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a(0) = 0, a(1) = 1, a(2n) = a(n), a(2n+1) = a(n) - a(n+1).
1

%I #11 Dec 21 2016 08:11:01

%S 0,1,1,0,1,1,0,-1,1,0,1,1,0,1,-1,-2,1,1,0,-1,1,0,1,1,0,-1,1,2,-1,1,-2,

%T -3,1,0,1,1,0,1,-1,-2,1,1,0,-1,1,0,1,1,0,1,-1,-2,1,-1,2,3,-1,-2,1,3,

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

%N a(0) = 0, a(1) = 1, a(2n) = a(n), a(2n+1) = a(n) - a(n+1).

%C Variation on Stern's Diatomic Series

%H Michael De Vlieger, <a href="/A145865/b145865.txt">Table of n, a(n) for n = 0..10000</a>

%F From _Chai Wah Wu_, Dec 20 2016: (Start)

%F a(2^k*n+1) = a(n+1) if k is even

%F a(2^k*n+1) = a(n)-a(n+1) = a(2n+1) if k is odd

%F a(2^k*n+2^k-1) = a(n) - k*a(n+1)

%F a(2^k*n+2^k-3) = a(n+1) for k >= 2

%F a(2^k*n+2^k-5) = (k-1)*a(n+1)-a(n) for k >= 3

%F a(2^k*n+2^k-7) = a(n) - (k-2)*a(n+1) for k >= 3

%F This implies that

%F a(2^k+1) = 1 if k is even

%F a(2^k+1) = 0 if k is odd

%F a(2^k-1) = 2 - k for k >= 1

%F a(2^k-3) = 1 for k >= 2

%F a(2^k-5) = k - 3 for k >= 3

%F a(2^k-7) = 4 - k for k >= 3

%F (End)

%t a[0] = 0; a[1] = 1; a[n_] := a[n] = If[EvenQ@ n, a[n/2], a[#] - a[# + 1] &[(n - 1)/2]]; Table[a@ n, {n, 0, 85}] (* _Michael De Vlieger_, Dec 21 2016 *)

%Y Cf. A002487, A005590.

%K easy,sign

%O 0,16

%A _Paolo P. Lava_ and _Giorgio Balzarotti_, Oct 22 2008