A008780 a(n) = (n-dimensional partitions of 6) + C(n,4) + C(n,3).

1, 11, 48, 141, 331, 672, 1232, 2094, 3357, 5137, 7568, 10803, 15015, 20398, 27168, 35564, 45849, 58311, 73264, 91049, 112035, 136620, 165232, 198330, 236405, 279981, 329616, 385903, 449471, 520986, 601152, 690712, 790449, 901187, 1023792, 1159173, 1308283

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• nice
• easy

Comments

These are the conjectured numbers of d-dimensional partitions for n=6, coming from a formula proposed by MacMahon in the general case that turned out to be wrong. Still, here for n=6, MacMahon's formula coincides for d < 3 with the first three terms of A042984. - Michel Marcus, Aug 16 2013

Binomial transform of [1,10,27,29,12,1,0,0,0,...], 6th row of A116672. - R. J. Mathar, Jul 18 2017

References

• G. E. Andrews, The Theory of Partitions, Add.-Wes. '76, p. 190.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..1000
• S. Balakrishnan, S. Govindarajan and N. S. Prabhakar, On the asymptotics of higher-dimensional partitions, arXiv:1105.6231 [cond-mat.stat-mech], 2011.
• Index entries for linear recurrences with constant coefficients, signature (6,-15,20,-15,6,-1).

Crossrefs

Cf. A007326, A007327, A042984, A116672.

Programs

GAP

List([0..40], n-> (120 + 404*n + 490*n^2 + 255*n^3 + 50*n^4 + n^5)/120); # G. C. Greubel, Sep 11 2019

Magma

R:=PowerSeriesRing(Integers(), 40); Coefficients(R!( (1+5*x-3*x^2-2*x^3)/(1-x)^6 )); // G. C. Greubel, Sep 11 2019

Maple

seq(1+10*n+27*binomial(n,2)+29*binomial(n,3)+12*binomial(n,4)+binomial(n,5), n=0..40);

Mathematica

Table[1+10n+27Binomial[n,2]+29Binomial[n,3]+12Binomial[n,4]+ Binomial[n,5], {n,0,40}] (* Harvey P. Dale, Jul 27 2011 *)
CoefficientList[Series[(1+5x-3x^2-2x^3)/(1-x)^6, {x, 0, 40}], x] (* Vincenzo Librandi, Aug 17 2013 *)
LinearRecurrence[{6,-15,20,-15,6,-1},{1,11,48,141,331,672},40] (* Harvey P. Dale, Aug 28 2019 *)

PARI

my(x='x+O('x^40)); Vec((1+5*x-3*x^2-2*x^3)/(1-x)^6) \\ G. C. Greubel, Sep 11 2019

Sage

[(120 + 404*n + 490*n^2 + 255*n^3 + 50*n^4 + n^5)/120 for n in (0..40)] # G. C. Greubel, Sep 11 2019

Formula

G.f.: (1 + 5*x - 3*x^2 - 2*x^3)/(1-x)^6. - Colin Barker, Sep 05 2012

From G. C. Greubel, Sep 11 2019: (Start)

a(n) = (120 + 404*n + 490*n^2 + 255*n^3 + 50*n^4 + n^5)/120.

E.g.f.: (120 + 1200*x + 1620*x^2 + 580*x^3 + 60*x^4 + x^5)*exp(x)/120. (End)

Extensions

Description corrected by Alford Arnold, Aug 1998

More terms added by G. C. Greubel, Sep 11 2019