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a(n) = f(f(n)), where f() = A035327().
5

%I #37 Apr 26 2024 06:23:57

%S 0,0,0,0,0,1,0,0,0,1,2,3,0,1,0,0,0,1,2,3,4,5,6,7,0,1,2,3,0,1,0,0,0,1,

%T 2,3,4,5,6,7,8,9,10,11,12,13,14,15,0,1,2,3,4,5,6,7,0,1,2,3,0,1,0,0,0,

%U 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17

%N a(n) = f(f(n)), where f() = A035327().

%H Alois P. Heinz, <a href="/A087734/b087734.txt">Table of n, a(n) for n = 0..16384</a>

%H J.-P. Allouche and J. Shallit, <a href="http://dx.doi.org/10.1016/S0304-3975(03)00090-2">The ring of k-regular sequences, II</a>, Theoret. Computer Sci., 307 (2003), 3-29.

%F From _Mikhail Kurkov_, Sep 29 2019: (Start)

%F Some conjectures:

%F a(n) = n - Sum_{k=A063250(n)..A000523(n)} 2^k = n - 2^(A000523(n)+1) + 2^A063250(n) for n>0 with a(0)=0.

%F G.f.: 1/(1-x) * Sum_{j>=0} (2^j)*((x^(2^j))/(1+x^(2^j)) - (1-x^(2^j)) * Sum_{k>=1} x^((2^j)*(2^k-1))).

%F a(n) = 2*a(floor(n/2)) + n mod 2 - A036987(n) for n>1 with a(0)=a(1)=0.

%F a(n) = (1 - A036987(n-1))*(1 + A063250(n) - A063250(n-1))*(1 + a(n-1)) for n>0 with a(0)=0. (End)

%p a:= n-> ((i->Bits[Nand](i$2))@@2)(n):

%p seq(a(n), n=0..100); # _Alois P. Heinz_, Sep 29 2019

%t {0}~Join~Array[Nest[BitXor[#, 2^IntegerPart[Log2@ # + 1] - 1] &, #, 2] /. -1 -> 0 &, 81] (* _Michael De Vlieger_, Sep 29 2019 *)

%K nonn

%O 0,11

%A _N. J. A. Sloane_, Oct 01 2003