Get nth prime number using sieve of Eratosthenes
First a small stylistic review, followed by an alternative algorithm.

You calculate math.sqrt(incEND) many times, just calculate it once outside of the inner while loop. Similarly, you calculate T**2 + (T*l) two to three times per loop. Do it once and save it to a variable.

Python has an official style-guide, PEP8, which recommends using a space in a comma-separated argument list, so use range(incST, incEND) and for k, v in ...

It also recommends using lower_case_with_underscores for variables and function names and using a space before and after the = in variable assignments.

You did not include the import math, which you should do for completeness.



It is probably more efficient to take a prime number generator and count up to the nth prime than keeping a list of all prime numbers in memory.

For this I would use something like this, which I got from a python cookbook site. Alternative algorithms can be found, for example in this Stack Overflow post. It is basically also a Sieve of Eratosthenes, but one where the yielding and the marking-off are interleaved.

def primes():
    '''Yields the sequence of prime numbers via the Sieve of Eratosthenes.'''
    D = {}
    yield 2
    # start counting at 3 and increment by 2
    for q in itertools.count(3, 2):
        p = D.pop(q, None)
        if p is None:
            # q not a key in D, so q is prime, therefore, yield it
            yield q
            # mark q squared as not-prime (with q as first-found prime factor)
            D[q*q] = q
        else:
            # let x &lt;- smallest (N*p)+q which wasn't yet known to be composite
            # we just learned x is composite, with p first-found prime factor,
            # since p is the first-found prime factor of q -- find and mark it
            x = p + q
            while x in D or x % 2 == 0:
                x += p
            D[x] = p


And then finding the nth prime becomes trivial:

def nth_prime(n):
    if n &lt; 1:
        raise ValueError("n must be &gt;= 1 for nth_prime")
    for i, p in enumerate(primes(), 1):
        if i == n:
            return p


This takes about 1 second to find the 200,000th prime (2750159) on my machine.