At 10:10 PM 6/21/2004, Simon Plouffe wrote:

There is this puzzling property of the B(4n+2),

I consider here only the fractional part, easily computable with the Von-Staudt-Clausen formula.

Also, the B(4n+2) are all positives.

Property ;

the fractional part of the positive Bernoulli Numbers is almost never : 1/6 < {Bernoulli(4*n+2)} < 2/3.

... except for 2 cases (so far), which is {B(2072070)} = .666443506838622... and {B(6216210)} = .6588649656359250...

That this is true for "small" n is easy to explain; essentially, it is a consequence of the von Staudt-Clausen theorem. I always have to look up the statement, so let me repeat it here: Given a positive even integer 2k, let S be the sum of the reciprocals of all the primes p such that p-1 divides 2k; then B(2k) + S is an integer. For example, if 2k = 6, then p = 2, 3 or 7, so S = 1/2 + 1/3 + 1/7 = 41/42; thus the fractional part of B(6) is 1/42 (in fact, B(6) = 1/42). No matter what 2k is, we always have to include p = 2 and 3, so S is at least 5/6. In order to have the fractional part of B(2k) between 1/6 and 2/3, S would have to exceed 4/3, which requires quite a few terms from the slow-growing series of reciprocals of primes. When 2k = 4n+2, all primes congruent to 1 mod 4 are excluded, so it requires a substantial n to get enough prime reciprocals into the sum. 2k = 2072070 works quite well because 2072070 has lots of factors: 2072070 = 2 3^2 5 7 11 13 23. Here, p-1 | 2k for all but two of the primes less than 100 that are not congruent to 1 mod 4; 29 scattered larger primes bring S up to 1.333556..., just over the threshold. 6216210 = 3*2072070, so it gets all the same primes, plus a few more. Pushing the fractional part down of B(4n+2) further is certainly possible, but it takes a while to get there. For 2k = 15322957650 the fractional part of B(2k) is just under 0.6. -- Fred W. Helenius <fredh@ix.netcom.com>