factor with a bound

edit

You can implement a small function

def factorize_small_primes(N,bound):
    res = []                  # list of pairs (prime,multiplicity)
    for p in Primes():
        if p >= bound:
            break
        k = N.valuation(p)    # the nb k such that p^k || N
        res.append((p,k))
        N /= p^k
    return Factorization(res + [(N,1)])

And use it as follows

sage: N = 2^3 * 13^5 * 523^2 * 541^3
sage: f = factorize_small_primes(N,20)
sage: f
2^3 * 13^5 * 43310697015709

You can also check that the answer is somewaht consistent

sage: f.expand() == N
True