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More likely, a(n)/n should tend to log_2(3)/2 = 0.79248+, since a bit is equally likely to be 0 or 1. (Well, the first and last bits are ones; I meant the other bits.)

The fours bit is always zero. Other bits (at least up to the 2^29s position) are exactly balanced.

to know the structure of the multiplicative group modulo 2^n , it is useful to consider the unit group in the ring of 2-adic integers. it is well-known that this group is generated by -1 and 5 . moreover, it is a direct product, with -1 having order 2 , and 5 having infinite order. when we work modulo 2^n (n > 2), this means that (Z / 2^n Z)^* is isomorphic to a product of two cyclic groups, C_2 x C_(2^n-2) with generators -1 and 5 . we're interested in the subgroup generated by 3 , so we express it as a product of our generators; we get 3 = -1 * 5^t mod 2^n , where t is an odd integer. (to see that t is odd, consider the congruence modulo 8 .) thus we see that 3 has order 2^n-2 , which means that it generates a subgroup of index 2 inside the multiplicative group (Z / 2^n Z)^* . dan hoey's observation that 3^t is never congruent to 5 or 7 mod 8 identifies which subgroup it is. therefore all bit patterns subject to this condition occur. mike