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a(n)= +2*a(n-2) +4*a(n-3), n>3.
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%I #14 Jan 27 2019 15:28:25

%S 1,-6,-13,-27,-50,-106,-208,-412,-840,-1656,-3328,-6672,-13280,-26656,

%T -53248,-106432,-213120,-425856,-851968,-1704192,-3407360,-6816256,

%U -13631488,-27261952,-54528000,-109049856,-218103808,-436211712,-872407040,-1744838656

%N a(n)= +2*a(n-2) +4*a(n-3), n>3.

%C Generated by scanning the diagonal of the table generated by A143025 in the top row followed by higher order differences in the other rows:

%C 1, 8, 2, 8, 1, 8, 2, 8, 1, 8, 2, 8, 1, 8, 2, 8,...

%C 7, -6, 6, -7, 7, -6, 6, -7, 7, -6, 6, -7, 7,...

%C -13, 12, -13, 14, -13, 12, -13, 14, -13, 12,..

%C 25, -25, 27, -27, 25, -25, 27, -27, 25, -25,..

%C -50, 52, -54, 52, -50, 52, -54, 52, -50, 52, ...

%C 102, -106, 106, -102, 102, -106, 106, -102,...

%H <a href="/index/Rec">Index entries for linear recurrences with constant coefficients</a>, signature (0,2,4).

%F a(n) = ( -13*2^n-2*A009116(n))/4, n>0.

%F a(n+1)-2*a(n) = -A137429(n-2), n>1.

%F G.f.: (6*x+15*x^2+19*x^3-1)/( (2*x-1) *(2*x^2+2*x+1)).

%t LinearRecurrence[{0,2,4},{1,-6,-13,-27},30] (* _Harvey P. Dale_, Jan 27 2019 *)

%K sign,easy

%O 0,2

%A _Paul Curtz_, Feb 21 2010