cp's OEIS Frontend

A164349 is generated as follows. Start with the string 01, and at each stage copy the previous string twice and remove the last symbol.

Since the number of symbols in the whole string is 2^n + 1, A164363 + A164362 = 2^n + 1.

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..