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Here is the PARI/GP program to check it numerically:

Here is the PARI/GP program to get $R(n,z)$ directly from $t$:

Unfortunately, I have not found an efficient way for computing $G(t,z)$ so that the values can be compared. By the way, my algorithm calculates the first $100$ values of $R(n,z)$ in $2$ seconds on a relatively old PC. Please write in the comments if you have an efficient way to compute $G(t,z)$.

• Let $T(n, k)$ be A140717 (i.e., triangle read by rows: $T(n,k)$ is the number of Dyck paths $d$ of semilength $n$ such that sum of peakheights of $d$ - number of peaks of $d$ equals $k$ ($n \geqslant 0, 0 \leqslant k \leqslant \left\lfloor\frac{n^2}{4}\right\rfloor$)). Here ordinary generating functions is $G(t,z) = H(t,\frac{1}{t},z)$ where $$ H(t,x,z) = 1 + zH(t,x,z)(H(t,tx,z)-1+tx). $$
• Let $R(n, z)$ be the $n$-th row polynomial of $T(n, k)$. In other words, $$ R(n, z) = \sum\limits_{k=0}^{\left\lfloor\frac{n^2}{4}\right\rfloor}T(n,k)z^k. $$
• Let $$ \sum\limits_{n=0}^{\infty}S(n,z)x^n = \cfrac{1}{1-\sum\limits_{n=0}^{\infty}R(n,z)x^{n+1}}. $$
• Start with vector $\nu$ of length $n$ with elements $\nu_i = 1$ (that is, $\nu = \{1, 1, \dotsc, 1\}$), then set $t = \nu$ and for $i$ from $1$ to $n-1$ and for $j$ from $i+1$ to $n$ apply $[\nu_i, \nu_j] = [\nu_i + z^{j-i}\nu_j, \nu_i + \nu_j]$ (here square brackets mean that instead of sequentially assigning $\nu_i$ and then $\nu_j$, we reserve their values (for example, as $A = \nu_i, B = \nu_j$) and then assign them in any order) and $t_{i+1} = \nu_{i+1}$ (after ending each cycle for $j$).

I conjecture that after the whole transform we have vector $t$ with elements $t_i = S(i, z)$.

Here is the PARI/GP program to generate $R(n,z)$ from $t$:

upto1(n) = my(v1); v1 = vector(n, i, 1); v2 = vector(n, i, 0); v2[1] = 1; for(i=1, n-1, for(j=i+1, n, A = v1[i]; B = v1[j]; v1[i] = A + z^(j-i)*B; v1[j] = A + B); v2[i+1] = v1[i+1]); v2
h(n, x) = my(v1); v1 = upto1(n); 1/(1 + sum(i=1, n, v1[i]*x^i)) + x*O(x^n)
upto2(n) = my(x = 'x, v1); v1 = Vec(abs(h(n, x)))
pr1(n) = my(v1); v1 = upto2(n); for(i=1, n, print(Vecrev(abs(v1[i+1]))))
pr1(20)

UPD1:

After a little inspection, I noticed that if we change $[\nu_i, \nu_j] = [\nu_i + z^{j-i}\nu_j, \nu_i + \nu_j]$ to $[\nu_i, \nu_j] = [\nu_i + z^{j-i}\nu_j, z\nu_i + \nu_j]$, then it conjecturally gives us vector $t$ with elements $t_i=R(i,z)$, so we even no need $S(n,z)$ here.

Is there a way to prove it?

• Let $T(n, k)$ be A140717 (i.e., triangle read by rows: $T(n,k)$ is the number of Dyck paths $d$ of semilength $n$ such that sum of peakheights of $d$ - number of peaks of $d$ equals $k$ ($n \geqslant 0, 0 \leqslant k \leqslant \left\lfloor\frac{n^2}{4}\right\rfloor$)). Here ordinary generating functions is $G(t,z) = H(t,\frac{1}{t},z)$ where $$ H(t,x,z) = 1 + zH(t,x,z)(H(t,tx,z)-1+tx). $$
• Let $R(n, z)$ be the $n$-th row polynomial of $T(n, k)$. In other words, $$ R(n, z) = \sum\limits_{k=0}^{\left\lfloor\frac{n^2}{4}\right\rfloor}T(n,k)z^k. $$
• Let $$ \sum\limits_{n=0}^{\infty}S(n,z)x^n = \cfrac{1}{1-\sum\limits_{n=0}^{\infty}R(n,z)x^{n+1}}. $$
• Start with vector $\nu$ of length $n$ with elements $\nu_i = 1$ (that is, $\nu = \{1, 1, \dotsc, 1\}$), then set $t = \nu$ and for $i$ from $1$ to $n-1$ and for $j$ from $i+1$ to $n$ apply $[\nu_i, \nu_j] = [\nu_i + z^{j-i}\nu_j, \nu_i + \nu_j]$ (here square brackets mean that instead of sequentially assigning $\nu_i$ and then $\nu_j$, we reserve their values (for example, as $A = \nu_i, B = \nu_j$) and then assign them in any order) and $t_{i+1} = \nu_{i+1}$ (after ending each cycle for $j$).

I conjecture that after the whole transform we have vector $t$ with elements $t_i = S(i, z)$.

Here is the PARI/GP program to generate $R(n,z)$ from $t$:

upto1(n) = my(v1); v1 = vector(n, i, 1); v2 = vector(n, i, 0); v2[1] = 1; for(i=1, n-1, for(j=i+1, n, A = v1[i]; B = v1[j]; v1[i] = A + z^(j-i)*B; v1[j] = A + B); v2[i+1] = v1[i+1]); v2
h(n, x) = my(v1); v1 = upto1(n); 1/(1 + sum(i=1, n, v1[i]*x^i)) + x*O(x^n)
upto2(n) = my(x = 'x, v1); v1 = Vec(abs(h(n, x)))
pr1(n) = my(v1); v1 = upto2(n); for(i=1, n, print(Vecrev(abs(v1[i+1]))))
pr1(20)

UPD1:

After a little inspection, I noticed that if we change $[\nu_i, \nu_j] = [\nu_i + z^{j-i}\nu_j, \nu_i + \nu_j]$ to $[\nu_i, \nu_j] = [\nu_i + z^{j-i}\nu_j, z\nu_i + \nu_j]$, then it conjecturally gives us vector $t$ with elements $t_i=R(i,z)$, so we even no need $S(n,z)$ here.

Here is the PARI/GP program to check it numerically:

upto1(n) = my(v1); v1 = vector(n, i, 1); v2 = vector(n, i, 0); v2[1] = 1; for(i=1, n-1, for(j=i+1, n, A = v1[i]; B = v1[j]; v1[i] = A + z^(j-i)*B; v1[j] = z*A + B); v2[i+1] = v1[i+1]); v2
pr1(n) = my(v1); v1 = upto1(n); for(i=1, n, print(Vecrev(v1[i])))
pr1(20)

Is there a way to prove it?

• Let $T(n, k)$ be A140717 (i.e., triangle read by rows: $T(n,k)$ is the number of Dyck paths $d$ of semilength $n$ such that sum of peakheights of $d$ - number of peaks of $d$ equals $k$ ($n \geqslant 0, 0 \leqslant k \leqslant \left\lfloor\frac{n^2}{4}\right\rfloor$)). Here ordinary generating functions is $G(t,z) = H(t,\frac{1}{t},z)$ where $$ H(t,x,z) = 1 + zH(t,x,z)(H(t,tx,z)-1+tx). $$
• Let $R(n, z)$ be the $n$-th row polynomial of $T(n, k)$. In other words, $$ R(n, z) = \sum\limits_{k=0}^{\left\lfloor\frac{n^2}{4}\right\rfloor}T(n,k)z^k. $$
• Let $$ \sum\limits_{n=0}^{\infty}S(n,z)x^n = \cfrac{1}{1-\sum\limits_{n=0}^{\infty}R(n,z)x^{n+1}}. $$
• Start with vector $\nu$ of length $n$ with elements $\nu_i = 1$ (that is, $\nu = \{1, 1, \dotsc, 1\}$), then set $t = \nu$ and for $i$ from $1$ to $n-1$ and for $j$ from $i+1$ to $n$ apply $[\nu_i, \nu_j] = [\nu_i + z^{j-i}\nu_j, \nu_i + \nu_j]$ (here square brackets mean that instead of sequentially assigning $\nu_i$ and then $\nu_j$, we reserve their values (for example, as $A = \nu_i, B = \nu_j$) and then assign them in any order) and $t_{i+1} = \nu_{i+1}$ (after ending each cycle for $j$).

I conjecture that after the whole transform we have vector $t$ with elements $t_i = S(i, z)$.

Here is the PARI/GP program to generate $R(n,z)$ from $t$:

upto1(n) = my(v1); v1 = vector(n, i, 1); v2 = vector(n, i, 0); v2[1] = 1; for(i=1, n-1, for(j=i+1, n, A = v1[i]; B = v1[j]; v1[i] = A + z^(j-i)*B; v1[j] = A + B); v2[i+1] = v1[i+1]); v2
h(n, x) = my(v1); v1 = upto1(n); 1/(1 + sum(i=1, n, v1[i]*x^i)) + x*O(x^n)
upto2(n) = my(x = 'x, v1); v1 = Vec(abs(h(n, x)))
pr1(n) = my(v1); v1 = upto2(n); for(i=1, n, print(Vecrev(abs(v1[i+1]))))
pr1(20)

Is there a way to prove it?

• Let $T(n, k)$ be A140717 (i.e., triangle read by rows: $T(n,k)$ is the number of Dyck paths $d$ of semilength $n$ such that sum of peakheights of $d$ - number of peaks of $d$ equals $k$ ($n \geqslant 0, 0 \leqslant k \leqslant \left\lfloor\frac{n^2}{4}\right\rfloor$)). Here ordinary generating functions is $G(t,z) = H(t,\frac{1}{t},z)$ where $$ H(t,x,z) = 1 + zH(t,x,z)(H(t,tx,z)-1+tx). $$
• Let $R(n, z)$ be the $n$-th row polynomial of $T(n, k)$. In other words, $$ R(n, z) = \sum\limits_{k=0}^{\left\lfloor\frac{n^2}{4}\right\rfloor}T(n,k)z^k. $$
• Let $$ \sum\limits_{n=0}^{\infty}S(n,z)x^n = \cfrac{1}{1-\sum\limits_{n=0}^{\infty}R(n,z)x^{n+1}}. $$
• Start with vector $\nu$ of length $n$ with elements $\nu_i = 1$ (that is, $\nu = \{1, 1, \dotsc, 1\}$), then set $t = \nu$ and for $i$ from $1$ to $n-1$ and for $j$ from $i+1$ to $n$ apply $[\nu_i, \nu_j] = [\nu_i + z^{j-i}\nu_j, \nu_i + \nu_j]$ (here square brackets mean that instead of sequentially assigning $\nu_i$ and then $\nu_j$, we reserve their values (for example, as $A = \nu_i, B = \nu_j$) and then assign them in any order) and $t_{i+1} = \nu_{i+1}$ (after ending each cycle for $j$).

I conjecture that after the whole transform we have vector $t$ with elements $t_i = S(i, z)$.

Here is the PARI/GP program to generate $R(n,z)$ from $t$:

upto1(n) = my(v1); v1 = vector(n, i, 1); v2 = vector(n, i, 0); v2[1] = 1; for(i=1, n-1, for(j=i+1, n, A = v1[i]; B = v1[j]; v1[i] = A + z^(j-i)*B; v1[j] = A + B); v2[i+1] = v1[i+1]); v2
h(n, x) = my(v1); v1 = upto1(n); 1/(1 + sum(i=1, n, v1[i]*x^i)) + x*O(x^n)
upto2(n) = my(x = 'x, v1); v1 = Vec(abs(h(n, x)))
pr1(n) = my(v1); v1 = upto2(n); for(i=1, n, print(Vecrev(abs(v1[i+1]))))
pr1(20)

UPD1:

After a little inspection, I noticed that if we change $[\nu_i, \nu_j] = [\nu_i + z^{j-i}\nu_j, \nu_i + \nu_j]$ to $[\nu_i, \nu_j] = [\nu_i + z^{j-i}\nu_j, z\nu_i + \nu_j]$, then it conjecturally gives us vector $t$ with elements $t_i=R(i,z)$, so we even no need $S(n,z)$ here.

Is there a way to prove it?