%I #13 Jul 22 2023 17:37:34
%S 1,0,0,2,0,2,2,2,0,2,2,2,2,2,2,6,0,2,2,2,2,2,2,6,2,2,2,6,2,6,6,10,0,2,
%T 2,2,2,2,2,6,2,2,2,6,2,6,6,10,2,2,2,6,2,6,6,10,2,6,6,10,6,10,10,22,0,
%U 2,2,2,2,2,2,6,2,2,2,6,2,6,6,10,2,2,2,6,2,6,6,10,2,6,6,10,6,10,10,22,2,2,2
%N A mod 2 related Jacobsthal sequence.
%C Conjecture: all the terms are in A078008.
%C The conjecture is true since a(n) = A078008(A000120(n)). - _Paul Barry_, Jan 07 2005
%H Amiram Eldar, <a href="/A102395/b102395.txt">Table of n, a(n) for n = 0..10000</a>
%F a(2^n-1) = A078008(n).
%F A001316(n) = a(n) + 2*A102396(n).
%F a(n) = Sum_{k=0..n} if(n+k == 0 (mod 3), C(n, k) mod 2, 0).
%F a(n) = Sum_{k=0..n} (1-ceiling((n+k)/3)+floor((n+k)/3))*(binomial(n,k) mod 2). - _Wesley Ivan Hurt_, Mar 01 2023
%t f[n_] := (2^n + 2*(-1)^n)/3; a[n_] := f[DigitCount[n, 2, 1]]; Array[a, 100, 0] (* _Amiram Eldar_, Jul 22 2023 *)
%Y Cf. A000120, A078008, A001316, A102396.
%K easy,nonn
%O 0,4
%A _Paul Barry_, Jan 06 2005