This is just an "elementarization" of @WhatsUp's elegant proof to help in giving some intuition.

Let there be two patterns of $n_i$ moves, respectively, each summing to a uniform increase of $k_i$ in each square. Let $\{x_{ij}\}$ the "cell count", i.e. the number of times square $j$ was chosen (as the center) in pattern $i$. Now multiply each cell count in pattern $1$ by each cell count in pattern $2$ that is within the "+"-pento centered at the first cell (this is, of course, symmteric, i.e., equivalently, the first cell is within the pento centered at the second cell) and form the sum: $S = \sum_{j,j' \text{"pento-connected"}} x_{1j}x_{2j'}$. Then $S = \sum_j x_{1j} \sum_{j\text{within pento at}j'} x_{2j'} = \sum_j x_{1j} k_2 = k_2 n_1$ and, similarly, $S = k_1 n_2$.

Substituting $k_1,n_1 = 11,69$ from pattern given by WhatsUp and $k_2 = 2020$ we find that a matching integer $n_2$ does not exist.

This is just an "elementarization" of @WhatsUp's elegant proof to help in giving some intuition.

Let there be two patterns of $n_i$ moves, respectively, each summing to a uniform increase of $k_i$ in each square. Let $\{x_{ij}\}$ the "cell count", i.e. the number of times square $j$ was chosen (as the center) in pattern $i$. Now multiply each cell count in pattern $1$ by each cell count in pattern $2$ that is within the "+"-pento centered at the first cell (this is, of course, symmteric, i.e., equivalently, the first cell is within the pento centered at the second cell) and form the sum: $S = \sum_{j,j' \text{"pento-connected"}} x_{1j}x_{2j'}$. Then $S = \sum_j x_{1j} \sum_{j\text{within pento at}j'} x_{2j'} = \sum_j x_{1j} k_2 = 25 k_2 n_1$ and, similarly, $S = 25 k_1 n_2$.

Substituting $k_1,n_1 = 11,69$ from pattern given by WhatsUp and $k_2 = 2020$ we find that a matching integer $n_2$ does not exist.

This is just an "elementarization" of @WhatsUp's elegant proof to help in giving some intuition.

Let there be two patterns of $n_i$ moves, respectively, each summing to a uniform increase of $k_i$ in each square. Let $\{x_{ij}\}$ the "cell count", i.e. the number of times square $j$ was chosen (as the center) in pattern $i$. Now multiply each cell count in pattern $1$ by each cell count in pattern $2$ that is within the "+"-pento centered at the first cell (this is, of course, symmteric, i.e., equivalently, the first cell is within the pento centered at the second cell) and form the sum: $S = \sum_{j,j' \text{"pento-connected"}} x_{1j}x_{2j'}$. Then $S = \sum_j x_{1j} \sum_{j\text{within pento at}j'} x_{2j'} = \sum_j x_{1j} k_2 = k_2 n_1$ and similarly, $S = k_1 n_2$.

Substituting $k_1,n_1 = 11,69$ from pattern given by WhatsUp and $k_2 = 2020$ we find that a matching integer $n_2$ does not exist.

This is just an "elementarization" of @WhatsUp's elegant proof to help in giving some intuition.

Let there be two patterns of $n_i$ moves, respectively, each summing to a uniform increase of $k_i$ in each square. Let $\{x_{ij}\}$ the "cell count", i.e. the number of times square $j$ was chosen (as the center) in pattern $i$. Now multiply each cell count in pattern $1$ by each cell count in pattern $2$ that is within the "+"-pento centered at the first cell (this is, of course, symmteric, i.e., equivalently, the first cell is within the pento centered at the second cell) and form the sum: $S = \sum_{j,j' \text{"pento-connected"}} x_{1j}x_{2j'}$. Then $S = \sum_j x_{1j} \sum_{j\text{within pento at}j'} x_{2j'} = \sum_j x_{1j} k_2 = k_2 n_1$ and, similarly, $S = k_1 n_2$.

Substituting $k_1,n_1 = 11,69$ from pattern given by WhatsUp and $k_2 = 2020$ we find that a matching integer $n_2$ does not exist.