Let $$S_n(x) = \log_{a^{1/2}} x + \log_{a^{1/3}} x + \log_{a^{1/6}} x + \log_{a^{1/11}} x + \log_{a^{1/18}} x + \log_{a^{1/27}} x + \ldots$$ up to $$n$$-terms, where $$a > 1$$. If $$S_{24}(x) = 1093$$ and $$S_{12}(2x) = 265$$, then value of $$a$$ is equal to ________.

We have $$S_n(x) = \log_{a^{1/2}} x + \log_{a^{1/3}} x + \log_{a^{1/6}} x + \log_{a^{1/11}} x + \log_{a^{1/18}} x + \log_{a^{1/27}} x + \ldots$$

Using the change of base formula, $$\log_{a^{1/k}} x = k \log_a x$$. The exponent denominators are $$2, 3, 6, 11, 18, 27, \ldots$$ and the differences are $$1, 3, 5, 7, 9, \ldots$$ (consecutive odd numbers). So the $$n$$-th term denominator is $$2 + \sum_{k=1}^{n-1}(2k-1) = 2 + (n-1)^2 = n^2 - 2n + 3$$.

Verification: for $$n=1$$: $$1-2+3=2$$ , $$n=2$$: $$4-4+3=3$$, $$n=3$$: $$9-6+3=6$$, $$n=4$$: $$16-8+3=11$$. All correct.

So $$S_n(x) = \log_a x \cdot \sum_{k=1}^{n}(k^2 - 2k + 3) = \log_a x \cdot \left(\frac{n(n+1)(2n+1)}{6} - n(n+1) + 3n\right)$$.

Simplifying: $$\frac{n(n+1)(2n+1)}{6} - n(n+1) + 3n = n\left(\frac{(n+1)(2n+1) - 6(n+1) + 18}{6}\right) = n\left(\frac{2n^2 - 3n + 13}{6}\right)$$.

For $$n=24$$: $$\sum = 24 \times \frac{2(576) - 72 + 13}{6} = 24 \times \frac{1093}{6} = 4 \times 1093 = 4372$$.

So $$S_{24}(x) = 4372 \log_a x = 1093$$, giving $$\log_a x = \frac{1}{4}$$, hence $$x = a^{1/4}$$.

For $$n=12$$: $$\sum = 12 \times \frac{2(144) - 36 + 13}{6} = 12 \times \frac{265}{6} = 530$$.

So $$S_{12}(2x) = 530 \log_a(2x) = 265$$, giving $$\log_a(2x) = \frac{1}{2}$$, hence $$2x = a^{1/2}$$.

Since $$x = a^{1/4}$$, we get $$2a^{1/4} = a^{1/2}$$. Let $$t = a^{1/4}$$, then $$2t = t^2$$, so $$t = 2$$, meaning $$a^{1/4} = 2$$ and $$a = 16$$.

The answer is $$16$$.