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A084100 Expansion of (1+x-x^2-x^3)/(1+x^2).

%I A084100 #21 May 08 2024 05:48:37
%S A084100 1,1,-2,-2,2,2,-2,-2,2,2,-2,-2,2,2,-2,-2,2,2,-2,-2,2,2,-2,-2,2,2,-2,
%T A084100 -2,2,2,-2,-2,2,2,-2,-2,2,2,-2,-2,2,2,-2,-2,2,2,-2,-2,2,2,-2,-2,2,2,
%U A084100 -2,-2,2,2,-2,-2,2,2,-2,-2,2,2,-2,-2,2,2,-2,-2,2,2,-2,-2,2,2,-2,-2,2,2,-2,-2,2,2,-2,-2,2,2,-2
%N A084100 Expansion of (1+x-x^2-x^3)/(1+x^2).
%C A084100 Partial sums are A084099.
%C A084100 The unsigned sequence 1,1,2,2,2,2,.. has g.f. (1+x^2)/(1-x) and a(n)=sum{k=0..n, binomial(1,k/2)(1+(-1)^k)/2}. Its partial sums are A004275(n+1). The sequence 1,-1,2,-2,2,-2,... has g.f. (1+x^2)/(1+x) and a(n)=sum{k=0..n, (-1)^(n-k)binomial(1,k/2)(1+(-1)^k)/2}. - _Paul Barry_, Oct 15 2004
%H A084100 <a href="/index/Rec#order_02">Index entries for linear recurrences with constant coefficients</a>, signature (0,-1).
%F A084100 Euler transform of length 4 sequence [1, -3, 0, 1]. - _Michael Somos_, Jan 05 2017
%F A084100 G.f.: (1 + x) * (1 - x^2) / (1 + x^2). - _Michael Somos_, Jan 05 2017
%F A084100 a(n) = a(1-n) for all n in Z. - _Michael Somos_, Jan 05 2017
%F A084100 a(2*n) = a(2*n + 1) = A280560(n) for all n in Z. - _Michael Somos_, Jan 05 2017
%e A084100 G.f. = 1 + x - 2*x^2 - 2*x^3 + 2*x^4 + 2*x^5 - 2*x^6 - 2*x^7 + 2*x^8 + 2*x^9 + ...
%t A084100 CoefficientList[Series[(1+x-x^2-x^3)/(1+x^2),{x,0,100}],x]  (* _Harvey P. Dale_, Apr 20 2011 *)
%t A084100 a[ n_] := (-1)^Quotient[n, 2] If[ Quotient[n, 2] != 0, 2, 1]; (* _Michael Somos_, Jan 05 2017 *)
%o A084100 (PARI) {a(n) = (-1)^(n\2) * if( n\2, 2, 1)}; /* _Michael Somos_, Jan 05 2017 */
%Y A084100 Cf. A084099, A280560.
%K A084100 easy,sign
%O A084100 0,3
%A A084100 _Paul Barry_, May 15 2003