ABCC, ABDD, ACCC, ACCD, ACDD, ADDD, BBBB, BBBC, BBBD, BBCC, BBCD, BBDD, BCCC, BCCD, BCDD, BDDD, CCCC, CCCD, CCDD, CDDD, and DDDD.
		

Crossrefs

Cf. A337883 (oriented), A337884 (unoriented), A337885 (chiral), A051168 (binary Lyndon words).
Other elements: A325001 (vertices), A327086 (edges).
Other polytopes: A337890 (orthotope), A337894 (orthoplex).
Rows 2-4 are A000027, A006003, A331353.

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[OddQ[Total[1-Mod[#, 2]]], pc[#] j^Total[CX[#, m+1]][[2]], 0] & /@ IntegerPartitions[n+1]]/((n+1)!/2)]
    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) = A337884(n,k) - A337883(n,k) = A337883(n,k) - 2*A337885(n,k) = A337884(n,k) - A337885(n,k).

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).

A327090 Triangle read by rows: T(n,k) is the number of achiral colorings of the edges of a regular n-dimensional simplex using exactly k colors. Row n has (n+1)*n/2 columns.

Original entry on oeis.org

1, 1, 2, 0, 1, 8, 18, 12, 0, 0, 1, 26, 306, 1400, 2800, 2520, 840, 0, 0, 0, 1, 126, 7971, 153660, 1268475, 5463990, 13534290, 20018880, 17478720, 8316000, 1663200, 0, 0, 0, 0
Offset: 1

Views

Author

Robert A. Russell, Aug 19 2019

Keywords

Comments

An n-dimensional simplex has n+1 vertices and (n+1)*n/2 edges. For n=1, the figure is a line segment with one edge. For n-2, the figure is a triangle with three edges. For n=3, the figure is a tetrahedron with six edges. The Schläfli symbol, {3,...,3}, of the regular n-dimensional simplex consists of n-1 threes. An achiral coloring is identical to its reflection.
T(n,k) is also the number of achiral colorings of (n-2)-dimensional regular simplices in an n-dimensional simplex using exactly k colors. Thus, T(2,k) is also the number of achiral colorings of the vertices (0-dimensional simplices) of an equilateral triangle.

Examples

			Triangle begins with T(1,1):
1
1  2   0
1  8  18   12    0    0
1 26 306 1400 2800 2520 840 0 0 0
For T(2,2) = 2, the two colorings of the triangle edges are AAB and ABB.
		

Crossrefs

Cf. A327087 (oriented), A327088 (unoriented), A327089 (chiral), A327086 (up to k colors), A325003 (vertices).

Programs

  • Mathematica
    CycleX[{2}] = {{1,1}}; (* cycle index for permutation with given cycle structure *)
    CycleX[{n_Integer}] := CycleX[n] = If[EvenQ[n], {{n/2,1}, {n,(n-2)/2}}, {{n,(n-1)/2}}]
    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)
    CycleX[p_List] := CycleX[p] = compress[Join[CycleX[Drop[p, -1]], If[Last[p] > 1, CycleX[{Last[p]}], ## &[]], If[# == Last[p], {#, Last[p]}, {LCM[#, Last[p]], GCD[#, Last[p]]}] & /@ Drop[p, -1]]]
    pc[p_List] := 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]]], 0, pc[#] j^Total[CycleX[#]][[2]]] & /@ IntegerPartitions[n+1]]/((n+1)!/2)]
    array[n_, k_] := row[n] /. j -> k
    Table[LinearSolve[Table[Binomial[i,j], {i,1,(n+1)n/2}, {j,1,(n+1)n/2}], Table[array[n,k], {k,1,(n+1)n/2}]], {n,1,6}] // 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. The last n-1 entries in row n are zero.
A327086(n,k) = Sum_{j=1..(n+1)*n/2} T(n,j) * binomial(k,j).
A(n,k) = 2*A327084(n,k) - A327083(n,k) = A327083(n,k) - 2*A327085(n,k) = A327084(n,k) - A327085(n,k).

A337410 Array read by descending antidiagonals: T(n,k) is the number of achiral 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, 18, 70, 1, 5, 40, 1407, 93024, 1, 6, 75, 12480, 294157089, 47823602694208, 1, 7, 126, 69050, 91983927296, 67514530382043163023924, 443077371786837979607993095063601152, 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 a cube with 12 edges. The number of edges is n*2^(n-1).
Also the number of achiral 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       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. A337407 (oriented), A337408 (unoriented), A337409 (chiral).
Rows 1-4 are A000027, A002411, A331351, A331361.
Cf. A327086 (simplex edges), A337414 (orthoplex edges), A325015 (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; 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,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) = 2*A337408(n,k) - A337407(n,k) = A337407(n,k) - 2*A337409(n,k) = A337408(n,k) - A337409(n,k).
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