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A mod 2 related Jacobsthal sequence.
2

%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