Divisible by 12: a^3 + b^3 = c^2

[James Buddenhagen]

Re sequences relating to a^3 + b^3 = c^2 where a and b are prime 
and c is a positive integer, I have submitted several new sequences, 
and added to yours (Hugo's), A098970. They are related as follows:

a:     A099806  these primes are -1 mod 12
b:     A099807  these primes are  1 mod 12
c/12:  A098970  your sequence.

sqrt((a+b)/48):        A099808
sqrt((a^2-a*b+b^2)/3)  A099809, so

corresponding to the factorization a^3+b^3 = (a+b)*(a^2-a*b+b^2), 
i.e. corresponding to  c^2 = (a+b)*(a^2-a*b+b^2) one has:

(12*A098970)^2 = (48*A099808^2) * (3*A099809^2).

The 48*3 gives 12^2, i.e. the 12 in your c/12 sequence.

Best regards,
Jim Buddenhagen

P.S. Please double check for correctness, typos etc.