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International Journal of Number Theory Vol. 4, No. 1 (2008) 73–106 c World Scientific Publishing Company
WOLSTENHOLME TYPE THEOREM FOR MULTIPLE HARMONIC SUMS
JIANQIANG ZHAO Department of Mathematics, Eckerd College St. Petersburg, Florida 33711, USA
[email protected]
Received 11 October 2006 Accepted 23 November 2006 In this paper, we will study the p-divisibility of multiple harmonic sums (MHS) which are partial sums of multiple zeta value series. In particular, we provide some generalizations of the classical Wolstenholme’s Theorem to both homogeneous and non-homogeneous sums. We make a few conjectures at the end of the paper and provide some very convincing evidence. Keywords: Multiple harmonic sum (MHS); multiple zeta values; Bernoulli numbers; irregular primes. Mathematics Subject Classification 2000: 11A07, 11Y40, 11M41
1. Introduction The Euler–Zagier multiple zeta functions of length one are nested generalizations of Riemann zeta function. They are defined as ζ(s) = ζ(s1 , . . . , sl ) = k1−s1 · · · kl−sl (1.1) 0 1 either one of sj = {1}d with odd d, or else all sj = {1}dj for even dj except one term sj = ({1}m , 2, {1}n ) where m + n < m + n − 2 is even. By induction on the length of s∗ (note that H(2) ≡ 0 (mod p)) we get S(s∗ ) ≡ H(s∗ )
(mod p2 ).
With (3.60) this proves case (a). For future reference we prove some congruences for arbitrary m and n in this case. Since |s| − l(s) = w − 2 we have by (2.11) (−1)w S(s) = (−1)|s|−l(s) S(s) = −H(s) + H(m + 1) · H(n + 1) ≡ −H(s) (mod p)
(3.62)
by Theorem 1.3. Now if l ≥ 2 then in (3.61) one of sj = {1} for some positive d so that H(sj ) ≡ 0 (mod p). Thus d
(−1)w S(s∗ ) ≡ H(s∗ ) ≡ (−1)w H(s∗ )
(mod p).
(3.63)
Combining this with (2.8) and (3.62) we see that whether w is even or odd it is always true that H(s) ≡ (−1)w H(s∗ )
(mod p).
(3.64)
In the rest of the proof we assume all congruences are modulo p. (b) Let s = ({1}n , 2, {1}n−1, 2, {1}n+1) where n ≥ 2 is even. Then s∗ = (n + 1, n + 1, n + 2). So |s| = |s∗ | = 3n + 4, l(s) = 3n + 2, and l(s∗ ) = 3. Since n is even we have by applying (2.11) to s∗ S(s∗ ) ≡ H(s∗ ) − H(2n + 2, n + 2) − H(n + 1, 2n + 3) + H(3n + 4) ≡ H(s∗ ) (3.65) by Theorems 3.2 and 1.3. Applying (2.11) to s and using the fact that H({1} ) ≡ 0 for any d we have d
S(s) ≡ (−1)|s|−l(s) S(s) ≡ −H(s) +
n−1
H({1}n , 2, {1}a) · H({1}n−1−a , 2, {1}n+1)
a=0
≡ −H(s) +
n−1 a=0
H(n + 1, a + 1) · H(n − a, n + 2)
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Wolstenholme Type Theorem for Multiple Harmonic Sums
101
by (3.64). Hence by (2.8) and (3.65) H(s) ≡ H(n + 1, n + 1, n + 2) +
n−1
H(n + 1, a + 1) · H(n − a, n + 2).
(3.66)
a=0
We know that for all j, k < p we have
p−2 0 a (k/j) ≡ −1
(mod p) (mod p)
a=0
if j = k, if j = k.
It follows that p−2
H(n + 1, a + 1) · H(n − a, n + 2) =
p−2
a=0 1≤i
