A006332 From the enumeration of corners.

0, 2, 28, 168, 660, 2002, 5096, 11424, 23256, 43890, 77924, 131560, 212940, 332514, 503440, 742016, 1068144, 1505826, 2083692, 2835560, 3801028, 5026098, 6563832, 8475040, 10829000, 13704210, 17189172, 21383208, 26397308, 32355010, 39393312, 47663616, 57332704

Offset: 0

Author

N. J. A. Sloane

Keywords

• nonn
• easy

References

• N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).

Links

• G. C. Greubel, Table of n, a(n) for n = 0..1000
• Germain Kreweras, Sur une classe de problèmes de dénombrement liés au treillis des partitions des entiers, Cahiers du Bureau Universitaire de Recherche Opérationnelle, Institut de Statistique, Université de Paris, 6 (1965), circa p. 82.
• Simon Plouffe, Approximations de séries génératrices et quelques conjectures, Dissertation, Université du Québec à Montréal, 1992; arXiv:0911.4975 [math.NT], 2009.
• Simon Plouffe, 1031 Generating Functions, Appendix to Thesis, Montreal, 1992.
• Index entries for linear recurrences with constant coefficients, signature (7,-21,35,-35,21,-7,1).

Crossrefs

Cf. A006858, A132339.

Programs

Magma

[Binomial(n+2, 3)*Binomial(2*n+3, 3)/5: n in [0..30]]; // G. C. Greubel, Dec 14 2021

Maple

A006332:=-2*(1+z)*(z**2+6*z+1)/(z-1)**7; # conjectured by Simon Plouffe in his 1992 dissertation

Mathematica

Table[(n(1+n)^2(2+n)(1+2n)(3+2n))/90, {n, 0, 30}] (* or *)
{0}~Join~CoefficientList[Series[2(x+1)(x^2 +6x +1)/(1-x)^7, {x, 0, 29}], x] (* Michael De Vlieger, Mar 26 2016 *)

PARI

my(x='x+O('x^99)); concat(0, Vec(2*(x+1)*(x^2+6*x+1)/(1-x)^7)) \\ Altug Alkan, Mar 26 2016

Sage

[binomial(n+2, 3)*binomial(2*n+3, 3)/5 for n in (0..30)] # G. C. Greubel, Dec 14 2021

Formula

a(n) = (n*(1 + n)^2*(2 + n)*(1 + 2*n)*(3 + 2*n))/90.

a(n) = 2*A006858(n).

a(n) = (-1)^(n+1)*A132339(3, n).

G.f.: 2*(1+x)*(1 + 6*x + x^2)/(1-x)^7.

From G. C. Greubel, Dec 14 2021: (Start)

E.g.f.: (1/90)*x*(180 + 1080*x + 1350*x^2 + 555*x^3 + 84*x^4 + 4*x^5)*exp(x).

a(n) = binomial(n+2, 3)*binomial(2*n+3, 3)/5. (End)

From Amiram Eldar, Jul 10 2023: (Start)

Sum_{n>=1} 1/a(n) = 15*Pi^2 - 295/2.

Sum_{n>=1} (-1)^(n+1)/a(n) = -15*Pi^2/2 + 120*Pi - 605/2. (End)