Primitive bit strings

In order for a bit string to repeatedly concatenate to another string of length n, the length of the string must be a proper divisor of n. We can develop a recurrence relation for the number of primitive strings: $F(n)={ 2 }^{ n }-\sum { F(d) }$ for each proper divisor d of n. Calculating only the values we need we get $F(1)={2}^{1}\\ F(2)={ 2 }^{ 2 }-F(1)=2\\ F(3)={ 2 }^{ 3 }-F(2)-F(1)=6\\ F(4)={ 2 }^{ 4 }-F(2)-F(1)=12\\ F(6)={ 2 }^{ 6 }-F(3)-F(2)-F(1)=54\\ F(12)={ 2 }^{ 12 }-F(6)-F(4)-F(3)-F(2)-F(1)=\boxed { 4020 }$

Let's compute the number of non-primitive strings first.

There are $2^6$ strings which can be divided into two equal parts, such as 101100101100, 100100100100, 010101010101, and 111111111111.

There are $2^4$ more strings which can be divided into three equal parts, such as 110011001100, 101110111011, 010101010101, and 111111111111.

However, $2^2$ of these strings actually count the same strings, those that can be divided into six equal parts, such as 010101010101 and 111111111111 above.

Thus, by Principle of Inclusion-Exclusion, there are $2^6 + 2^4 - 2^2 = 76$ non-primitive strings. None of the remaining strings can be non-primitive, since the length of the string is $12$ and the number of parts must thus be a divisor of $12$ , which thus must have either $2$ or $3$ as a prime factor and thus is counted above.

There are $2^{12}$ bit strings in total. Thus, by complementary counting, there are $2^{12} - 76 = \boxed{4020}$ primitive strings.