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a(n) = 4^n*(4^n - 1)/2.
3

%I #23 Oct 02 2024 07:55:25

%S 0,6,120,2016,32640,523776,8386560,134209536,2147450880,34359607296,

%T 549755289600,8796090925056,140737479966720,2251799780130816,

%U 36028796884746240,576460751766552576,9223372034707292160,147573952581086478336,2361183241400462868480

%N a(n) = 4^n*(4^n - 1)/2.

%H Vincenzo Librandi, <a href="/A026337/b026337.txt">Table of n, a(n) for n = 0..200</a>

%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (20,-64).

%F a(n) = binomial(4^n, 2), n >= 0. - _Zerinvary Lajos_, Jan 07 2008

%F From _R. J. Mathar_, Mar 20 2009: (Start)

%F a(n) = 20*a(n-1) - 64*a(n-2).

%F G.f.: 6*x/((1-4*x)*(1-16*x)). (End)

%F a(n) = 6*A166984(n-1). - _R. J. Mathar_, Jun 23 2013

%F E.g.f.: exp(10*x)*sinh(6*x). - _G. C. Greubel_, Oct 02 2024

%p seq(binomial(4^n,2),n=0..18); # _Zerinvary Lajos_, Jan 07 2008

%t Table[4^n (4^n-1)/2,{n,0,30}] (* or *) LinearRecurrence[{20,-64},{0,6},30] (* _Harvey P. Dale_, Nov 05 2023 *)

%o (Magma) [4^n*(4^n-1)/2: n in [0..30]]; // _Vincenzo Librandi_, May 01 2011

%o (SageMath) [binomial(4^n, 2) for n in range(21)] # _G. C. Greubel_, Oct 02 2024

%Y Cf. A166984.

%K nonn

%O 0,2

%A _N. J. A. Sloane_