To gain insight into why some of the remaining odd diagonals may have a higher concentration of primes than others, consider formula_4 and formula_5. Compute remainders upon division by 3 as "n" takes successive values 0, 1, 2, ... For the first of these polynomials, the sequence of remainders is 1, 2, 2, 1, 2, 2, ..., while for the second, it is 2, 0, 0, 2, 0, 0, ... This implies that in the sequence of values taken by the second polynomial, two out of every three are divisible by 3, and hence certainly not prime, while in the sequence of values taken by the first polynomial, none are divisible by 3. Thus it seems plausible that the first polynomial will produce values with a higher density of primes than will the second. At the very least, this observation gives little reason to believe that the corresponding diagonals will be equally dense with primes. One should, of course, consider divisibility by primes other than 3. Examining divisibility by 5 as well, remainders upon division by 15 repeat with pattern 1, 11, 14, 10, 14, 11, 1, 14, 5, 4, 11, 11, 4, 5, 14 for the first polynomial, and with pattern 5, 0, 3, 14, 3, 0, 5, 3, 9, 8, 0, 0, 8, 9, 3 for the second, implying that only three out of 15 values in the second sequence are potentially prime (being divisible by neither 3 nor 5), while 12 out of 15 values in the first sequence are potentially prime (since only three are divisible by 5 and none are divisible by 3).