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A131292 a(2n+1)=3a(2n)-3a(2n-1)+2a(2n-2), a(2n+2)=3a(2n+1)-3a(2n), a(0)=a(1)=a(2)=1.

%I A131292 #10 Sep 23 2015 15:37:16
%S A131292 1,1,1,2,3,5,6,9,9,12,9,9,0,-9,-27,-54,-81,-135,-162,-243,-243,-324,
%T A131292 -243,-243,0,243,729,1458,2187,3645,4374,6561,6561,8748,6561,6561,0,
%U A131292 -6561,-19683,-39366,-59049,-98415,-118098,-177147,-177147,-236196,-177147,-177147,0,177147,531441,1062882,1594323
%N A131292 a(2n+1)=3a(2n)-3a(2n-1)+2a(2n-2), a(2n+2)=3a(2n+1)-3a(2n), a(0)=a(1)=a(2)=1.
%C A131292 Sequence is identical to its third differences in absolute value.
%H A131292 <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (0, 3, 0, -3).
%F A131292 G.f.: (2*x^3-x^2-x+1)*(1+x)^2/(1-3*x^2+3*x^4). - _R. J. Mathar_, Jul 17 2009
%p A131292 A131292 := proc(n) option remember ; if n <= 2 then 1; elif n mod 2 = 1 then 3*(A131292(n-1)-A131292(n-2))+2*A131292(n-3) ; else 3*(A131292(n-1)-A131292(n-2)) ; fi ; end: seq(A131292(n),n=0..80) ; # _R. J. Mathar_, Oct 18 2007
%t A131292 Join[{1, 1},LinearRecurrence[{0, 3, 0, -3},{1, 2, 3, 5},51]] (* _Ray Chandler_, Sep 23 2015 *)
%Y A131292 Cf. A131665 (0, 0, 1, 3, 6, 11).
%Y A131292 Cf. A057083 (bisection). - _R. J. Mathar_, Jul 17 2009
%K A131292 sign,easy
%O A131292 0,4
%A A131292 _Paul Curtz_, Sep 29 2007
%E A131292 More terms from _R. J. Mathar_, Oct 18 2007