login

Year-end appeal: Please make a donation to the OEIS Foundation to support ongoing development and maintenance of the OEIS. We are now in our 61st year, we have over 378,000 sequences, and we’ve reached 11,000 citations (which often say “discovered thanks to the OEIS”).

A008414
Coordination sequence for 6-dimensional cubic lattice.
8
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
J. H. Conway and N. J. A. Sloane, Low-Dimensional Lattices VII: Coordination Sequences, Proc. Royal Soc. London, A453 (1997), 2369-2389 (pdf).
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.
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
KEYWORD
nonn,easy
STATUS
approved