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%I #3 Mar 31 2012 14:40:24
%S 37,2137,8929,1801,48817,6637,57241,133597,151477,334717,3889,127717,
%T 786697,735781,1154017,38557,1662229,2446777,3882661,3811669,2747449,
%U 3716701,5634637,3600097,9836221,10591849,7139569,9473161,11395309
%N If a,b are prime numbers satisfying the Diophantine equation a^3+b^3=c^2, then a is -1 mod 12 and b is 1 mod 12, or vice versa. Choose 'b' to be 1 mod 12. This is the sequence of 'b' values, sorted by the magnitude of c.
%C All terms of this sequence are of the form -3*M^4+N^4+6*M^2*N^2 for some pair M,N of relatively prime positive integers of opposite parity. For each n, a=A099806[n], b=A099807[n] are prime numbers and a^3 + b^3 = c^2, for some integer c. c is divisible by 12 and A098970 gives the values of c/12.
%H James Buddenhagen, <a href="http://www.buddenbooks.com/jb/num_theory/sum_of_2_cubes_a_square.htm">Two Primes Cubed which Sum to a Square</a>.
%e 37 is in the sequence because 37 is a prime congruent to 1 mod 12 and 11^3+37^3=228^2.
%Y Cf. A099806, A098970, A099808, A099809.
%K nonn
%O 0,1
%A _James R. Buddenhagen_, Oct 26 2004
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