1. **State the problem:** We need to find integers $n$, $a$, $b$, and $c$ such that the first three terms in descending powers of $x$ in the expansion of $$\left(2x^{2} - \frac{1}{4x}\right)^n$$ are $$256x^{16} + ax^{13} + bx^{c}.$$\n\n2. **Recall the binomial expansion formula:** For any integer $n$, $$(A + B)^n = \sum_{k=0}^n \binom{n}{k} A^{n-k} B^k.$$\nHere, $A = 2x^2$ and $B = -\frac{1}{4x}$.\n\n3. **Write the first three terms explicitly:**\n$$T_0 = \binom{n}{0} (2x^2)^n \left(-\frac{1}{4x}\right)^0 = (2x^2)^n = 2^n x^{2n}.$$\n$$T_1 = \binom{n}{1} (2x^2)^{n-1} \left(-\frac{1}{4x}\right)^1 = n \cdot 2^{n-1} x^{2(n-1)} \cdot \left(-\frac{1}{4x}\right) = -\frac{n}{4} 2^{n-1} x^{2n-2-1} = -\frac{n}{4} 2^{n-1} x^{2n-3}.$$\n$$T_2 = \binom{n}{2} (2x^2)^{n-2} \left(-\frac{1}{4x}\right)^2 = \frac{n(n-1)}{2} 2^{n-2} x^{2(n-2)} \cdot \frac{1}{16 x^2} = \frac{n(n-1)}{2} 2^{n-2} \frac{1}{16} x^{2n-4-2} = \frac{n(n-1)}{2} \frac{2^{n-2}}{16} x^{2n-6}.$$\n\n4. **Match the powers and coefficients with the given terms:**\nThe first term is $$256 x^{16}.$$\nFrom $T_0$, the power of $x$ is $2n$, so $$2n = 16 \implies n = 8.$$\nCheck coefficient: $$2^n = 2^8 = 256,$$ which matches perfectly.\n\n5. **Find $a$ from $T_1$:**\nSubstitute $n=8$ into $T_1$ coefficient:\n$$a = -\frac{8}{4} 2^{7} = -2 \times 128 = -256.$$\nPower of $x$ in $T_1$ is $$2n - 3 = 16 - 3 = 13,$$ so the term is $$-256 x^{13}.$$\n\n6. **Find $b$ and $c$ from $T_2$:**\nCoefficient:\n$$b = \frac{8 \times 7}{2} \times \frac{2^{6}}{16} = 28 \times \frac{64}{16} = 28 \times 4 = 112.$$\nPower of $x$ is $$2n - 6 = 16 - 6 = 10,$$ so the term is $$112 x^{10}.$$\n\n7. **Final answer:**\n$$n = 8, \quad a = -256, \quad b = 112, \quad c = 10.$$