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%I #18 Jun 13 2015 00:51:38
%S 0,1,1,1,0,-3,-9,-18,-27,-27,0,81,243,486,729,729,0,-2187,-6561,
%T -13122,-19683,-19683,0,59049,177147,354294,531441,531441,0,-1594323,
%U -4782969,-9565938,-14348907,-14348907,0,43046721,129140163,258280326,387420489,387420489,0,-1162261467,-3486784401
%N A transform of the Jacobsthal numbers.
%C Apply the Chebyshev transform (1/(1+x^2), x/(1+x^2)) followed by the binomial involution (1/(1-x),-x/(1-x)) (expressed as Riordan arrays) to -A001045(n). All elements are multiples of a power of 3. - _Ralf Stephan_, Jan 28 2005
%H Vincenzo Librandi, <a href="/A103312/b103312.txt">Table of n, a(n) for n = 0..1000</a>
%H <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (3,-3).
%F G.f.: x(1-x)^2/(1-3x+3x^2); a(n)=-sum{j=0..n, (-1)^j*C(n, j)*sum{k=0..floor(j/2), (-1)^k*C(n-k, k)A001045(j-2k)}}.
%F Recurrence: a(n+2) = 3a(n-1) - 3a(n), starting with 0, 1, 1, 1. - _Ralf Stephan_, Jan 28 2005
%t Join[{0,1},LinearRecurrence[{3,-3},{1,1},50]] (* _Harvey P. Dale_, Apr 12 2014 *)
%t CoefficientList[Series[x (1 - x)^2/(1 - 3 x + 3 x^2), {x, 0, 50}], x] (* _Vincenzo Librandi_, Apr 13 2014 *)
%o (PARI) {a(n) = if( n<0, 0, polcoeff( x * (1 - x)^2 / (1 - 3*x + 3*x^2) + x * O(x^n), n))} /* _Michael Somos_, Sep 29 2007 */
%o (PARI) {a(n) = if(n<2, n>0, 3^(n\2-1) * (-1)^((n+1)\6) * (1 + (-1)^((n-1)\3) * (n%3==1)))} /* _Michael Somos_, Sep 29 2007 */
%Y Cf. A057681.
%K easy,sign
%O 0,6
%A _Paul Barry_, Jan 30 2005
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