Von Staudt–Clausen theorem explained

In number theory, the von Staudt–Clausen theorem is a result determining the fractional part of Bernoulli numbers, found independently by and .

Specifically, if is a positive integer and we add to the Bernoulli number for every prime such that divides, then we obtain an integer; that is,

B _{2n} + \sum _{(p-1)|2n} (1 p)/(\in) \Z .

This fact immediately allows us to characterize the denominators of the non-zero Bernoulli numbers as the product of all primes such that divides ; consequently, the denominators are square-free and divisible by 6.

These denominators are

6, 30, 42, 30, 66, 2730, 6, 510, 798, 330, 138, 2730, 6, 870, 14322, 510, 6, 1919190, 6, 13530, ... .

The sequence of integers

B _{2n} + \sum _{(p-1)|2n} (1 p)/() is

1, 1, 1, 1, 1, 1, 2, -6, 56, -528, 6193, -86579, 1425518, -27298230, ... .

Proof

A proof of the Von Staudt–Clausen theorem follows from an explicit formula for Bernoulli numbers which is:

B _{2n} = \sum ^{2 n} _{j =0} { (1)/(j +1) }\sum_^ and as a corollary:B _{2n} = \sum ^{2 n} _{j =0} { (j!)/(j +1) }(-1)^jS(2n,j) where are the Stirling numbers of the second kind.

Furthermore the following lemmas are needed:

Let be a prime number; then

1. If divides , then

\sum ^{p -1} _{m =0} {( -1 ) ^{m{p-1\choose} m} m ^{2n} }\equiv\pmod p.

2. If does not divide , then

\sum ^{p -1} _{m =0} {( -1 ) ^{m{p-1\choose} m} m ^{2n} }\equiv0\pmod p.

Proof of (1) and (2): One has from Fermat's little theorem,

m ^{p-1} \equiv 1 \pmod{p}

for .

If divides , then one has

m ^{2n} \equiv 1 \pmod{p}

for . Thereafter, one has

\sum ^{p -1} _{m =1} ( -1 ) ^{m} \binom{p -1 }{m} m ^{2n} \equiv \sum ^{p -1} _{m =1} ( -1 ) ^{m} \binom{p -1 }{m} \pmod{p},

from which (1) follows immediately.

If does not divide , then after Fermat's theorem one has

m ^{2n} \equiv m ^{2n-(p-1)} \pmod{p}.

If one lets, then after iteration one has

m ^{2n} \equiv m ^{2n-\wp(p-1)} \pmod{p}

for and .

Thereafter, one has

\sum ^{p -1} _{m =0} ( -1 ) ^{m} \binom{p -1 }{m} m ^{2n} \equiv \sum ^{p -1} _{m =0} ( -1 ) ^{m} \binom{p -1 }{m} m ^{2n-\wp(p-1)} \pmod{p}.

Lemma (2) now follows from the above and the fact that for .

(3). It is easy to deduce that for and, divides .

(4). Stirling numbers of the second kind are integers.

Now we are ready to prove the theorem.

If is composite and, then from (3), divides .

For,

\sum ^{3} _{m =0} ( -1 ) ^{m} \binom{ 3 }{m} m ^{2n} = 3 ⋅ 2 ^{2n} - 3 ^{2n} - 3 \equiv 0 \pmod{ 4 }.

If is prime, then we use (1) and (2), and if is composite, then we use (3) and (4) to deduce

B _{2n} = I _{n} - \sum _{(p-1)|2n} (1)/(p) ,

where is an integer, as desired.^{[1] [2]}

See also

• Kummer's congruence

References

1. H. Rademacher, Analytic Number Theory, Springer-Verlag, New York, 1973.
2. T. M. Apostol, Introduction to Analytic Number Theory, Springer-Verlag, 1976.