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G.f. A(x) satisfies: A(x) = 2*x*(1 - x)*A(x^2) + x/(1 - x).
1

%I #41 Aug 29 2019 18:58:40

%S 0,1,1,3,-1,3,-1,7,-5,-1,3,7,-5,-1,3,15,-13,-9,11,-1,3,7,-5,15,-13,-9,

%T 11,-1,3,7,-5,31,-29,-25,27,-17,19,23,-21,-1,3,7,-5,15,-13,-9,11,31,

%U -29,-25,27,-17,19,23,-21,-1,3,7,-5,15,-13,-9,11,63,-61,-57,59

%N G.f. A(x) satisfies: A(x) = 2*x*(1 - x)*A(x^2) + x/(1 - x).

%H Alois P. Heinz, <a href="/A309790/b309790.txt">Table of n, a(n) for n = 0..65534</a>

%H Ilya Gutkovskiy, <a href="/A309790/a309790.jpg">Scatter plot of a(n) up to n=1000</a>

%F a(0) = 0; a(2*n+2) = -2*a(n) + 1, a(2*n+1) = 2*a(n) + 1.

%p a:= proc(n) option remember; `if`(n=0, 0, 2*

%p `if`(irem(n, 2, 'r')=0, -a(r-1), a(r))+1)

%p end:

%p seq(a(n), n=0..2^7-2); # _Alois P. Heinz_, Aug 29 2019

%t nmax = 66; A[_] = 0; Do[A[x_] = 2 x (1 - x) A[x^2] + x/(1 - x) + O[x]^(nmax + 1) // Normal, nmax + 1]; CoefficientList[A[x], x]

%t a[0] = 0; a[n_] := If[EvenQ[n], -2 a[(n - 2)/2] + 1, 2 a[(n - 1)/2] + 1]; Table[a[n], {n, 0, 66}]

%Y Cf. A000225 (fixed points), A006257.

%Y Compare also to the scatter plots of A117966, A317825.

%K sign,look,hear

%O 0,4

%A _Ilya Gutkovskiy_, Aug 28 2019