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%I #39 Sep 08 2022 08:44:44
%S 0,4,32,192,1024,5120,24576,114688,524288,2359296,10485760,46137344,
%T 201326592,872415232,3758096384,16106127360,68719476736,292057776128,
%U 1236950581248,5222680231936,21990232555520
%N a(n) = n*4^n.
%C Bisection of A001787. That is, a(n) = A001787(2*n). - _Graeme McRae_, Jul 12 2006
%C All numbers of the form n*4^n+(4^n-1)/3 have the property that they are sums of two squares and also their indices are the sum of two squares. This follows from the identity n*4^n+(4^n-1)/3=4*(4*(..4*(4*n+1)+1)+1)+1..)+1. - _Artur Jasinski_, Nov 12 2007
%H Vincenzo Librandi, <a href="/A018215/b018215.txt">Table of n, a(n) for n = 0..300</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (8,-16).
%F G.f.: 4*x/(1-4*x)^2.
%F E.g.f.: 4*x*exp(4*x).
%F From _Amiram Eldar_, Jul 20 2020: (Start)
%F Sum_{n>=1} 1/a(n) = log(4/3) = A083679.
%F Sum_{n>=1} (-1)^(n+1)/a(n) = log(5/4). (End)
%t Table[n 4^n,{n,0,20}] (* or *) LinearRecurrence[{8,-16},{0,4},30] (* _Harvey P. Dale_, Apr 22 2018 *)
%o (Magma) [n*4^n: n in [0..25]]; // _Vincenzo Librandi_, Jun 01 2011
%o (PARI) a(n) = n<<(2*n) \\ _David A. Corneth_, Apr 22 2018
%Y Cf. A000302 (4^n), A001787, A002450, A083679.
%Y Row n=4 of A258997.
%K nonn,easy
%O 0,2
%A _N. J. A. Sloane_, Peter Winkler (pw(AT)bell-labs.com)
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