Number of terms in the expansion of $(1+x^3+x^5) ^n$? How can we find the number of terms in the expansion of $(1+x^3+x^5)^n$ ?

Note that
$$(1+x^3+x^5)^n = \sum_{k=0}^n \binom{n}{k} x^{3 k} (1+ x^2)^k$$
There are $n+1$ terms.  For the $k$th term, there are $k+1$ terms.  Therefore, there are
$$\sum_{k=0}^n (k+1) = \frac12 (n+1)(n+2)$$
terms before combining. However, there are many terms that will combine (i.e., like powers).  To get an accurate count, then, we must figure out what those powers are.
Let's write, ignoring the coefficients, the powers by value of $k$:
$$(1+x^3+x^5)^n = \begin{array}\\ 1  \\ x^3 & x^5\\x^6 & x^8 & x^{10}\\ x^9 & x^{11} & x^{13} & x^{15}\\ x^{12} & x^{14} & x^{16} & x^{18} & x^{20}\\  x^{15} & x^{17} & x^{19} & x^{21} & x^{23} & x^{25} \end{array}$$
and so on.  In general, we want to find all values of $3 k+2 \ell$ that overlap for different lattice points $(k,\ell)$.  Note that we have a first combining when $k=5$ (which explains why my test for small $n$ worked); $(5,0)$ and $(3,3)$.
Unfortunately, I know of no closed for expression for the number, $h(m)$, of pairs of lattice points $(k,\ell)$ that have $3 k+2 \ell$ being equal to a given integer $m$.  The best that I can see written is
$$\frac12 (n+1)(n+2) - \sum_{m=0, m=3 k+2 \ell}^{5 n} [h(m)-1]$$
