%I #19 Jul 22 2023 17:35:38
%S 0,1,1,1,1,1,1,3,1,1,1,3,1,3,3,5,1,1,1,3,1,3,3,5,1,3,3,5,3,5,5,11,1,1,
%T 1,3,1,3,3,5,1,3,3,5,3,5,5,11,1,3,3,5,3,5,5,11,3,5,5,11,5,11,11,21,1,
%U 1,1,3,1,3,3,5,1,3,3,5,3,5,5,11,1,3,3,5,3,5,5,11,3,5,5,11,5,11,11,21,1,3
%N A mod 2 related Jacobsthal sequence.
%C Conjecture: all the terms are Jacobsthal numbers.
%C The conjecture is true since a(n) = A001045(A000120(n)). - _Paul Barry_, Jan 07 2005
%H Amiram Eldar, <a href="/A102396/b102396.txt">Table of n, a(n) for n = 0..10000</a>
%F a(2^n-1) = A001045(n).
%F A001316(n) = A102395(n) + 2*a(n).
%F a(n) = Sum_{k=0..n} if(n+k == 1 (mod 3), C(n, k) mod 2, 0).
%F a(n) = Sum_{k=0..n} if(n+k == 2 (mod 3), C(n, k) mod 2, 0).
%F a(n) = (1/2) * Sum_{k=0..n} ({F(n+k)*binomial(n, k)} mod 2) where F = A000045.
%F Recurrence : a(0) = 0 then a(2n) = a(n) and a(2n+1) = 2*a(n) + 1 - 2*t(n) where t(n) = A010060(n) is the Thue-Morse sequence. - _Benoit Cloitre_, May 15 2005
%t j[n_] := (2^n - (-1)^n) / 3; a[n_] := j[DigitCount[n, 2, 1]]; Array[a, 100, 0] (* _Amiram Eldar_, Jul 22 2023 *)
%Y Cf. A000045, A000120, A001045, A001316, A010060, A102395.
%K easy,nonn
%O 0,8
%A _Paul Barry_, Jan 06 2005