TITLE: Explicit expression for recursive sums
QUESTION [10 upvotes]: Let $t_1,t_2,\dots,t_k$ be non-negative integers. Can the following sum
$$f_k(t_1,t_2,\dots,t_k):=\sum_{j_1=0}^{t_1} \sum_{j_2=0}^{t_2+j_1} \sum_{j_3=0}^{t_2+j_2} \dots \sum_{j_k=0}^{t_k+j_{k-1}} 1$$
be explicitly expressed as a polynomial in $t_1,t_2,\dots,t_k$ or via known combinatorial entities?
We surely have a recurrence formula:
$$f_{k+1}(t_1,t_2,\dots,t_{k+1}) = \sum_{j=0}^{t_1} f_k(j+t_2,\dots,t_{k+1}),$$
which does not seem to easily unroll.
Just in case, first few terms are
\begin{split}
f_0 &= 1,\\
f_1(t_1) &= 1+t_1,\\
f_2(t_1,t_2) &= (1+t_1)(1+t_2) + \frac{t_1(1+t_1)}2,\\
f_3(t_1,t_2,t_3) &= \left[ (1+t_1)(1+t_2) + \frac{t_1(1+t_1)}2 \right](1+t_3) + \frac{t_2(1+t_2)}2 + \frac{3t_2^2 + 6t_2 + 2}6t_1 + \frac{1+t_2}2t_1^2 + \frac16t_1^3.
\end{split}

UPDATED. Billy Joe found that
$$f_k(n,d,d,\dots, d) = \frac{n+1}k \binom{n+k(d+1)}{k-1}.$$
In particular, at $f_k(1,1,\dots,1)$ gives $(k+1)$-st Catalan number.

REPLY [9 votes]: $f_k(t_1,\dots,t_k)$ is counting the number of integer points in the
"Pitman-Stanley polytope" $\Pi_k(t_1,\dots,t_k)$ defined here. The
notation $N(\Pi_k(\mathbf{t}))$ is used in this paper, which has the
determinantal formula given by Hugh Denoncourt, as well as a
combinatorial formula. The combinatorial formula is equivalent to
$$ f_k(t_1,\dots,t_k)=\sum_{\mathbf{h}\in K_k} \binom{t_1+h_1}{h_1}
     \prod_{i=2}^k \binom{t_i+h_i-1}{h_i}, $$
where
$$ K_k := \{\mathbf{h}\in\mathbb{N}^k\colon \sum_{i=1}^j h_i\geq
  j\ \mathrm{for\ all}\ 1\leq j\leq k-1\ \mathrm{and}\ \sum_{i=1}^k
  h_i=k \}. $$
The set $K_k$ has a Catalan number $C_k$ of elements.