A164364 a(n) = A164349(2^n).

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, 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, 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

Offset

0,1

Comments

This is the last symbol at each stage of the method for generating A164349 using string operations.

The number of 1's in the string is given by A164363, and this number is given by the recurrence

A164363(n+1) = 2 * A164363(n) - A164364(n).

This leads to the formula A164363(n+1) = 2^n - 2^(n-1) * A164364(1) - 2^(n-2) * A164364(2) - ... - A164364(n);

for example,

A164363(5) = 16 - 8 A164364(1) - 4 A164364(2) - 2 A164364(3) - A164364(4).

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..

Links

Paul Tek, Table of n, a(n) for n = 0..10000

Maple

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

Mathematica

t = Nest[ Most@ Flatten@ {#, #} &, {0, 1}, 25]; Table[ t[[2^n + 1]], {n, 0, 25}] (* Robert G. Wilson v, Aug 17 2009 *)

Crossrefs

Cf. A164349, A164362, A164363.

Sequence in context: A186518 A127829 A127831 * A198517 A105385 A190227

Adjacent sequences: A164361 A164362 A164363 * A164365 A164366 A164367

Keyword

nonn

Author

Jack W Grahl, Aug 14 2009

Extensions

More terms from R. J. Mathar, Aug 17 2009

Incorrect comments removed by Jack W Grahl, Dec 26 2014

Status

approved