Speed is everything

Let's say that $x$ is an almostprime so it has 3 divisors. If we write x as the product of it's prime factors, it will be $x = p_1 ^{n_1} p_2^{n_2} p_3^{n_3} \ldots$ where $p_i$ is a unique prime and $n_i$ is natural. If $x$ is the product of 2 or more unique primes then it will have factors $1, p_1, p_2, p_1 p_2 \ldots$ . However, this means it will have more than 3 factors so it is not almost prime so therefore $x$ is a power of a single prime. The only power this will work for is $p^2$ where $p$ is prime because it will have 3 factors $1, p, p^2$ . Hence, $x$ is almostprime iff $x = p^2$ for prime $p$

The sum of the first 6 almost primes is $2^2 + 3^2 + 5^2 + 7^2 + 11^2 + 13^2 = \boxed{337}$

Note that the 6 smallest value of almost prime (p) numbers are the squares of the 6 smallest prime numbers (which are 2,3,5,7,11,13). Therefore the sum is: $2^2+3^2+5^2+7^2+11^2+13^2=4+9+25+49+121+169=\boxed{377}$

Being almostprime means that the number has only three divisors, and the only possible way for that configuration is when the number is the square of a prime. Thus, the sum of the 6 smallest almostprime numbers is 2^2+3^2+5^2+7^2+11^2+13^2=377

In order for a number to be an almost prime, it must have only one other divisor other than 1 or itself.

This means that the other divisor must be a prime, for otherwise the candidate number would also be divisible by the divisors of that divisor. For example, 4 is an almost prime because its divisors are 1, 2, and 4. Note that 2 is a prime number. However, 8 is not an almost prime, because its divisors are 1, 2, 4, and 8. 4 is not a prime number, so it is divisible by 2, and thus 8 is also divisible by 2, making it not almost prime.

Furthermore, the almost prime has to be a square of a prime number, because there can be no other divisors. For example, 6 is not an almost prime, because even though its other divisors are primes, (2 and 3) only one other divisor is allowed in the definition of an almost prime.

Thus, the sum of the 6 smallest almost primes are 2^2 + 3^2 + 5^2 + 7^2 + 11^2 + 13^2 = 4+9+25+49+121+169 = 377

Python:

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def almost_p(n):
    divisors = 2
    for i in xrange(2,n):
        if n % i == 0:
            divisors += 1
    return divisors == 3
test = 0
count = 0
total = 0
while count < 6:
    if almost_p(test):
        count += 1
        total += test
    test += 1
print "Answer:", total

The answer will be $377$ . Since the numbers that are almost prime are the square of prime numbers. So, $2^2+3^2+5^2+7^2+11^2+13^2=377$

nice problem

Any number will have just three divisor only if it is a square of a prime number. So this sum will be : $2^{2} + 3^{2} + 5^{2}+7^{2}+11^{2}+13^{2}$