A100381 a(n) = 2^n*binomial(n,2).

0, 0, 4, 24, 96, 320, 960, 2688, 7168, 18432, 46080, 112640, 270336, 638976, 1490944, 3440640, 7864320, 17825792, 40108032, 89653248, 199229440, 440401920, 968884224, 2122317824, 4630511616, 10066329600, 21810380800, 47110422528

Offset: 0

Author

Jorge Coveiro, Dec 30 2004

Keywords

• nonn
• easy

Comments

From Enrique Navarrete, Jun 13 2023: (Start)

a(n) is the number of ways to partition the set [n]={1,2,...,n} into two sets S,T and select 2 elements in total (from either S or T or both).

Example. For n=4, sample partitions are given (where S(i),T(j) means i elements are selected from S, j elements are selected from T):

S={ }, T={1,2,3,4}: partition [4] in 1 way, S(0),T(2) (6 ways);

S={1}, T={2,3,4}: partition [4] in 4 such ways, S(1),T(1) or S(0),T(2) (24 ways);

S={1,2}, T={3,4}: partition [4], in such 6 ways, S(1),T(1) or S(0),T(2) or S(2),T(0) (36 ways);

S={1,2,3}, T={4}: partition [4] in 4 such ways, S(1),T(1) or S(2),T(0) (24 ways);

S={1,2,3,4}, T={ }: partition [4] in 1 way, S(2),T(0) (6 ways). (End)

References

• Jolley, Summation of Series, Dover (1961), eq (214) page 40.

Links

• Vincenzo Librandi, Table of n, a(n) for n = 0..3000
• The Ramanujan Machine, Using algorithms to discover new mathematics.
• Index entries for linear recurrences with constant coefficients, signature (6,-12,8).

Crossrefs

Cf. A001788, A001815, A162007, A244009.

Programs

Maple

seq(2^n*binomial(n,2),n=0..20);

Mathematica

Range[0,20]! CoefficientList[Series[2x^2 Exp[2x],{x,0,20}],x]
Table[2^n Binomial[n,2],{n,0,30}] (* or *) LinearRecurrence[{6,-12,8},{0,0,4},30] (* Harvey P. Dale, Aug 15 2020 *)

PARI

a(n)=binomial(n,2)<Charles R Greathouse IV, Oct 16 2015

Formula

Sum_{n>=2} 1/a(n) = 1 - log(2) = 0.3068528.... - Graeme McRae, Jul 28 2006

a(n) = Sum_{k=0..n} k*2^k = 2*A001815(n). - Zerinvary Lajos, Oct 09 2006

E.g.f.: 2*x^2*exp(2x).

a(n) = 4*A001788(n-1). - Johannes W. Meijer, Jun 27 2009

Sum_{j=1..k} (j+2)/a(j+1) = 1 - 1/((k+1)*2^k). [Jolley]

G.f.: -4*x^2 / (2*x-1)^3. - R. J. Mathar, Oct 05 2011

Sum_{n>=2} (-1)^n/a(n) = 3*log(3/2) - 1. - Amiram Eldar, Jul 20 2020

From Peter Bala Mar 04 2024: (Start)

Sum_{k = 2..n+2} 1/a(k) = 1/(4 - 4/(7 - 12/(10 - ... - 2*n*(n + 1)/(3*n + 4)))).

Sum_{k = 2..n+2} (-1)^k/a(k) = 1/(4 + 4/(5 + 12/(6 + ... + 2*n*(n + 1)/(n + 4)))).

Letting n -> oo in the above gives the continued fraction representations

1 - log(2) = Sum_{k >= 2} 1/a(k) = 1/(4 - 4/(7 - 12/(10 - ... - 2*n*(n + 1)/((3*n + 4) - ... )))) (an equivalent continued fraction for 1 - log(2) was conjectured by the Ramanujan machine) and

3*log(3/2) - 1 = Sum_{k >= 2} (-1)^k/a(k) = 1/(4 + 4/(5 + 12/(6 + ... + 2*n*(n + 1)/((n + 4) + ... )))). (End)