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Even bisection gives sequence a itself, n->a(2*(2*n+k)-1) gives k-th differences of a for k=1..2 with a(n)=n for n<2.
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%I #12 Feb 22 2017 09:47:16

%S 0,1,1,-1,1,0,-1,-2,1,-2,0,4,-1,2,-2,-3,1,-1,-2,0,0,-1,4,0,-1,-1,2,4,

%T -2,3,-3,-6,1,-3,-1,5,-2,2,0,2,0,4,-1,-9,4,-5,0,8,-1,3,-1,-7,2,-4,4,3,

%U -2,-1,3,5,-3,4,-6,-6,1,-2,-3,1,-1,-1,5,3,-2,2,2

%N Even bisection gives sequence a itself, n->a(2*(2*n+k)-1) gives k-th differences of a for k=1..2 with a(n)=n for n<2.

%H Alois P. Heinz, <a href="/A229817/b229817.txt">Table of n, a(n) for n = 0..10000</a>

%F a(2*n) = a(n),

%F a(4*n+1) = a(n+1) - a(n),

%F a(4*n+3) = a(n+2) - 2*a(n+1) + a(n).

%p a:= proc(n) option remember; local m, q, r;

%p m:= (irem(n, 4, 'q')+1)/2;

%p `if`(n<2, n, `if`(irem(n, 2, 'r')=0, a(r),

%p add(a(q+m-j)*(-1)^j*binomial(m, j), j=0..m)))

%p end:

%p seq(a(n), n=0..100);

%t a[n_] := a[n] = Module[{m, q, r}, {q, m} = QuotientRemainder[n, 4]; m = (m + 1)/2; If[n<2, n, If[Mod[n, 2]==0, a[Quotient[n, 2]], Sum[a[q+m-j] * (-1)^j * Binomial[m, j], {j, 0, m}]]]]; Table[a[n], {n, 0, 100}] (* _Jean-François Alcover_, Feb 22 2017, translated from Maple *)

%Y Cf. A005590, A229818, A229819, A229820, A229821, A229822, A229823, A229824, A229825.

%K sign,eigen

%O 0,8

%A _Alois P. Heinz_, Sep 30 2013