Quote:
Originally Posted by sweety439
Thus, for R4: (we should find n such that all of these formulas take prime values)

e.g. for k=5, 5*4^n1 has neither trivial factors nor algebraic factors, thus we need n such that 5*4^n1 takes prime value.
e.g. for k=7, 7*4^n1 has trivial factor of 3, thus we should take out this factor, and this formula is (7*4^n1)/3, we need n such that this formula takes prime value. (7*4^n1 has no algebra factors)
e.g. for k=9, 9*4^n1 = (3*2^n1) * (3*2^n+1), thus we need n such that both these two formulas take prime values. (9*4^n1 has no trivial factors)
e.g. for k=25, 25*4^n1 has trivial factor of 3, thus we should take out this factor, and this formula is (25*4^n1)/3, besides, (25*4^n1)/3 = (5*2^n1) * (5*2^n+1)/3 (for even n) or (5*2^n1)/3 * (5*2^n+1) (for odd n), thus we need n such that both these two formulas take prime values.