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

%I #64 Oct 31 2022 07:32:56

%S 1,-1,1,-1,2,0,4,3,11,17,40,74,154,301,609,1211,2430,4852,9712,19415,

%T 38839,77669,155348,310686,621382,1242753,2485517,4971023,9942058,

%U 19884104,39768220,79536427,159072867,318145721,636291456,1272582898,2545165810

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

%H Harvey P. Dale, <a href="/A348405/b348405.txt">Table of n, a(n) for n = 0..1000</a>

%H <a href="/index/Rec#order_05">Index entries for linear recurrences with constant coefficients</a>, signature (1,2,-1,1,2).

%F a(n+1) = 2*a(n) - A104581(n+6).

%F a(n) + a(n+1) = A113405(n).

%F a(n) + a(n+3) = A001045(n).

%F a(n+2) = a(n) + A131666(n).

%F From _Thomas Scheuerle_, Oct 18 2021: (Start)

%F G.f.: (x^4-x^3+2x-1)/((2*x^3-3*x^2+3*x-1)*(x+1)^2).

%F 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 A175656(n) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*a(2+2*n-k).

%F A136298(n+1) = Sum_{k=0..n} (-1)^(n-k)*binomial(n, k)*a(4+2*n-k).

%F 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 (End)

%t 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 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 Cf. A139797 (a(n) + a(n+1) = round(2^n/9) too, but a(0) = 0).

%Y Cf. A001045, A104581, A113405, A131666.

%Y Cf. A136298, A172481, A175656, A348407.

%K sign

%O 0,5

%A _Paul Curtz_, Oct 17 2021

%E a(22)-a(36) from _Thomas Scheuerle_, Oct 17 2021