The sequence $a_n$ satisfies $$a_1=1,\\a_n+a_{n+12}=2a_{n+6},\\a_n+a_{n+14}=2a_{n+7}.$$
I know that these conditions imply that $a_n$ consists of several arithmetic subsequences with index interval as $6$ and $7.$ Because $6$ and $7$ are coprime, I guess this implies that the whole sequence is arithmetic.
But how do we prove that the sequence is indeed arithmetic?
Also: is the sequence arithmetic regardless of the value of $a_1\,?$
On
Please ignore this flawed proof -- see the comments by @Hans J.
$$a_n + a_{n+12} = 2a_{n+6} \Longleftrightarrow$$ $$ a_{n+12}-a_{n+6} = a_{n+6} -a_n \Longleftrightarrow$$ $$ \exists p: a_{6m+i} = a_i + pm\Longleftrightarrow$$ $$ a_{i-6m} = a_i - pm$$ Similarly: $$ \exists q: a_{7m+j} = a_j + qm$$
$$a_{m+1} = a_{7m-6m+1} = a_{7m+1} - pm = a_1 + qm - pm \Longrightarrow$$ $$a_{m+1} = a_1 + (q-p)m$$
I.e., sequence $\{a_i\}$ is an arithmetic progression with a common difference of $q-p$.
On
(Too long for comment)
Eliminating $a_n$ gives the relation
$$a_{n+14} - a_{n+12} = (a_{n+14} - a_{n+13}) + (a_{n+13} - a_{n+12}) = 2(a_{n+7} - a_{n+6})$$
If $b_n=a_{n+1}-a_n$ is the sequence of first-order differences of $a_n$, then this is equivalent for $n\ge1$ to
$$b_{n+13}+b_{n+12}=2b_{n+6} \implies b_{n+7} + b_{n+6} - 2b_n = 0$$
with solution
$$b_n = c_1 + \sum_{i=2}^7 c_i {r_i}^n$$
where $r_i$ are the roots to the characteristic polynomial
$$r^7 + r^6 - 2 = (r-1) (r^6 + 2r^5 + 2r^4 + 2r^3 + 2r^2 + 2r + 2)$$
Solving for $a_n$ by substitution gives
$$\begin{align*} a_n &= a_1 + \sum_{i=1}^n b_i \\[1ex] &= 1 + \sum_{i=1}^n \left(c_1 + \sum_{j=2}^7 c_j {r_j}^n\right) \\[1ex] &= 1 + c_1n + \left(c_2 \rho^n + c_3 \bar\rho^n + c_4 \sigma^n + c_5 \bar\sigma^n + c_6 \tau^n + c_7 \bar\tau^n\right)n \end{align*}$$
where $\rho,\sigma,\tau$ and their conjugates $\bar\rho,\bar\sigma,\bar\tau$ are the remaining complex characteristic roots.
Then it suffices to show the exponential terms will cancel or reduce to a constant to be combined with $c_1$. This would be trivial if we knew $\gcd(c_{2i},c_{2i+1})\neq1$ for $i\in\{1,2,3\}$...
The $a_n$ sequence is always arithmetic with the two provided recurrence relations of
$$a_{n} + a_{n+12} = 2a_{n+6} \tag{1}\label{eq1A}$$
$$a_{n} + a_{n+14} = 2a_{n+7} \tag{2}\label{eq2A}$$
Choose any $n = n_1 \ge 1$. There exist $e_1$ and $e_2$ with $a_{n_{1}+6} = a_{n_{1}} + 6e_{1}$ and $a_{n_{1}+12} = a_{n_{1}} + 12e_{2}$. Using these in \eqref{eq1A} gives
$$\begin{equation}\begin{aligned} a_{n_{1}} + (a_{n_{1}} + 12e_{2}) & = 2(a_{n_{1}} + 6e_{1}) \\ 2a_{n_{1}} + 12e_{2} & = 2a_{n_{1}} + 12e_{1} \\ e_{2} & = e_{1} \end{aligned}\end{equation}\tag{3}\label{eq3A}$$
This means $a_{n_{1} + 12} = a_{n_{1}} + 12e_{1} \; \to \; a_{n_{1} + 12} - a_{n_{1} + 6} = a_{n_{1} + 6} - a_{n_{1}} = 6e_{1}$, i.e., the change in values of $a_n$ when the index increases by $6$ is the same. Note we can repeat this using $n = n_1 + 6$ to get that $a_{n_{1} + 18} - a_{n_{1} + 12} = 6e_{1}$, as well as going up or down by $6$ any number of times, as can be proven using induction. This means for all integers $k_1$ with $n_1 + 6k_1 \ge 1$, we've that
$$a_{n_1 + 6k_1} = a_{n_1} + 6k_{1}e_{1} \tag{4}\label{eq4A}$$
We can use a similar procedure with \eqref{eq2A} to get that, for some $f_1$, we have for all integers $k_2$ with $n_1 + 7k_2 \ge 1$ that
$$\color{blue}{a_{n_1 + 7k_2} = a_{n_1} + 7k_{2}f_{1}} \tag{5}\label{eq5A}$$
Using $k_1 = 7$ in \eqref{eq4A} and $k_2 = 6$ in \eqref{eq5A} gives
$$a_{n_1 + 42} = a_{n_1} + 6(7)e_1 = \color{blue}{a_{n_1} + 7(6)f_1} \; \; \to \; \; e_1 = f_1 = d_1 \tag{6}\label{eq6A}$$
We can next use $n = n_1 + 7$ as a starting point, and repeat the procedure above to get that there exists $e_{3}$ and $f_{3}$ where, for any integers $k_3$ and $k_4$, we have that
$$\color{red}{a_{n_1 + 7 + 6k_3} = a_{n_1 + 7} + 6k_{3}e_{3}} \tag{7}\label{eq7A}$$
$$a_{n_1 + 7 + 7k_4} = a_{n_1 + 7} + 7k_{4}f_{3} \tag{8}\label{eq8A}$$
with $e_{3} = f_{3}$. Since \eqref{eq8A} has a base point of $n_1 + 7$, and \eqref{eq5A} applies for all $k_2$, we get $f_{3} = f_{1}$. An alternate way to show this is to use $k_2 = k_4 + 1$ and $k_2 = 1$ in \eqref{eq5A} to therefore get from \eqref{eq8A} that
$$\begin{equation}\begin{aligned} a_{n_1 + 7(k_4 + 1)} & = a_{n_1 + 7} + 7k_{4}f_{3} \\ \color{blue}{a_{n_1} + 7(k_4 + 1)f_{1}} & = (\color{blue}{a_{n_1} + 7f_{1}}) + 7k_{4}f_{3} \\ 7k_{4}f_{1} & = 7k_{4}f_{3} \\ f_{1} & = f_{3} \end{aligned}\end{equation}\tag{9}\label{eq9A}$$
i.e., $e_{3} = f_{3} = d_{1}$ as well. Thus, using $k_3 = -1$ in \eqref{eq7A}, plus $k_2 = 1$ in \eqref{eq5A}, we next have
$$\begin{equation}\begin{aligned} \color{red}{a_{n_{1} + 7 + 6(-1)}}\text{ } & \color{red}{= a_{n_1 + 7} + 6(-1)d_{1}} \\ a_{n_{1} + 1} & = (\color{blue}{a_{n_1} + 7(1)d_{1}}) - 6d_{1} \\ a_{n_{1} + 1} & = a_{n_1} + d_{1} \end{aligned}\end{equation}\tag{10}\label{eq10A}$$
Note we can use a procedure similar to the above with a starting point of $n = n_1 + 7k$ for any integer $k$ where $n \ge 1$ (actually, it's also true for $n \le 0$ as well) to get that
$$a_{n_{1} + k} = a_{n_1} + kd_{1} \tag{11}\label{eq11A}$$
i.e., $a_{n}$ is an arithmetic sequence with a common difference of $d_1$. This shows the value of $a_{1}$ doesn't affect whether or not it's an arithmetic sequence but, instead, just determines the actual values of $a_n$ based on the multiples of $d_1$ offsets from $a_1$.
Note: This works not only for index differences of $6$ and $7$, but also for any $2$ integers which are coprime to each other. The basic underlying reason for this is Bézout's identity which states that, if the index differences were $x$ and $y$ with $\gcd(x,y) = d$, then there are integers $m$ and $n$ with
$$mx + ny = d \tag{12}\label{eq12A}$$
This allows matching up the value differences with index offsets of $d$ for the sequences determined by integer index differences of $x$ and $y$ to each other, as done above for $x = 6$ and $y = 7$ with $d = 1$. However, if $d \gt 1$, then the best we can state is that there are multiple (possibly different) arithmetic sequences where, for each $0 \le i \le d - 1$, we have $a_{kd + i}$ forms an arithmetic sequence for all integers $k$.