This is really messy :(

Written here

Computational Remarks (regarding "main.c", the best version that needs less RAM):

1. In order to factorize the numbers, I kept a bitset where bitset[n] is false iff n is prime. Also, I kept an unsigned char array (0-256) of the prime gaps. from Wikipedia, we know that the largest prime gap before 303 billion is 500, and the prime gaps are always even, so we can divide every gap by 2 and store it in the array, sufficent to our unsigned char memory limit. Surprisingly enough, because there aren't that many primes (therefore there aren't that many prime gaps), We know that the gaps unsigned char array uses less memory than the bitset.
2. We could improve the memory usage on the bitset by using wheel factorization. That is, because each prime is uneven (except 2), instead of defining, "a bitset where bitset[n] is false iff n is prime" we could store it in such a way where it is "a bitset where bitset[n] is false iff 2·n+1 is prime". This cuts off 50% of the memory cost. We can take this further. Because each prime is in the form of 6·n+1 or 6·n-1 (except 2,3), we can make two bitsets "bitset1[n] is false iff 6·n+1 is prime" and "bitset2[n] is false iff 6·n-1 is prime" (two bitsets). This cuts off 67% of memory usage. Again, surprisingly enough, we could do this until very high precentages (despite diminishing returns) to consume as little memory as we could computationaly afford.
3. If p^alpha divides n and p^alpha > sqrt(2n), we can rule out n. Can we identify all sequences of 11 integers in a row where this holds? those are less than 1% of sequences.

Define b_n as a_n + n

b_n computed:

b_1 = 1

b_2 = 210

b_3 = 3478

b_4 = 8178

b_5 = 252970

b_6 = 3648841

b_7 = 72286099

b_8 = 159329615

b_9 = 2935782898

b_10 = 19295451546

So far all of those are squarefree.