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%I #17 Jan 27 2015 11:22:22
%S 1,0,1,0,0,1,0,1,0,0,1,0,0,0,1,0,1,0,0,1,0,0,1,0,1,0,0,1,0,0,0,1,0,1,
%T 0,0,1,0,1,0,1,0,0,1,0,0,0,1,0,1,0,0,1,0,0,1,0,1,0,0,1,0,0,0,1,0,1,0,
%U 0,1,0,0,1,0,0,1,0,0,0,1,0,1,0,0,1,0,0,1,0,1,0,0,1,0,0,0,1,0,1,0,0,1,0,1,0
%N a(n) = A164349(2^n).
%C This is the last symbol at each stage of the method for generating A164349 using string operations.
%C The number of 1's in the string is given by A164363, and this number is given by the recurrence
%C A164363(n+1) = 2 * A164363(n) - A164364(n).
%C This leads to the formula A164363(n+1) = 2^n - 2^(n-1) * A164364(1) - 2^(n-2) * A164364(2) - ... - A164364(n);
%C for example,
%C A164363(5) = 16 - 8 A164364(1) - 4 A164364(2) - 2 A164364(3) - A164364(4).
%C This means that since the total number of symbols in the n-th string is 2**n + 1, the proportion of 0's in the first k terms of A164349, as n tends to infinity, is given by the number whose binary expansion is exactly this sequence. This number is approximately 0.6450588..
%H Paul Tek, <a href="/A164364/b164364.txt">Table of n, a(n) for n = 0..10000</a>
%p A053645 := proc(n) local dgs ; dgs := convert(n,base,2) ; add(op(i,dgs)*2^(i-1),i=1..nops(dgs)-1) ; end: A164349 := proc(n) option remember; if n <= 1 then n; else a := A053645(n-1) ; while a > 1 do a := A053645(a-1) ; od: a ; fi; end: A164364 := proc(n) A164349(2^n) ; end: seq(A164364(n),n=0..120) ; # _R. J. Mathar_, Aug 17 2009
%t t = Nest[ Most@ Flatten@ {#, #} &, {0, 1}, 25]; Table[ t[[2^n + 1]], {n, 0, 25}] (* _Robert G. Wilson v_, Aug 17 2009 *)
%Y Cf. A164349, A164362, A164363.
%K nonn
%O 0,1
%A _Jack W Grahl_, Aug 14 2009
%E More terms from _R. J. Mathar_, Aug 17 2009
%E Incorrect comments removed by _Jack W Grahl_, Dec 26 2014
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