Here a combinatorics problem. I offer 30 euro for a proof and 100 bounty points for a counterexample: Let $n \geq 2$.

An $n$-*Kupisch series* is a list of $n$ numbers $c:=[c_1,c_2,...,c_n]$ with $c_n=1$, $c_i \ge 2$ for $i \neq n$ and $c_i-1 \leq c_{i+1}$ for all $i=1,...,n-1$ and setting $c_0:=c_n$.
The number of such $n$-Kupisch series is equal to $C_{n-1}$ (Catalan numbers).
The *CoKupisch series* $d$ of $c$ is defined as $d=[d_1,d_2,...,d_n]$ with $d_i:= \min \{k | k \geq c_{i-k} \} $ and $d_1=1$. One can show that the $d_i$ are a permutation of the $c_i$. A number $a \in \{1,...,n \}$ is a *descent* if $a=1$ or $c_a >c_{a-1}$. Define a corresponding set, indexed by descents: $X_1 := \{1,2,...,c_1-1 \}$, and $X_a := \{ c_{a-1}, c_{a-1}+1 ,..., c_a -1 \}$ for descents $a > 1$.

A $n$-Kupisch series is called $2-$Gorenstein if it satisfies the following condition:

for each descent $a$, and each $b \in X_a$: either $c_{a+b} \geq c_{a+b-1}$ or $d_{a+b-1} = d_{a+b + c_{a+b}-1} - c_{a+b}$ is satisfied.

Conjecture: The number of $n$-Kupisch series that are $2$-Gorenstein for $n \geq 2$ equals 1, 2, 4, 9, 21, 51, 127, 323, 835, 2188, 5798, 15511, 41835, ... which are the Motzin numbers. https://oeis.org/A001006

My computer says it is true for $n=2,3,...,14$. (also for n=1 if [1] is the unique 1-Kupisch series, by a representation-theoretic interpretation)

Here an example which visualises the c and d sequences in a picture of a Dyck path: https://drive.google.com/file/d/0B9hKtvQe-4-bQlpjcURfYnNzUGs/view

Background: This has a representation theoretic background (see http://www.sciencedirect.com/science/article/pii/0022404994900442 )and would give a certain classification result together with a new categorification of the Motzkin numbers. There is a very natural bijection of n-Kupisch series to Dyck paths from (0,0) to (2n-2,0) and probably the 2-Gorenstein algebras among them might give a new combinatorial interpretation of Motzkin paths as subpaths of Dyck paths.

I found in fact many such things and translated them into elementary problems, but I have no talent in complicated combinatorics :(

example for n=5:

All 5-Kupisch series (number is 14): [ [ 2, 2, 2, 2, 1 ], [ 3, 2, 2, 2, 1 ], [ 2, 3, 2, 2, 1 ], [ 3, 3, 2, 2, 1 ], [ 4, 3, 2, 2, 1 ], [ 2, 2, 3, 2, 1 ], [ 3, 2, 3, 2, 1 ], [ 2, 3, 3, 2, 1 ], [ 3, 3, 3, 2, 1 ], [ 4, 3, 3, 2, 1 ], [ 2, 4, 3, 2, 1 ], [ 3, 4, 3, 2, 1 ], [ 4, 4, 3, 2, 1 ], [ 5, 4, 3, 2, 1 ] ]

All 2-Gorenstein 5-Kupisch series (number is 9): [ [ 2, 2, 2, 2, 1 ], [ 3, 2, 2, 2, 1 ], [ 2, 3, 2, 2, 1 ], [ 4, 3, 2, 2, 1 ], [ 2, 2, 3, 2, 1 ], [ 3, 2, 3, 2, 1 ], [ 3, 3, 3, 2, 1 ], [ 2, 4, 3, 2, 1 ], [ 5, 4, 3, 2, 1 ] ]

Representation theoretic conjecture/background:

Conjecture:

The number of 2-Gorenstein algebras that are Nakayama algebras with n simple modules with a linear oriented line as a quiver is equal to the sequence of Motzkin numbers.

Background/explanations:

Here the $c_i$ are the dimension of the indecomposable projective modules which determine the algebra uniquely (assuming that the algebras are connected quiver algebras over a field). The $d_i$ are the dimension of the indecomposable injective modules at point $i$. 2-Gorenstein means that the dual of the regular module $D(A)$ has a projective presentation $P_1 \rightarrow P_0 \rightarrow D(A) \rightarrow 0$ with $P_1$ having injective dimension bound by 1. ($P_0$ is always projective-injective for Nakayama algebras)

Here a link for the program I used to test things (copy all programs, and the finaltest program does the job then. 1 means the Kupisch series is 2-Gorenstein and 0 means it is not): https://docs.google.com/document/d/1U9mriuvCEE9FeXY1TfY_yJ1mi05TCdHRfppwCSwi1S8/pub

edit: Since we have 2 people with proofs now, I award 30 Euro each to them in case their proof is correct (I still have to check) and then no more money for new proofs.

I award additionally 200 bounty points to the nicest proof (which might be one of the two first posted proofs or another not yet posted proof), which can be decided by the community in terms of upvotes near the end when the bounty expires (16.08.)

edit 2: Now at 16.08. I give the bounty points to Anton, since it is the most complete answer and the details of findstats answer are not posted yet. I will give 30 Euro to each after checking findstats proof (when it is posted and correct).

edit 3: Since findstat updated the answer and now has a full proof of an extension of the conjecture just a little time after I awarded the first bounty, I also will give a bounty to them and accept their answer (second bounty had to be 400 or higher and can be awarded after 24 hours).