I assume the "between" is inclusive. The transition matrix $P$ is thus lower triangular with all entries $1/n$ in the $n$'th row, and you want $(P^t)_{nj}$ for $1 \le j \le n$. The ordinary generating function with respect to $t$ is $$ g_{nj}(z) = \sum_{t=0}^\infty z^t (P^t)_{nj} = (I - z P)^{-1}_{nj}$$ For $j=n$ it is easy to see that $g_{nn}(z) = \dfrac{n}{n - z}$, i.e. $(P^t)_{nn} = 1/n^t$. It appears that for $j < n$,

$$g_{nj}(z) = \dfrac{(n-1)!\; z}{(j-1)! \prod_{k=j}^n (k - z)}$$

EDIT: Expand this in partial fractions:

$$ g_{nj}(z) = \dfrac{(n-1)!}{(j-1)!} \sum_{i=j}^n \dfrac{(1-z/i)^{-1}}{\prod_{k \ne i} (k-i)} $$

(the product in the denominator being over all $k$ from $j$ to $n$ except $i$). And then I get

$$ (P^t)_{nj} = \sum_{i=j}^n \dfrac{(-1)^{i-j} (n-1)!}{(j-1)!\; (i-j)!\; (n-i)! \; i^t}$$

which is still not quite closed-form, but better than a sum over paths.

I assume the "between" is inclusive. The transition matrix $P$ is thus lower triangular with all entries $1/n$ in the $n$'th row, and you want $(P^t)_{nj}$ for $1 \le j \le n$. The ordinary generating function with respect to $t$ is $$ g_{nj}(z) = \sum_{t=0}^\infty z^t (P^t)_{nj} = (I - z P)^{-1}_{nj}$$ For $j=n$ it is easy to see that $g_{nn}(z) = \dfrac{n}{n - z}$, i.e. $(P^t)_{nn} = 1/n^t$. It appears that for $j < n$,

$$g_{nj}(z) = \dfrac{(n-1)!\; z}{(j-1)! \prod_{k=j}^n (k - z)}$$

EDIT: Expand this in partial fractions:

$$ g_{nj}(z) = \dfrac{(n-1)!}{(j-1)!} \sum_{i=j}^n \dfrac{(1-z/i)^{-1}}{\prod_{k \ne i} (k-i)} $$

(the product in the denominator being over all $k$ from $j$ to $n$ except $i$). And then I get

$$ (P^t)_{nj} = \sum_{i=j}^n \dfrac{(-1)^{i-j} (n-1)!}{(j-1)!\; (i-j)!\; (n-i)! \; i^t}$$

which is still not quite closed-form, but better than a sum over paths. It can be written (for fixed $t > 1$) using a hypergeometric function

$$ \dfrac{{}_{t+1}F_t(j,\ldots,j,j-n;\; j+1,\ldots,j+1;\;1)\; (n-1)!} {j^{t}\; (j-1)!\; (n-j)!}$$

I assume the "between" is inclusive. The transition matrix $P$ is thus lower triangular with all entries $1/n$ in the $n$'th row, and you want $(P^t)_{nj}$ for $1 \le j \le n$. The ordinary generating function with respect to $t$ is $$ g_{nj}(z) = \sum_{t=0}^\infty z^t (P^t)_{nj} = (I - z P)^{-1}_{nj}$$ For $j=n$ it is easy to see that $g_{nn}(z) = \dfrac{n}{n - z}$, i.e. $(P^t)_{nn} = 1/n^t$. It appears that for $j < n$,

$$g_{nj}(z) = \dfrac{(n-1)!\; z}{(j-1)! \prod_{k=j}^n (k - z)}$$

I assume the "between" is inclusive. The transition matrix $P$ is thus lower triangular with all entries $1/n$ in the $n$'th row, and you want $(P^t)_{nj}$ for $1 \le j \le n$. The ordinary generating function with respect to $t$ is $$ g_{nj}(z) = \sum_{t=0}^\infty z^t (P^t)_{nj} = (I - z P)^{-1}_{nj}$$ For $j=n$ it is easy to see that $g_{nn}(z) = \dfrac{n}{n - z}$, i.e. $(P^t)_{nn} = 1/n^t$. It appears that for $j < n$,

$$g_{nj}(z) = \dfrac{(n-1)!\; z}{(j-1)! \prod_{k=j}^n (k - z)}$$

EDIT: Expand this in partial fractions:

$$ g_{nj}(z) = \dfrac{(n-1)!}{(j-1)!} \sum_{i=j}^n \dfrac{(1-z/i)^{-1}}{\prod_{k \ne i} (k-i)} $$

(the product in the denominator being over all $k$ from $j$ to $n$ except $i$). And then I get

$$ (P^t)_{nj} = \sum_{i=j}^n \dfrac{(-1)^{i-j} (n-1)!}{(j-1)!\; (i-j)!\; (n-i)! \; i^t}$$

which is still not quite closed-form, but better than a sum over paths.