How to prove that these conditions generate an arithmetic sequence? 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 $a_1\,?$
 A: 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 $q-p$.
A: (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 $i\in\{1,2,3\}$...
