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%I #12 Feb 02 2023 17:16:11
%S 1,1,0,2,-3,8,-16,35,-72,150,-307,628,-1276,2587,-5228,10546,-21235,
%T 42704,-85784,172179,-345344,692286,-1387155,2778492,-5563748,
%U 11138443,-22294596,44617850,-89282067,178639160,-357399712,714995843,-1430309496,2861133222,-5723098483,11447543236
%N The trinomial transform (A027907) gives powers of 2, while the trinomial transform of this sequence shift one place left gives powers of 3.
%H G. C. Greubel, <a href="/A100321/b100321.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_04">Index entries for linear recurrences with constant coefficients</a>, signature (-2,2,3,-2).
%F G.f.: (1 + 3*x - 3*x^3) / (1 + 2*x - 2*x^2 - 3*x^3 + 2*x^4).
%F 2^n = Sum_{k=0..2*n} A027907(n, k)*a(k).
%F 3^n = Sum_{k=0..2*n} A027907(n, k)*a(k+1).
%F a(n) = (1/3)*((-1)^n*(3*Fibonacci(n-1) - 2^n) + 1). - _Ralf Stephan_, May 15 2007
%e 2^3 = 1*(1) + 3*(1) + 6*(0) + 7*(2) + 6*(-3) + 3*(8) + 1*(-16).
%e 3^3 = 1*(1) + 3*(0) + 6*(2) + 7*(-3) + 6*(8) + 3*(-16) + 1*(35).
%t LinearRecurrence[{-2,2,3,-2}, {1,1,0,2}, 41] (* _G. C. Greubel_, Feb 01 2023 *)
%o (PARI) a(n)=polcoeff((1+3*x-3*x^3)/(1+2*x-2*x^2-3*x^3+2*x^4+x*O(x^n)),n)
%o (Magma) [((-1)^n*(3*Fibonacci(n-1) -2^n) +1)/3: n in [0..40]]; // _G. C. Greubel_, Feb 01 2023
%o (SageMath)
%o def A100321(n): return ((-1)^n*(3*fibonacci(n-1) -2^n) +1)/3
%o [A100321(n) for n in range(41)] # _G. C. Greubel_, Feb 01 2023
%Y Cf. A000045, A027907.
%Y Cf. A000079, A000244.
%K sign
%O 0,4
%A _Paul D. Hanna_, Nov 15 2004
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