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A114482 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).

%I A114482 #16 Mar 20 2015 18:09:43
%S A114482 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 A114482 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 A114482 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 A114482 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 A114482 Number of terms in S(n) is A062318(n).
%C A114482 Interpreting S(n) in binary and converting to decimal gives 1,2,20,164,84296,43159880,5792821120672400,...,.
%e A114482 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 A114482 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 A114482 Cf. A114483, A062318, A112361.
%K A114482 easy,nonn
%O A114482 1,1
%A A114482 _Leroy Quet_, Nov 30 2005
%E A114482 More terms from _Robert G. Wilson v_, Jan 01 2006
%E A114482 Edited by _N. J. A. Sloane_, Jan 03 2006