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a(0)=2, a(1)=7, and a(n) = (3*n+1)*2^(n-1) if n > 1.
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%I #24 Oct 05 2019 14:06:40

%S 2,7,14,40,104,256,608,1408,3200,7168,15872,34816,75776,163840,352256,

%T 753664,1605632,3407872,7208960,15204352,31981568,67108864,140509184,

%U 293601280,612368384,1275068416,2650800128,5502926848,11408506880,23622320128,48855252992

%N a(0)=2, a(1)=7, and a(n) = (3*n+1)*2^(n-1) if n > 1.

%C The sequence appears on the main diagonal of the array defined by A123167 in the first row and successive differences in followup rows:

%C 2, 3, 10, 7, 18, 11, 26, 15, 34, 19, ... A123167

%C 1, 7, -3, 11, -7, 15, -11, 19, -15, 23, ... first diff

%C 6, -10, 14, -18, 22 -26, 30, -34, 38, ... second diff

%C -16, 24, -32, 40, -48, 56, -64, 72, -80, ... third diff

%H Vincenzo Librandi, <a href="/A176662/b176662.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec">Index entries for linear recurrences with constant coefficients</a>, signature (4,-4).

%F a(n) mod 9 = A010710(n-1), n > 2.

%F a(2n) + a(2n+1) = 9, 54, 360, 2016, ...

%F a(n) - 2*a(n-1) = 12*A131577(n-2), n > 1.

%F a(n) = 4*a(n-1) - 4*a(n-2), n > 3.

%F G.f.: (-6*x^2+12*x^3+2-x)/(1-2*x)^2.

%t LinearRecurrence[{4,-4},{2,7,14,40},40] (* or *) Join[{2,7},Table[ (3n+1) 2^(n-1),{n,2,40}]] (* _Harvey P. Dale_, Oct 05 2019 *)

%K nonn,easy

%O 0,1

%A _Paul Curtz_, Apr 23 2010

%E Edited by _R. J. Mathar_, Jun 30 2010