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Let S(1)=1, S(2)=10; S(2n)=concatenation of S(2n-1), S(2n-2) and 0; and S(2n+1)=concatenation of S(2n), S(2n) and 0. Sequence gives S(infinity).
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%I #16 Mar 20 2015 18:09:43

%S 1,0,1,0,0,1,0,0,1,0,1,0,0,1,0,0,0,1,0,1,0,0,1,0,0,0,1,0,1,0,0,1,0,0,

%T 1,0,1,0,0,1,0,0,0,1,0,1,0,0,1,0,0,0,0,1,0,1,0,0,1,0,0,1,0,1,0,0,1,0,

%U 0,0,1,0,1,0,0,1,0,0,0,0,1,0,1,0,0,1,0,0,1,0,1,0,0,1,0,0,0,1,0,1,0,0,1,0,0

%N Let S(1)=1, S(2)=10; S(2n)=concatenation of S(2n-1), S(2n-2) and 0; and S(2n+1)=concatenation of S(2n), S(2n) and 0. Sequence gives S(infinity).

%C Number of terms in S(n) is A062318(n).

%C Interpreting S(n) in binary and converting to decimal gives 1,2,20,164,84296,43159880,5792821120672400,...,.

%e S(3) = {1,0,1,0,0}, S(4) = {1,0,1,0,0,1,0,0}, S(5) = {1,0,1,0,0,1,0,0,1,0,1,0,0,1,0,0,0}, ...

%t a[1] = {1}; a[2] = {1, 0}; a[n_] := a[n] = If[EvenQ[n], Join[a[n - 1], a[n - 2], {0}] // Flatten, Join[a[n - 1], a[n - 1], {0}] // Flatten]; a[8] (* _Robert G. Wilson v_ *)

%Y Cf. A114483, A062318, A112361.

%K easy,nonn

%O 1,1

%A _Leroy Quet_, Nov 30 2005

%E More terms from _Robert G. Wilson v_, Jan 01 2006

%E Edited by _N. J. A. Sloane_, Jan 03 2006