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%I #27 Mar 18 2024 12:04:43
%S 1,4,20,120,816,5984,45760,357760,2829056,22500864,179481600,
%T 1433753600,11461636096,91659526144,733141975040,5864598896640,
%U 46914643623936,375308558925824,3002434111406080,24019335451770880,192154133857304576,1537230871833083904,12297838178567454720
%N a(n) = binomial(2+2^n,3).
%C a(n) = Sum_{i=1...(2^n)} i*(i+1)/2, this sequence is thus similar to A016131 as it is a sum of triangular numbers on the interval <1,2^n>, A016131 is a sum of triangular numbers on the interval <1,2^n - 1>. - _Ctibor O. Zizka_, Mar 03 2009
%C a(n) is the number of unordered (not necessarily distinct) triples of subsets taken from the power set of {1,2,...,n}. Cf. A007582 (pairs of such subsets). - _Geoffrey Critzer_, Jul 10 2013
%H <a href="/index/Rec#order_03">Index entries for linear recurrences with constant coefficients</a>, signature (14,-56,64).
%F a(n) = (2^(3n-1) +3*2^(2n-1) + 2^n)/3 = A092056(3, n) = A007581(n)*A000079(n) = 2*a(n-1)+4^(n-1)+8^(n-1).
%F a(n) = [x^3] 1/(1-x)^(2^n). - _Geoffrey Critzer_, Jul 11 2013
%F a(n) = 14*a(n-1)-56*a(n-2)+64*a(n-3). - _Colin Barker_, Sep 13 2014
%F G.f.: -(20*x^2-10*x+1) / ((2*x-1)*(4*x-1)*(8*x-1)). - _Colin Barker_, Sep 13 2014
%e a(5) = C(2+2^5,3) = C(34,3) = 5984.
%p seq(binomial(2+2^n, 3), n=0..25); # _Zerinvary Lajos_, Feb 22 2008
%t nn=20;Table[Coefficient[Series[1/(1-x)^(2^n),{x,0,nn}],x^3],{n,0,nn}] (* _Geoffrey Critzer_, Jul 10 2013 *)
%o (PARI) Vec(-(20*x^2-10*x+1)/((2*x-1)*(4*x-1)*(8*x-1)) + O(x^100)) \\ _Colin Barker_, Sep 13 2014
%Y Cf. A016131. - _Ctibor O. Zizka_, Mar 03 2009
%K nonn,easy
%O 0,2
%A _Henry Bottomley_, Feb 19 2004
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