- Let $a(n)$ be A258173 i.e. sum over all Dyck paths of semilength $n$ of products over all peaks $p$ of $y_p$, where $y_p$ is the $y$-coordinate of peak $p$.
- A Dyck path of semilength $n$ is a $(x,y)$-lattice path from $(0,0)$ to $(2n,0)$ that does not go below the $x$-axis and consists of steps $U=(1,1)$ and $D=(1,-1)$. A peak of a Dyck path is any lattice point visited between two consecutive steps $UD$.
- The sequence begins with $$ 1, 1, 3, 12, 58, 321, 1975, 13265, 96073, 743753, 6113769, 53086314, 484861924, 4641853003 $$
- Here $T(0)$ is a generating function for $a(n)$ where $$ T(k) = 1 - \cfrac{x}{(k + 2)x - \cfrac{1}{T(k+1)}} $$
- Start with vector $\nu$ of fixed length $m$ with elements $\nu_i=1$ (that is, $\nu=\{1,1,\dotsc,1\}$), reserve $t$ as an empty vector of fixed legnth $m$ and for $i$ from $1$ to $m-1$ apply $t:=\nu$ and for $j$ from $i+1$ to $m$ apply $$ \nu_j := \nu_j + \nu_{j-1} + (j-i)t_{j-1}. $$
I conjecture that after the whole transform we have $$ \nu_n = a(n). $$
Here is the PARI/GP program to check it numerically:
upto1(n) = my(CF = 1); for(i = 0, n, CF = 1 - x / ((n - i + 2) * x - 1 / CF) + x * O(x ^ n)); Vec(CF-1)
upto2(n) = my(v1); v1 = vector(n, i, 1); for(i=1, n-1, v2 = v1; for(j=i+1, n, v1[j] += v1[j-1] + (j-i)*v2[j-1])); v1
test1(n) = upto1(n) == upto2(n)
In addition, this question can be rephrased as follows:
- Let $P(n,k)$ be a triangle read by rows such that $$ P(n, k) = \begin{cases} 0 & \textrm{if } k > n \\ 1 & \textrm{if } k = 0 \\ P(n-1, k) + P(n, k-1) + (n-k)P(n-1, k-1) & \textrm{otherwise} \end{cases} $$
I conjecture that $$ P(n,n)=a(n). $$
Is there a way to prove it?
