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It appears that the zeta map which I used to answer Vince Vatter's initial question and which I describe in my answer there, see also page 50 of Jim Haglund's book, indeed solves also this problem:

As discussed in a comment above, the zeta map has the property that it sends

• the number of $0$'s in an "area sequence" $a = (a_1,\ldots,a_n)$ (item $u$ in Stanley's list, see also Property $D$ in my answer to the other question) to the number of initial north (or up) steps, and
• the number of $i$'s for which all $i$'s appear before all $i+1$'s within $a$ to the number of returns to $0$ (excluding the very last return).

On the other hand, David's bijection between rooted planar trees and Dyck paths sends

• the number of children of the root to the number of $0$'s in the corresponding area sequence, and
• the number of crucial vertices in a tree (excluding the root which is always, except for $n=1$, crucial) to the number of $i$'s for which all $i$'s appear before all $i+1$'s within the corresponding area sequence.

Combining his bijection with the zeta map yields therefore a bijection between rooted planar trees and Dyck paths that sends the bistatistic

(number of crucial vertices, number of children of the root)

to the bistatistic

(number of returns to $0$, number of initial north steps).

To extend David's initial example of the $5$ rooted planar trees of size $4$, the $5$ area sequences corresponding to them (in the ordering $$(1,2), (2,1), (3,1), (2,2), (1,3)$$ as above) are $$010, 011, 012, 001, 000.$$ Applying the zeta map yields the area sequences $$011, 001, 000, 010, 012.$$ For those, the (sequence of the) number of returns to $0$ is $1,2,3,2,1$ and the (sequence of the) number of initial north steps is $2,1,1,2,3$. We thus obtain the same bistatistical sequence.

We thus proved that $\sum c(p,q,n)x^py^q$ is indeed equal to the Tutte polynomial $T_n(x,y)$.

It appears that the zeta map which I used to answer Vince Vatter's initial question and which I describe in my answer there, see also page 50 of Jim Haglund's book, indeed solves also this problem:

As discussed in a comment above, the zeta map has the property that it sends

• the number of $0$'s in an "area sequence" $a = (a_1,\ldots,a_n)$ (item $u$ in Stanley's list, see also Property $D$ in my answer to the other question) to the number of initial north (or up) steps, and
• the number of $i$'s for which all $i$'s appear before all $i+1$'s within $a$ to the number of returns to $0$ (excluding the very last return).

On the other hand, David's bijection between rooted planar trees and Dyck paths sends

• the number of children of the root to the number of $0$'s in the corresponding area sequence, and
• the number of crucial vertices in a tree (excluding the root which is always, except for $n=1$, crucial) to the number of $i$'s for which all $i$'s appear before all $i+1$'s within the corresponding area sequence.

Combining his bijection with the zeta map yields therefore a bijection between rooted planar trees and Dyck paths that sends the bistatistic

(number of crucial vertices, number of children of the root)

to the bistatistic

(number of returns to $0$, number of initial north steps).

To extend David's initial example of the $5$ rooted planar trees of size $4$, the $5$ area sequences corresponding to them (in the ordering $$(1,2), (2,1), (3,1), (2,2), (1,3)$$ as above) are $$010, 011, 012, 001, 000.$$ Applying the zeta map yields the area sequences $$011, 001, 000, 010, 012.$$ For those, the (sequence of the) number of returns to $0$ is $1,2,3,2,1$ and the (sequence of the) number of initial north steps is $2,1,1,2,3$. We thus obtain the same bistatistical sequence.

We thus proved that $\sum c(p,q,n)x^py^q$ is indeed equal to the Tutte polynomial $T_n(x,y)$.

It appears that the zeta map which I used to answer Vince Vatter's initial question and which I describe in my answer there, see also page 50 of Jim Haglund's book, indeed solves also this problem:

As discussed in a comment above, the zeta map has the property that it sends

• the number of $0$'s in an "area sequence" $a = (a_1,\ldots,a_n)$ (item $u$ in Stanley's list, see also Property $D$ in my answer to the other question) to the number of initial north (or up) steps, and
• the number of $i$'s for which all $i$'s appear before all $i+1$'s within $a$ to the number of returns to $0$ (excluding the very last return).

On the other hand, David's bijection between rooted planar trees and Dyck paths sends

• the number of children of the root to the number of $0$'s in the corresponding area sequence, and
• the number of crucial vertices in a tree (excluding the root which is always, except for $n=1$, crucial) to the number of $i$'s for which all $i$'s appear before all $i+1$'s within the corresponding area sequence.

Combining his bijection with the zeta map yields therefore a bijection between rooted planar trees and Dyck paths that sends the bistatistic

(number of children of the root, number of crucial vertices)

to the bistatistic

(number of initial north steps, number of returns to $0$).

We thus proved that $\sum c(p,q,n)x^py^q$ is indeed equal to the Tutte polynomial $T_n(x,y)$.

It appears that the zeta map which I used to answer Vince Vatter's initial question and which I describe in my answer there, see also page 50 of Jim Haglund's book, indeed solves also this problem:

As discussed in a comment above, the zeta map has the property that it sends

• the number of $0$'s in an "area sequence" $a = (a_1,\ldots,a_n)$ (item $u$ in Stanley's list, see also Property $D$ in my answer to the other question) to the number of initial north (or up) steps, and
• the number of $i$'s for which all $i$'s appear before all $i+1$'s within $a$ to the number of returns to $0$ (excluding the very last return).

On the other hand, David's bijection between rooted planar trees and Dyck paths sends

• the number of children of the root to the number of $0$'s in the corresponding area sequence, and
• the number of crucial vertices in a tree (excluding the root which is always, except for $n=1$, crucial) to the number of $i$'s for which all $i$'s appear before all $i+1$'s within the corresponding area sequence.

Combining his bijection with the zeta map yields therefore a bijection between rooted planar trees and Dyck paths that sends the bistatistic

(number of crucial vertices, number of children of the root)

to the bistatistic

(number of returns to $0$, number of initial north steps).

To extend David's initial example of the $5$ rooted planar trees of size $4$, the $5$ area sequences corresponding to them (in the ordering $$(1,2), (2,1), (3,1), (2,2), (1,3)$$ as above) are $$010, 011, 012, 001, 000.$$ Applying the zeta map yields the area sequences $$011, 001, 000, 010, 012.$$ For those, the (sequence of the) number of returns to $0$ is $1,2,3,2,1$ and the (sequence of the) number of initial north steps is $2,1,1,2,3$. We thus obtain the same bistatistical sequence.

We thus proved that $\sum c(p,q,n)x^py^q$ is indeed equal to the Tutte polynomial $T_n(x,y)$.

It appears that the zeta map which I used to answer Vince Vatter's initial question and which I describe in my answer there, see also page 50 of Jim Haglund's book, indeed solves also this problem:

As discussed in a comment above, the zeta map has the property that it sends

• the number of $0$'s in an "area sequence" $a = (a_1,\ldots,a_n)$ (item $u$ in Stanley's list, see also Property $D$ in my answer to the other question) to the number of initial north (or up) steps, and
• the number of $i$'s for which all $i$'s appear before all $i+1$'s within $a$ to the number of returns to $0$.

On the other hand, David's bijection between rooted planar trees and Dyck paths sends

• the number of children of the root to the number of $0$'s in the corresponding area sequence, and
• the number of crucial vertices in a tree to the number of $i$'s for which all $i$'s appear before all $i+1$'s within the corresponding area sequence.

Combining his bijection with the zeta map yields therefore a bijection between rooted planar trees and Dyck paths that sends the bistatistic

(number of children of the root, number of crucial vertices)

to the bistatistic

(number of initial north steps, number of returns to $0$).

We thus proved that $\sum c(p,q,n)x^py^q$ is indeed equal to the Tutte polynomial $T_n(x,y)$.

It appears that the zeta map which I used to answer Vince Vatter's initial question and which I describe in my answer there, see also page 50 of Jim Haglund's book, indeed solves also this problem:

As discussed in a comment above, the zeta map has the property that it sends

• the number of $0$'s in an "area sequence" $a = (a_1,\ldots,a_n)$ (item $u$ in Stanley's list, see also Property $D$ in my answer to the other question) to the number of initial north (or up) steps, and
• the number of $i$'s for which all $i$'s appear before all $i+1$'s within $a$ to the number of returns to $0$ (excluding the very last return).

On the other hand, David's bijection between rooted planar trees and Dyck paths sends

• the number of children of the root to the number of $0$'s in the corresponding area sequence, and
• the number of crucial vertices in a tree (excluding the root which is always, except for $n=1$, crucial) to the number of $i$'s for which all $i$'s appear before all $i+1$'s within the corresponding area sequence.

Combining his bijection with the zeta map yields therefore a bijection between rooted planar trees and Dyck paths that sends the bistatistic

(number of children of the root, number of crucial vertices)

to the bistatistic

(number of initial north steps, number of returns to $0$).

We thus proved that $\sum c(p,q,n)x^py^q$ is indeed equal to the Tutte polynomial $T_n(x,y)$.