%I #11 Mar 30 2023 16:43:39
%S 0,0,1,0,2,0,2,2,1,0,4,0,2,8,0,0,1,8,4,8,2,8,2,2,1,8,4,8,2,8,2,2,1,8,
%T 4,8,2,8,2,2,1,8,4,8,2,8,2,2,1,8,4,8,2,8,2,2,1,8,4,8,2,8,2,2,1,8,4,8,
%U 2,8,2,2,1,8,4,8,2,8,2,2,1,8,4,8,2,8,2
%N A variant of Van Eck's sequence where we only consider powers of 2: for n >= 0, if a(n) = a(n-2^e) for some e, take the least such e and set a(n+1) = 2^e; otherwise a(n+1) = 0. Start with a(1) = 0.
%C The sequence is eventually 8-periodic.
%C The variant with powers of 4 is 3-periodic: (0 0 1)*.
%H <a href="/index/Rec#order_08">Index entries for linear recurrences with constant coefficients</a>, signature (0,0,0,0,0,0,0,1).
%F a(n) = a(n-8) for n >= 25.
%e a(1) = 0 by definition.
%e a(2) = 0 as there is only one occurrence of a(1) = 0 so far.
%e a(3) = 2^0 = 1 as a(2) = a(2-2^0).
%e a(4) = 0 as there is only one occurrence of a(3) = 1 so far.
%e a(5) = 2^1 = 2 as a(4) = a(4-2^1).
%e a(6) = 0 as there is only one occurrence of a(5) = 2 so far.
%e a(7) = 2^1 = 2 as a(6) = a(6-2^1).
%e a(8) = 2^1 = 2 as a(7) = a(7-2^1).
%e a(9) = 2^0 = 1 as a(8) = a(8-2^0).
%e a(10) = 0 as a(9) <> a(9-2^e) for any admissible e.
%o (PARI) { for (n=1, #a=vector(87), for (e=0, oo, m = n-1-d=2^e; if (m<1, break, a[n-1]==a[m], a[n]=d; break)); print1 (a[n]", ")) }
%Y Cf. A181391, A353323 (variant for powers of 3).
%K nonn
%O 1,5
%A _Rémy Sigrist_, Apr 12 2022