A008414 Coordination sequence for 6-dimensional cubic lattice.

1, 12, 72, 292, 912, 2364, 5336, 10836, 20256, 35436, 58728, 93060, 142000, 209820, 301560, 423092, 581184, 783564, 1038984, 1357284, 1749456, 2227708, 2805528, 3497748, 4320608, 5291820, 6430632, 7757892, 9296112, 11069532

Offset

0,2

Comments

If Y_i (i=1,2,3,4,5,6) are 2-blocks of a (n+6)-set X then a(n-5) is the number of 11-subsets of X intersecting each Y_i (i=1,2,3,4,5,6). - Milan Janjic, Oct 28 2007

Links

Seiichi Manyama, Table of n, a(n) for n = 0..10000

J. H. Conway and N. J. A. Sloane, Low-Dimensional Lattices VII: Coordination Sequences, Proc. Royal Soc. London, A453 (1997), 2369-2389 (pdf).

Milan Janjic, Two Enumerative Functions

Milan Janjić, On Restricted Ternary Words and Insets, arXiv:1905.04465 [math.CO], 2019.

Ross McPhedran, Numerical Investigations of the Keiper-Li Criterion for the Riemann Hypothesis, arXiv:2311.06294 [math.NT], 2023. See p. 6.

Index entries for linear recurrences with constant coefficients, signature (6, -15, 20, -15, 6, -1).

Formula

G.f.: ((1+x)/(1-x))^6.

a(n) = 4*n*(2/15*n^4+4/3*n^2+23/15) for n > 0. - S. Bujnowski (slawb(AT)atr.bydgoszcz.pl), Nov 26 2002

n*a(n) = 12*a(n-1) + (n-2)*a(n-2) for n > 1. - Seiichi Manyama, Jun 06 2018

Maple

for n from 1 to 8 do eval(4*n*(2/15*n^4+4/3*n^2+23/15)) od;

Mathematica

{1}~Join~Table[4 n (2/15 n^4 + 4/3 n^2 + 23/15), {n, 29}] (* or *)
CoefficientList[Series[((1 + x)/(1 - x))^6, {x, 0, 29}], x] (* Michael De Vlieger, Oct 04 2016 *)
LinearRecurrence[{6, -15, 20, -15, 6, -1}, {1, 12, 72, 292, 912, 2364, 5336}, 30] (* Harvey P. Dale, Jul 01 2020 *)

Prog

(PARI) a(n)=if(n, 4*n*(2*n^4+20*n^2+23)/15, 1) \\ Charles R Greathouse IV, Oct 04 2016

Crossrefs

Sequence in context: A235870 A008533 A010024 * A052181 A118979 A014970

Adjacent sequences: A008411 A008412 A008413 * A008415 A008416 A008417

Keyword

nonn,easy

Author

N. J. A. Sloane

Status

approved