Simple Complex Quadratic Fields:
For any odd prime p compute the sequence of squares k^2 (mod p) with k=0,1,2,3,...,p. For example, with p=19 the squares are; 0,1,4,9,16,6,17,11,7,5,5,7,11,17,6,16,9,4,1,0.
Therefore, I can say with a degree of certainty that:
1 ==> times 2, plus 1 ==> = 3
3 ==> times 4, minus 1 ==> = 11
11 ==> times 6, plus 1 ==> = 67
And so the next number in the sequence will be
67 ==> times 8, minus 1 ==> = 535