The answer is yes. We show that not all large enough odd integers are expressible in this way. Suppose this were not the case and all odd $n \geqslant M$ are expressible in this way. Pick some $n > \max(10^{100}, M)$ and suppose $a, b, c$ are such that $n$ equals the expression. Obviously, we must have $b+c = 2^k$, for some positive integer $k$. We split into three cases: 1). $k < c+a$ In this case, $n = a+2^k-c+k-2^{c+a} \leqslant 2a-2^{a+c-1}$ which is negative for $a > 2$, which implies that the expression, is clearly always bounded, by in fact $20$, but here we assume $10^{100}$ for completeness. Thus, this case is not possible. 2). $k = c+a$ In this case, the expression is $2a$, contradicting the fact that $n$ is odd. This case is also not possible. 3). $k > c+a$ Suppose $k = c+a+\ell$ for some $\ell > 0$. The expression becomes $2a + 2^{a+c+\ell} - 2^{a+c} + \ell$. Let us count the number of values this expression can take which are at most $C$ for some large $C$. Because $C \geqslant 2^{a+c+\ell-1}$, we obtain that $a+c+\ell = \mathcal O(\log C)$. Because $a, c, \ell$ are all $\mathcal O(\log C)$ the number of values this expression takes which are at most $C$ is $\mathcal O(\log^3 C) = o(C)$. In particular, this contradicts the fact that we have a positive density of such expressible positive integers.