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A348405 a(0) = 1, a(n) + a(n+1) = round(2^n/9), n >= 0.

%I A348405 #64 Oct 31 2022 07:32:56
%S A348405 1,-1,1,-1,2,0,4,3,11,17,40,74,154,301,609,1211,2430,4852,9712,19415,
%T A348405 38839,77669,155348,310686,621382,1242753,2485517,4971023,9942058,
%U A348405 19884104,39768220,79536427,159072867,318145721,636291456,1272582898,2545165810
%N A348405 a(0) = 1, a(n) + a(n+1) = round(2^n/9), n >= 0.
%H A348405 Harvey P. Dale, <a href="/A348405/b348405.txt">Table of n, a(n) for n = 0..1000</a>
%H A348405 <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,-1,1,2).
%F A348405 a(n+1) = 2*a(n) - A104581(n+6).
%F A348405 a(n) + a(n+1) = A113405(n).
%F A348405 a(n) + a(n+3) = A001045(n).
%F A348405 a(n+2) = a(n) + A131666(n).
%F A348405 From _Thomas Scheuerle_, Oct 18 2021: (Start)
%F A348405 G.f.: (x^4-x^3+2x-1)/((2*x^3-3*x^2+3*x-1)*(x+1)^2).
%F A348405 A172481(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*a(2*n-k). With negative sign for ...*a(1+2*n-k) and ...*a(3+2*n-k) too.
%F A348405 A175656(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*a(2+2*n-k).
%F A348405 A136298(n+1) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*a(4+2*n-k).
%F A348405 A348407(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*(a(2+2*n-k) - 2*a(1+2*n-k) - a(2*n-k)).
%F A348405 (End)
%t A348405 CoefficientList[ Series[(x^4-x^3+2x-1)/((2*x^3-3*x^2+3*x-1)*(x+1)^2), {x, 0, 40}], x] (* _Thomas Scheuerle_, Oct 17 2021 *)
%t A348405 nxt[{n_,a_}]:={n+1,Round[(2^n)/9]-a}; NestList[nxt,{0,1},40][[All,2]] (* or *) LinearRecurrence[{1,2,-1,1,2},{1,-1,1,-1,2},40] (* _Harvey P. Dale_, Apr 28 2022 *)
%Y A348405 Cf. A139797 (a(n) + a(n+1) = round(2^n/9) too, but a(0) = 0).
%Y A348405 Cf. A001045, A104581, A113405, A131666.
%Y A348405 Cf. A136298, A172481, A175656, A348407.
%K A348405 sign
%O A348405 0,5
%A A348405 _Paul Curtz_, Oct 17 2021
%E A348405 a(22)-a(36) from _Thomas Scheuerle_, Oct 17 2021