Count the number of shortest paths to n This code challenge will have you compute the number of ways to reach \$n\$ starting from \$2\$ using maps of the form \$x \mapsto x + x^j\$ (with \$j\$ a non-negative integer), and doing so in the minimum number of steps.
(Note, this is related to OEIS sequence A307092.)
Example
So for example, \$f(13) = 2\$ because three maps are required, and there are two distinct sequences of three maps that will send \$2\$ to \$13\$:
$$\begin{array}{c} x \mapsto x + x^0 \\ x \mapsto x + x^2 \\ x \mapsto x + x^0\end{array} \qquad \textrm{or} \qquad \begin{array}{c}x \mapsto x + x^2 \\ x \mapsto x + x^1 \\ x \mapsto x + x^0\end{array}$$
Resulting in \$2 \to 3 \to 12 \to 13\$ or \$2 \to 6 \to 12 \to 13\$.
Example values
f(2)   = 1 (via [])
f(3)   = 1 (via [0])
f(4)   = 1 (via [1])
f(5)   = 1 (via [1,0])
f(12)  = 2 (via [0,2] or [2,1])
f(13)  = 2 (via [0,2,0] or [2,1,0], shown above)
f(19)  = 1 (via [4,0])
f(20)  = 2 (via [1,2] or [3,1])
f(226) = 3 (via [2,0,2,1,0,1], [3,2,0,0,0,1], or [2,3,0,0,0,0])
f(372) = 4 (via [3,0,1,0,1,1,0,1,1], [1,1,0,2,0,0,0,1,1], [0,2,0,2,0,0,0,0,1], or [2,1,0,2,0,0,0,0,1])

Challenge
The challenge is to produce a program which takes an integer \$n \ge 2\$ as an input, and outputs the number of distinct paths from \$2\$ to \$n\$ via a minimal number of maps of the form \$x \mapsto x + x^j\$.
This is code-golf, so fewest bytes wins.
JavaScript (ES6),  111 ... 84  80 bytes
Returns true rather than \$1\$ for \$n=2\$.

f=(n,j)=&gt;(g=(i,x,e=1)=&gt;i?e&gt;n?g(i-1,x):g(i-1,x+e)+g(i,x,e*x):x==n)(j,2)||f(n,-~j)

Try it online!
Commented
f = (                     // f is the main recursive function taking:
  n,                      //   n = input
  j                       //   j = maximum number of steps
) =&gt; (                    //
  g = (                   // g is another recursive function taking:
    i,                    //   i = number of remaining steps
    x,                    //   x = current sum
    e = 1                 //   e = current exponentiated part
  ) =&gt;                    //
    i ?                   // if there's still at least one step to go:
      e &gt; n ?             //   if e is greater than n:
                          //     add the result of a recursive call with:
        g(i - 1, x)       //       i - 1, x unchanged and e = 1
      :                   //   else:
                          //     add the sum of recursive calls with:
        g(i - 1, x + e) + //       i - 1, x + e and e = 1
        g(i, x, e * x)    //       i unchanged, x unchanged and e = e * x
    :                     // else:
      x == n              //   stop recursion; return 1 if x = n
)(j, 2)                   // initial call to g with i = j and x = 2
|| f(n, -~j)              // if it fails, try again with j + 1