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

%I A100381 #73 Mar 09 2024 09:59:21
%S A100381 0,0,4,24,96,320,960,2688,7168,18432,46080,112640,270336,638976,
%T A100381 1490944,3440640,7864320,17825792,40108032,89653248,199229440,
%U A100381 440401920,968884224,2122317824,4630511616,10066329600,21810380800,47110422528
%N A100381 a(n) = 2^n*binomial(n,2).
%C A100381 From _Enrique Navarrete_, Jun 13 2023: (Start)
%C A100381 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).
%C A100381 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):
%C A100381 S={ }, T={1,2,3,4}: partition [4] in 1 way, S(0),T(2) (6 ways);
%C A100381 S={1}, T={2,3,4}: partition [4] in 4 such ways, S(1),T(1) or S(0),T(2) (24 ways);
%C A100381 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);
%C A100381 S={1,2,3}, T={4}: partition [4] in 4 such ways, S(1),T(1) or S(2),T(0) (24 ways);
%C A100381 S={1,2,3,4}, T={ }: partition [4] in 1 way, S(2),T(0) (6 ways). (End)
%D A100381 Jolley, Summation of Series, Dover (1961), eq (214) page 40.
%H A100381 Vincenzo Librandi, <a href="/A100381/b100381.txt">Table of n, a(n) for n = 0..3000</a>
%H A100381 The Ramanujan Machine, <a href="http://www.ramanujanmachine.com/">Using algorithms to discover new mathematics</a>.
%H A100381 <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (6,-12,8).
%F A100381 Sum_{n>=2} 1/a(n) = 1 - log(2) = 0.3068528.... - _Graeme McRae_, Jul 28 2006
%F A100381 a(n) = Sum_{k=0..n} k*2^k = 2*A001815(n). - _Zerinvary Lajos_, Oct 09 2006
%F A100381 E.g.f.: 2*x^2*exp(2x).
%F A100381 a(n) = 4*A001788(n-1). - _Johannes W. Meijer_, Jun 27 2009
%F A100381 Sum_{j=1..k} (j+2)/a(j+1) = 1 - 1/((k+1)*2^k). [Jolley]
%F A100381 G.f.: -4*x^2 / (2*x-1)^3. - _R. J. Mathar_, Oct 05 2011
%F A100381 Sum_{n>=2} (-1)^n/a(n) = 3*log(3/2) - 1. - _Amiram Eldar_, Jul 20 2020
%F A100381 From _Peter Bala_ Mar 04 2024: (Start)
%F A100381 Sum_{k = 2..n+2} 1/a(k) = 1/(4 - 4/(7 - 12/(10 - ... - 2*n*(n + 1)/(3*n + 4)))).
%F A100381 Sum_{k = 2..n+2} (-1)^k/a(k) = 1/(4 + 4/(5 + 12/(6 + ... + 2*n*(n + 1)/(n + 4)))).
%F A100381 Letting n -> oo in the above gives the continued fraction representations
%F A100381 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
%F A100381 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)
%p A100381 seq(2^n*binomial(n,2),n=0..20);
%t A100381 Range[0,20]! CoefficientList[Series[2x^2 Exp[2x],{x,0,20}],x]
%t A100381 Table[2^n Binomial[n,2],{n,0,30}] (* or *) LinearRecurrence[{6,-12,8},{0,0,4},30] (* _Harvey P. Dale_, Aug 15 2020 *)
%o A100381 (PARI) a(n)=binomial(n,2)<<n \\ _Charles R Greathouse IV_, Oct 16 2015
%Y A100381 Cf. A001788, A001815, A162007, A244009.
%K A100381 nonn,easy
%O A100381 0,3
%A A100381 _Jorge Coveiro_, Dec 30 2004