Why don't we know how many knots there are with 9 crossings? Much of the focus on knot tabulation has focused on prime knots, because the composite knots can be built up from the prime knots. So for example, people have enumerated all the prime knots up to 16 crossings (https://oeis.org/A002863). But, the OEIS entry for all knots, prime and composite, only goes up to 8 crossings (https://oeis.org/A086825), and I can't find any other lists online that go further. Shouldn't this be easy to calculate given the enumerations of prime knots we have?
 A: KnotInfo tabulates knots up to 11 crossings, and you can download its database as an Excel spreadsheet. According to this page, there are 49 knots with crossing number 9.
One paper about knot tabulation is

Hoste, Jim; Thistlethwaite, Morwen; Weeks, Jeff, The first
1,701,936 knots, Math.
Intell. 20, No. 4, 33-48 (1998).
ZBL0916.57008.

Appendix 1 summarizes the numbers of knots for each crossing number, split up into, for example, 9a and 9n for alternating knots and nonalternating knots with 9 crossings.  It goes up to 16 crossings.
Ben Burton extended this enumeration to 19 crossings. (I saw him give a talk about it in 2018, and it appears he must still be working on the paper!  In the meantime, https://regina-normal.github.io/data.html has all the data.)

Now, regarding your question about enumerating composite knots, an issue is that it's still an open problem whether the crossing number of a connect sum is the sum of the crossing numbers.  Supposing this conjecture is true, then we can count composite knots by using the symmetry information in KnotInfo and utilizing the fact that connect sums are commutative.  I wrote a Python program to calculate this from the KnotInfo database, and I got this sequence:
1, 0, 0, 1, 1, 2, 5, 8, 26, 62, 194, 623
At first I thought that 26 was in error, but it looks like the OEIS sequence might be incorrect!  If I'm not mistaken, they are missing $4_1\mathbin{\#}4_1$.
Take the remaining numbers with a grain of salt, though, since I'm not guaranteeing I did the calculation correctly, and in fact I think I'm overcounting because I'm not certain I'm dealing with knot symmetries correctly in case the knot contains multiple connect summands of the same prime knot, potentially with different symmetries applied. Edit: In fact, I checked by hand that the number of 9-crossing knots is 58, not 62.
Here's some source code if you want to work on it yourself:
import xlrd

symmetry_count = {
    "chiral": 4,
    "positive amphicheiral": 2,
    "negative amphicheiral": 2,
    "reversible": 2,
    "fully amphicheiral": 1
}

print("Processing knotinfo")
knots = [[] for i in range(13)]
with xlrd.open_workbook("knotinfo_data_complete.xls") as book:
    sheet = book.sheet_by_index(0)
    cols = {}
    for i, cell in enumerate(sheet.row(0)):
        cols[cell.value] = i

    for row_idx in range(2, sheet.nrows):
        def get(colname):
            return sheet.cell(row_idx, cols[colname])
        
        entry={}
        entry['name'] = get('name').value
        if entry['name'] == "0_1":
            continue
        entry['crossing_number'] = int(get('crossing_number').value)
        entry['symmetry'] = symmetry_count[get("symmetry_type").value.strip()]

        knots[entry['crossing_number']].append(entry)

print("Done loading", len(knots), "knots")

# the number of knots (modulo symmetry) with a given crossing number
n_options = [len(lst) for lst in knots]

# the number of knots (incl. symmetry) with a given crossing number
n_symm_options = [sum(k['symmetry'] for k in lst) for lst in knots]

memo = {}
def count(crossings, i=3, first=True):
    #print("count(%r, %r, %r)" %(crossings, i, first))
    if crossings == 0:
        return 1
    if crossings < i:
        return 0
    options = n_options[i] if first else n_symm_options[i]
    c = 0
    c += options * count(crossings-i, i=i, first=False)
    c += count(crossings, i=i+1, first=first)
    memo[(crossings, i, first)] = c
    #print("count(%r, %r, %r) = %r" %(crossings, i, first, c))
    return c

print([count(i) for i in range(12)])

