A132366 Partial sum of centered tetrahedral numbers A005894.

1, 6, 21, 56, 125, 246, 441, 736, 1161, 1750, 2541, 3576, 4901, 6566, 8625, 11136, 14161, 17766, 22021, 27000, 32781, 39446, 47081, 55776, 65625, 76726, 89181, 103096, 118581, 135750, 154721, 175616, 198561, 223686, 251125, 281016, 313501, 348726, 386841

Offset

0,2

Comments

From Robert A. Russell, Oct 09 2020: (Start)

a(n-1) is the number of achiral colorings of the 5 tetrahedral facets (or vertices) of a regular 4-dimensional simplex using n or fewer colors. An achiral arrangement is identical to its reflection. The 4-dimensional simplex is also called a 5-cell or pentachoron. Its Schläfli symbol is {3,3,3}.

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 class of the permutation. The first formula for a(n-1) 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_1x_4^1

32 20 x_2^1x_3^1

2111 10 x_1^3x_2^1 (End)

Links

Nathaniel Johnston, Table of n, a(n) for n = 0..5000

Index entries for linear recurrences with constant coefficients, signature (5,-10,10,-5,1).

Formula

a(n) = (n^4 + 4*n^3 + 11*n^2 + 14*n + 6)/6.

G.f.: -(x+1)*(x^2+1) / (x-1)^5. - Colin Barker, May 04 2013

From Robert A. Russell, Oct 09 2020: (Start)

a(n-1) = n^2 * (5 + n^2) / 6.

a(n-1) = binomial(n+4,5) - binomial(n,5).

a(n-1) = 1*C(n,1) + 4*C(n,2) + 6*C(n,3) + 4*C(n,4), where the coefficient of C(n,k) is the number of achiral colorings using exactly k colors.

a(n-1) = 2*A000389(n+4) - A337895(n) = A337895(n) - 2*A000389(n) = A000389(n+4) - A000389(n).

G.f. for a(n-1): x * (x+1) * (x^2+1) / (1-x)^5. (End)

From Amiram Eldar, Feb 14 2023: (Start)

Sum_{n>=0} 1/a(n) = Pi^2/5 + 3/25 - 3*Pi*coth(sqrt(5)*Pi)/(5*sqrt(5)).

Sum_{n>=0} (-1)^n/a(n) = Pi^2/10 - 3/25 + 3*Pi*cosech(sqrt(5)*Pi)/(5*sqrt(5)). (End)

Mathematica

Do[Print[n, " ", (n^4 + 4 n^3 + 11 n^2 + 14 n + 6)/6 ], {n, 0, 10000}]
Accumulate[Table[(2n+1)(n^2+n+3)/3, {n, 0, 40}]] (* or *) LinearRecurrence[ {5, -10, 10, -5, 1}, {1, 6, 21, 56, 125}, 40] (* Harvey P. Dale, Feb 26 2020 *)

Crossrefs

Cf. A000292, A005894, A063488, A001845, A063489, A005898, A063490, A057813, A063491, A005902, A063492, A005917, A063493, A063494, A063495, A063496.

Cf. A337895 (oriented), A000389(n+4) (unoriented), A000389 (chiral), A331353 (5-cell edges, faces), A337955 (8-cell vertices, 16-cell facets), A337958 (16-cell vertices, 8-cell facets), A338951 (24-cell), A338967 (120-cell, 600-cell).

a(n-1) = A325001(4,n).

Sequence in context: A162539 A259474 A002817 * A015641 A292950 A282845

Adjacent sequences: A132363 A132364 A132365 * A132367 A132368 A132369

Keyword

easy,nonn

Author

Jonathan Vos Post, Nov 09 2007

Extensions

Corrected offset, Mathematica program by Tomas J. Bulka (tbulka(AT)rodincoil.com), Sep 02 2009

Status

approved