6. Given the following conditions for the sequence $\left\{a_{n}\right\}$:
(1) $\left\{2^{a_{n}}\right\}$ is a geometric sequence;
(2) The sum of the first $n$ terms $S_{n}=n^{2}+1$;
(3) $a_{1}>0$, and $a_{k}=\frac{2}{k-1}\left(a_{1}+a_{2}+\cdots+\right.$ $\left.a_{k-1}\right), k \geqslant 2$.
The condition that makes $\left\{a_{n}\right\}$ an arithmetic sequence is $\qquad$ (fill in the sequence number).
6. (1)(3).

First, consider (1). Let $b_{n}=2^{a_{n}}, n \geqslant 1, n \in \mathbf{N}$, then $a_{n}=\log _{2} b_{n}$. Since $\left\{2^{a_{n}}\right\}$ is a geometric sequence, we have
$2^{a_{n}}=b_{1} \times q^{n-1}\left(b_{1}, q\right.$ are positive constants $)$,
thus, $a_{n}=\log _{2} b_{n}=\log _{2} b_{1}+(n-1) \log _{2} q$.

Here, $\log _{2} b_{1}$ and $\log _{2} q$ are constants.
Therefore, $\left\{a_{n}\right\}$ is an arithmetic sequence.
Consider (2). Since $a_{n}=S_{n}-S_{n-1}=2 n-1(n \geqslant 2)$, and $a_{1}=S_{1}=2$ does not fit $a_{n}=2 n-1$.
Thus, $\left\{a_{n}\right\}$ is not an arithmetic sequence.
Consider (3). From the given equation, we have
$$
\begin{array}{l}
2\left(a_{1}+a_{2}+\cdots+a_{k-2}+a_{k-1}\right)=(k-1) a_{k} . \\
2\left(a_{1}+a_{2}+\cdots+a_{k-2}\right)=(k-2) a_{k-1} .
\end{array}
$$
(1) - (2), we get
$$
2 a_{k-1}=(k-1) a_{k}-(k-2) a_{k-1} \text {. }
$$

Thus, $a_{k}=\frac{k}{k-1} a_{k-1}(k \geqslant 2)$.
By repeatedly applying this formula, we get
$$
\begin{aligned}
a_{k} & =\frac{k}{k-1} a_{k-1}=\frac{k}{k-1} \cdot \frac{k-1}{k-2} a_{k-2} \\
& =\cdots \\
& =\frac{k}{k-1} \cdot \frac{k-1}{k-2} \cdots \cdots \cdot \frac{3}{2} \cdot \frac{2}{1} a_{1} \\
& =k a_{1} .
\end{aligned}
$$

From this, we know that $\left\{a_{n}\right\}$ is an arithmetic sequence (with the common difference equal to the first term):
$$
a_{1}, 2 a_{1}, 3 a_{1}, \cdots, n a_{1}, \cdots \text {. }
$$

In summary, (1) and (3) are the required results.