Sequences generated by sum & product of terms (with rotating indices): combinatorial? Fix an integer $t\geq0$ and consider the sequence $T_{0,t}=1$ and for $n\geq1$, with $k*=n-1-k+t\mod n$,
$$T_{n,t}=\sum_{k=0}^{n-1}T_{k,t}T_{k*,t}.$$
EXAMPLES. Some initial terms:
(a) the case $t=0$: $1, 1, 2, 5, 14, 42, 132, 429, 1430,\dots$ (restores the Catalan numbers)
(b) the case $t=1$: $1, 1, 2, 5, 15, 51, 188, 731, 2950, \dots$ (results in number of Schroeder paths)
(c) the case $t=2$: $1, 1, 2, 6, 26, 142, 882, 5910, 41610, \dots$ (unavailable on OEIS)

QUESTION. Is there a common approach for a combinatorial interpretation of $T_{n,t}$, for all $t$?

 A: For simplicity, I assume that $t$ is fixed and omit it as the second index in $T$.
First, let's focus on initial terms with $n\leq t$, for which
$$T_n = \sum_{k=0}^{n-1} T_k T_{(t-1-k)\bmod n},$$
implying in particular that $T_1 = 1$ and
$$T_t = \sum_{k=0}^{t-1} T_k T_{t-1-k}.$$
Let
$$p(x) := \sum_{n=0}^{t-1} T_n x^n$$
be the generating polynomial for the initial terms.

For $n\geq t$, we have
\begin{split}
T_n &= \sum_{k=0}^{t-1} T_k T_{t-1-k} + \sum_{k=t}^{n-1} T_k T_{n-1+t-k} \\
& = T_t + \sum_{k=0}^{n+t-1} T_k T_{n-1+t-k} - \sum_{k=0}^{t-1} T_k T_{n-1+t-k} - \sum_{k=n}^{n+t-1} T_k T_{n-1+t-k}\\
& = T_t + \sum_{k=0}^{n+t-1} T_k T_{n-1+t-k} - 2\sum_{k=0}^{t-1} T_k T_{n-1+t-k}\\
\end{split}
Let $F(x):=\sum_{n\geq 0} T_n x^n$ be the generating function, and $q(x)$ be the polynomial formed by the initial terms of degree $<2t-1$ from $F(x)^2$ and $r(x)$ be a similar polynomial from $F(x)p(x)$.
Considering the terms of degree $<2t-1$ in
$$F(x)^2 = (F(x)-p(x)+p(x))^2 = (F(x)-p(x))^2 + 2(F(x)-p(x))F(x) + p(x)^2,$$
we obtain the identity:
$$(\star)\qquad q(x) - 2r(x) + p(x)^2\equiv 0.$$
Then the recurrence for $T_n$ with $n\geq t$ translates into
$$x^{t-1}(F(x) - p(x)) = T_t\frac{x^{2t-1}}{1-x} + (F(x)^2 - q(x)) - 2(F(x)p(x)-r(x)),$$
which is a quadratic equation for $F(x)$. Solving it and using $(\star)$, we get an explicit expression for the generating function:
\begin{split}
F(x) &= p(x)+\frac{1}{2}x^{t-1} - \sqrt{p(x)^2+\frac{1}{4}x^{2t-2}+q(x)-2r(x)-T_t\frac{x^{2t-1}}{1-x}} \\
&= p(x)+\frac{1}{2}x^{t-1}\big(1 - \sqrt{\frac{1-(4T_t+1)x}{1-x}}\big).
\end{split}

Example for $t=1$. In this case, $T_t=1$ and $p(x)=1$, implying that
$$F(x) = \frac12\big(3-\sqrt{\frac{1-5x}{1-x}}\big).$$
Example for $t=2$. In this case, $T_t=2$ and $p(x)=1+x$, implying that
$$F(x) = 1 + \frac{x}{2}\big( 3 - \sqrt{\frac{1-9x}{1-x}}\big).$$
Example for $t=3$. In this case, $T_t=5$ and $p(x)=1+x+2x^2$, implying that
$$F(x) = 1 + x + \frac{x^2}2\big(5-\sqrt{\frac{1-21x}{1-x}}\big).$$

Just for the record, the sequence of $T_t$ for $t=0,1,2,\dots$ starts with
$$1, 1, 2, 5, 14, 50, 194, 1319, 8834, 117869, 1269734, 560616089, \dots$$
