NEW PROPERTIES OF MULTIPLE HARMONIC SUMS MODULO p AND p-ANALOGUES OF LESHCHINER’S SERIES
arXiv:1206.0407v3 [math.NT] 22 Mar 2013
KH. HESSAMI PILEHROOD, T. HESSAMI PILEHROOD, AND R. TAURASO Abstract. In this paper we present some new binomial identities for multiple harmonic sums whose indices are the sequences ({1}a , c, {1}b ), ({2}a , c, {2}b ) and prove a number of congruences for these sums modulo a prime p. The congruences obtained allow us to find nice p-analogues of Leshchiner’s series for zeta values and to refine a result due to M. Hoffman and J. Zhao about the set of generators of the multiple harmonic sums of weight 7 and 9 modulo p. As a further application we provide a new proof of Zagier’s formula for ζ ∗ ({2}a , 3, {2}b ) based on a finite identity for partial sums of the zeta-star series.
1. Introduction In the last few years there has been a growing attention in the study of p-adic analogues of various binomial series related to multipleP zeta values, which are nested generalizations of the classical Riemann zeta function ζ(s) = sn=1 1/ns . The main reason of interest of such p-analogues is that they are related to divisibility properties of multiple harmonic sums which can be considered as elementary “bricks” for expressing complicated congruences. Before discussing this further we recall the precise definitions of such objects. For r ∈ N, s = (s1 , s2 , . . . , sr ) ∈ (Z∗ )r , and a non-negative integer n, the alternating multiple harmonic sum is defined by r X Y sgn (si )ki Hn (s1 , s2 , . . . , sr ) = |s | ki i 1≤k1 0
(m) n X Hk−1 (b, s)An,k k=1
(m) n X Hk−1 (b, s)An,k
kj
k=1
j+|s|=a+c j≥0,s1 >a+1
kc−1
(m) n X Hk−1 (b)An,k k=1
+
k=1
(m) n X Hk−1 (b, a + 1)An,k k=1
=
+
n m X (m) Hk−1 (b, a + 1)An,k nc−1
+
l(s)
(m) n X Hk−1 (b, a + 1, s)An,k k=1
kj
(m) n X Hk−1 (b, s)An,k k=1
kj
.
NEW PROPERTIES OF MULTIPLE HARMONIC SUMS
9
Theorem 2.1. For a positive integer n and non-negative integers a, b, and c ≥ 2,
a
b
Sn ({2} , c, {2} ) =
n X (−1)k−1
n k 2 2a+2b+c n+k k k k=1
X
+4
l(s)
2
i+j+|s|=c i≥1,j≥2,|s|≥0
n X Hk−1 (2a + i, s)(−1)k−1 k2b+j n+k k k=1
where s = (s1 , s2 , . . . , sr ) ∈ Nr for r ≥ 0, and |s| =
Pr
i=1 si ,
n k
(21) ,
l(s) = r.
Proof. The proof is by induction on n. For n = 1 we have S1 ({2}a , c, {2}b ) = 1, and the formula is true. For n > 1 we proceed as follows. If c = 2 then we should prove that
m
Sn ({2} ) =
n X (−1)k−1
n k 2 n+k 2m k k k=1
(22)
where m = a + b + 1. First note that m
X
Sn ({2}m ) =
1≤k1 ≤k2 ≤···≤km ≤n
X 1 1 = Sn−1 ({2}l ). 2 2 2 2(m−l) k1 k2 · · · km n l=0
Then by the induction hypothesis, we have that
m
Sn ({2} ) = 2
n X l=0
= =
1 n2(m−l)
2 n2m+2
n−1 X k=1
n−1 k k2l n+k−1 k
(−1)k−1
n X (−1)k−1
k=1 n X (−1)k−1 n k 2 2m n+k k k k=1
−
n k
=
n−1 2 X (−1)k−1
n2m
k=1
(n2m+2 − k2m+2 ) , n+k
k2m 2
n2m+2
m n−1 X k n+k−1 k l=0
n2l k2l
k
n X (−1)k−1 k2 n+k k=1
k
n k
and formula (22) easily follows by the equation (13). To prove (21) for c > 2 we note that for n > 1, Sn ({2}a , c, {2}b ) =
=
1
X
c k2 2 k2 · · · ka2 ka+1 a+2 · · · ka+b+1 1≤k1 ≤···≤ka ≤ka+1 ≤ka+2 ≤···≤ka+b+1 ≤n 1 b X l=0
1 n2(b−l)
Sn−1 ({2}a , c, {2}l ) +
1 n2b+c
Sn ({2}a ).
10
KH. HESSAMI PILEHROOD, T. HESSAMI PILEHROOD, AND R. TAURASO
Then by the induction hypothesis and formula (22), we have a
b X
Sn ({2} , c, {2} ) = 2 +4
b X l=0
+
n−1 X
n−1 k n+k−1 2(b−l) 2(a+l)+c n k k l=0 k=1
b
1
1
n2(b−l)
(−1)k−1
X
n2b+c
k=1
2
i+j+|s|=c i≥1,j≥2,|s|≥0
n X (−1)k−1
2
l(s)
k2a
n k
n−1 X
Hk−1 (2a + i, s)(−1)k−1 k2l+j n−1+k k k=1
n−1 k
.
n+k k
Changing the order of summation and summing the inner sum b n−1 X k n−1+k k l=0
we obtain
a
n−1 k n−1+k k
n2b+2 − k2b+2 n2l = k2l (n2 − k2 )k2b
b
Sn ({2} , c, {2} ) =
n X (−1)k−1
n k 2 n+k 2a+2b+c k k k=1 n X X Hk−1 (2a l(s)
+ 4
+ −
=
2
+ i, s)(−1)k−1 n+k
n k , n+k k
k
n k
n X (−1)k−1
n k − n+k n+k 2b+2 2a 2a+c−2 n2b+c n k k k k k=1 k=1 n X Hk−1 (2a + i, s)(−1)k−1 n X 4 l(s) k 2 . j−2 n+k n2b+2 k k k=1 i+j+|s|=c i≥1,j≥2,|s|≥0
(−1)k−1
2
n k
k2 n2b − k2b n2
k2b+j
k=1
i+j+|s|=c i≥1,j≥2,|s|≥0 n X
2
X
2l(s)
The final result follows as soon as we show that 1 nc−2
(2) n X An,k
k2a
k=1
(2)
where An,k = (−1)k−1 2
X
(2) n X An,k k=1
k2a+c−2
n n+k k / k .
l(s)
2
i+j+|s|=c i≥1,j≥2,|s|≥0
=
+2
kj−2
k=1
i+j+|s|=c i≥1,j≥2,|s|≥0
This identity holds by Lemma 2.2 because
(2) n X Hk−1 (2a + i, s)An,k k=1
(2) n X Hk−1 (2a + i, s)An,k
kj−2
=
X
l(s)
2
j+|s|=2a+c−2 j≥0,s1 >2a
(2) n X Hk−1 (s)An,k k=1
kj
.
NEW PROPERTIES OF MULTIPLE HARMONIC SUMS
11
Corollary 2.1. For a positive integer n and non-negative integers a, b, n k−1 n X (−1) Sn ({2}a ) = 2 k , 2a n+k k k k=1 n n k−1 n X X (−1)k−1 n H (2a + 1)(−1) k−1 k k . Sn ({2}a , 3, {2}b ) = 2 +4 2(a+b)+3 n+k 2b+2 n+k k k k k k=1 k=1
(23)
Theorem 2.2. For a positive integer n and non-negative integers a, b, c, n n X X X (−1)k−1 nk Hk−1 (a + i, s)(−1)k−1 nk a b Sn ({1} , c, {1} ) = + . ka+b+c kb+j k=1
(24)
i+j+|s|=c k=1 i≥1,j≥1,|s|≥0
Proof. Note that the case c = 1 is well known (see [6, Corollary 3]). It also follows from the polynomial identity n X (1 + x)k1 − 1 X n xk = k km k1 · · · km k=1
1≤k1 ≤...≤km ≤n
which appeared in [27, Lemma 5.5]. For c > 1 the identity can be proved by induction on n. By a similar argument as in the proof of Theorem 2.1, we have b X 1 1 Sn ({1} , c, {1} ) = Sn−1 ({1}a , c, {1}l ) + b+c Sn ({1}a ) b−l n n l=0 b n n−1 n n−1 X 1 X (−1)k−1 k 1 X (−1)k−1 k + = b+c n ka nb−l ka+c+l l=0 k=1 k=1 n−1 b X Hk−1 (a + i, s)(−1)k−1 n−1 X X 1 k . + nb−l kl+j a
b
l=0
i+j+|s|=c k=1 i≥1,j≥1,|s|≥0
Changing the order of summation and summing the inner sum b b n − 1 X nl n k n = − l b k k k n k l=0
we obtain a
b
Sn ({1} , c, {1} ) =
n X (−1)k−1
+ −
+
X
n X Hk−1 (a + i, s)(−1)k−1
kb+j i+j+|s|=c k=1 i≥1,j≥1,|s|≥0 n n n n 1 X (−1)k−1 k 1 X (−1)k−1 k − b+1 nb+c ka n ka+c−1 k=1 k=1 n X X Hk−1 (a + i, s)(−1)k−1 nk 1 . nb+1 kj−1 i+j+|s|=c k=1 i≥1,j≥1,|s|≥0 k=1
ka+b+c
n k
n k
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KH. HESSAMI PILEHROOD, T. HESSAMI PILEHROOD, AND R. TAURASO
(1) Now the result easily follows from Lemma 2.2 with An,k = (−1)k−1 nk and the equality n n X X X X Hk−1 (s)(−1)k−1 nk Hk−1 (a + i, s)(−1)k−1 nk = . kj−1 kj j+|s|=a+c−1 k=1 j≥0,s1 >a
i+j+|s|=c k=1 i≥1,j≥1,|s|≥0
Theorem 2.3. Let a, b be integers satisfying a ≥ 1, b ≥ 0. Then for any positive integer n, n n X b k Sn (1, {2} ) = 2 , (25) n+k 2b+1 k k k=1 n n n k−1 X X (−1) Hk−1 (−2a) nk a b k (26) Sn ({2} , 1, {2} ) = 2 −4 . 2(a+b)+1 n+k 2b+1 n+k k k k k k=1 k=1 Proof. To prove the first identity we proceed by induction on b. For b = 0 its validity follows from (11). Assume the formula holds for b > 0. Then we easily obtain from (12), n n k k X X Sk (1, {2}b ) 1 X b+1 l Sn (1, {2} ) = =2 k2 k2 l2m+1 k+l l k=1 k=1 l=1 n n n n k X X 1 X l l =2 =2 , 2m+3 n+l l2m+1 k2 k+l l l l l=1
l=1
k=l
as required. To prove the second identity of our theorem we proceed by induction on n. Obviously, it is valid for n = 1. For n > 1 we use the equality Sn ({2}a , 1, {2}b ) =
b X l=0
1 1 Sn−1 ({2}a , 1, {2}l ) + 2b+1 Sn ({2}a ) n n2(b−l)
and apply the same arguments as in the proof of Theorem 2.1. Then with the help of the formula (10) we may easily deduce the result. Remark 2.3.1. Note that the identities for the finite multiple harmonic sums found in Theorems 2.1, 2.3 can be used for evaluating multiple zeta-star values. For example, letting n tend to infinity in (23), (25), (26) we get expressions for ζ ∗ ({2}a , 3, {2}b ), ζ ∗ ({2}a , 1, {2}b ) in terms of alternating Euler sums: ζ ∗ ({2}a , 3, {2}b ) = 2ζ(2a + 2b + 3) + 4
∞ X (−1)k−1 Hk−1 (2a + 1) k=1
ζ ∗ ({2}a , 1, {2}b ) = 2ζ(2a + 2b + 1) − 4
∞ X Hk−1 (−2a) k=1
∗
b
ζ (1, {2} ) = 2ζ(2b + 1), where ζ(s) :=
∞ X (−1)k−1 k=1
ks
k2b+2
k2b+1
b ≥ 1,
,
,
(27)
a, b ≥ 1,
(28) (29)
= (1 − 21−s )ζ(s),
with ζ(1) = log 2, is the alternating zeta function. Formula (29) was proved previously by several authors (see [23, p. 292, Ex. b], [39], [29]). Evaluations for length 2 Euler sums of
NEW PROPERTIES OF MULTIPLE HARMONIC SUMS
13
odd weight appearing on the right-hand sides of (27) and (28) in terms of zeta values are well known (see, for example, [7, Theorem 7.2]). Therefore by [7, Theorem 7.2], we obtain K X 2r 1 2r ∗ a b ζ ({2} , 3, {2} ) = 4 1 − 2r + δr,a − ζ(2r + 1)ζ(2K − 2r), 2b + 1 2 2a r=1
where ζ(0) := 1, K = a + b + 1, which gives another proof of Zagier’s formula (1) for the zeta-star value ζ ∗ ({2}a , 3, {2}b ), and a+b X 2r 2r 1 ∗ a b ζ ({2} , 1, {2} ) = 4 − 1 − 2r ζ(2r + 1)ζ(2a + 2b − 2r) (30) 2b 2 2a − 1 r=1 for a, b ≥ 1. Note that formula (30) was also proved in [25, Theorem 1.6] by another method. 3. Auxiliary results on congruences In this section we collect several congruences that will be required later in this paper. (i) ([35, Theorem 1.6]) for positive integers a, r and for any prime p > ar + 2, ( 2 3 (−1)r a(ar+1) 2(ar+2) p Bp−ar−2 (mod p ) if ar is odd, Hp−1 ({a}r ) ≡ a p Bp−ar−1 (mod p2 ) if ar is even; (−1)r−1 ar+1 (ii) ([35, Theorems 3.1, 3.2]) for positive integers a1 , a2 and for any prime p ≥ a1 + a2 , (−1)a2 a1 + a2 Hp−1 (a1 , a2 ) ≡ Bp−a1 −a2 (mod p), a1 + a2 a1 moreover, if a1 + a2 is even, then for any prime p > a1 + a2 + 1, a1 + a2 + 1 a1 + a2 + 1 a1 a1 Hp−1 (a1 , a2 ) ≡ p (−1) a2 − (−1) a1 − a1 − a2 a1 a2 ×
Bp−a1 −a2 −1 2(a1 + a2 + 1)
(mod p2 );
(iii) ([35, Theorem 3.5] if a1 , a2 , a3 ∈ N and w := a1 + a2 + a3 is odd, then for any prime p > w, we have Bp−w a1 w a3 w (mod p); Hp−1 (a1 , a2 , a3 ) ≡ (−1) − (−1) 2w a1 a3 (iv) ([24, Theorem 5.2]) for a positive integer a and for any prime p ≥ a + 2, we have 7 −2qp (2) + pqp2 (2) − p2 32 qp3 (2) + 12 Bp−3 (mod p3 ) if a = 1, a+1 −1) H p−1 (a) ≡ a(22(a+1) p Bp−a−1 (mod p2 ) if a is even, 2 a 2 −2 − a Bp−a (mod p) if a > 1 is odd, where qp (2) = (2p−1 − 1)/p is the so-called Fermat quotient;
(v) ([15, Lemma 1]) if a, b are positive integers and a + b is odd, then for any prime p > a + b, Bp−a−b b a+b a+b (−1) +2 −2 (mod p); H p−1 (a, b) ≡ 2 2(a + b) a
14
KH. HESSAMI PILEHROOD, T. HESSAMI PILEHROOD, AND R. TAURASO
(vi) by (i) and (iv), for any positive integer a and any prime p ≥ a + 2, we have Hp−1 (−a) = 21−a H p−1 (a) − Hp−1 (a) 2 −2q (2) + pqp2 (2) − p2 32 qp3 (2) + 14 Bp−3 (mod p3 ) if a = 1, p −a ) ≡ a(1−2 p Bp−a−1 (mod p2 ) if a is even, a+1 2(1−21−a ) Bp−a (mod p) if a > 1 is odd; − a
(vii) ([27, Theorem 3.1]) for positive integers a, b of distinct parity and a prime p ≥ a+b+1, 1 − 21−a−b Bp−a−b (mod p), a+b 21−a−b − 1 b a+b Hp−1 (−a, −b) ≡ (−1) Bp−a−b (mod p); a+b b (viii) ([27, Theorem 4.1]) if a, b, c ∈ N and w := a + b + c is odd, then for any prime p > w, Hp−1 (−a, b) ≡ Hp−1 (a, −b) ≡
2Hp−1 (a, −b, −c) ≡ Hp−1 (c + b, a) + Hp−1 (−c, −b − a) − Hp−1 (−c)Hp−1 (−b, a)
(mod p),
2Hp−1 (−a, b, −c) ≡ Hp−1 (−c)Hp−1 (b, −a) − Hp−1 (−c, b)Hp−1 (−a) + Hp−1 (−c − b, −a) + Hp−1 (−c, −b − a) (mod p). In [5, Lemma 6.2] it was proved that for positive integers a, b and a prime p > 2a + 2b + 1, Hp−1 (−2a, −2b) ≡ 0 (mod p). However, we will need a stronger version of this congruence. Lemma 3.1. Let a, b be positive integers and a prime p > 2a + 2b + 1. Then the following congruence holds modulo p2 : a+b (a − b)(1 − 2−2a−2b ) 2a + 2b pBp−2a−2b−1 . − Hp−1 (−2a, −2b) ≡ (2a + 1)(2b + 1) 2a + 2b + 1 2a Proof. Clearly (see [27, Lemma 7.1]), X X (−1)j+k (−1)p−j+p−k Hp−1 (−2a, −2b) = = 2a 2b j k (p − j)2a (p − k)2b 1≤j 2a + 2b + 1. Then 4(b − a)(1 − 4−a−b ) 2a + 2b a b Bp−2a−2b−1 (mod p), Sp−1 ({2} , 1, {2} ) ≡ (2a + 1)(2b + 1) 2a 4(−1)a+b (a − b)(1 − 4−a−b ) 2a + 2b a b Bp−2a−2b−1 (mod p). Hp−1 ({2} , 1, {2} ) ≡ (2a + 1)(2b + 1) 2a
(37)
NEW PROPERTIES OF MULTIPLE HARMONIC SUMS
17
Proof. We begin by considering the case a ≥ 1. Then from identity (26) with n = p − 1 and congruence (34) we have Sp−1 ({2}a , 1, {2}b ) ≡ − − By the obvious equality
4 p
p−1 p 2X 1 1 − 2pH (1) − k−1 p k2a+2b k
k=1 p−1 X k=1
Hk−1 (−2a)(−1)k p 1 − 2pH (1) − k−1 k2b k
Hn (a)Hn (b) = Hn (a, b) + Hn (b, a) + Hn sgn(ab)(|a| + |b|)
(mod p).
a, b ∈ Z \ {0},
(38)
we obtain
2 4 Sp−1 ({2}a , 1, {2}b ) ≡ − Hp−1 (2a + 2b) + 4Hp−1 (1, 2a + 2b) − Hp−1 (−2a, −2b) p p + 8Hp−1 (−2a, 1, −2b) + 8Hp−1 (1, −2a, −2b) + 8Hp−1 (−2a − 1, −2b) + 4Hp−1 (−2a, −2b − 1)
(mod p).
Applying (viii), (38), (vi) and simplifying we deduce that 4 2 Sp−1 ({2}a , 1, {2}b ) ≡ − Hp−1 (2a + 2b) − Hp−1 (−2a, −2b) p p
(mod p).
Now the required congruence follows from (i) and Lemma 3.1. The case a = 0 is handled in the same way with the help of identity (25) and congruences (vi), (vii). The corresponding congruence for Hp−1 ({2}a , 1, {2}b ) easily follows from relation (35). Corollary 4.1. Let p be a prime greater than 7. Then 27 Bp−7 (mod p), 16 1 3 1889 Sp−1 (1, 1, 1, 6) ≡ Bp−3 + Bp−9 (mod p). 54 648
Sp−1 (1, 1, 1, 4) ≡ Sp−1 (1, 2, 2, 2) ≡
(39) (40)
Proof. The first congruence in (39) follows from [19, Theorem 7.3] and the second one is a consequence of Theorem 4.2 with a = 0, b = 3. To prove congruence (40) we set a = 0, b = 4 in Theorem 4.2 to obtain 85 Bp−9 (mod p). (41) Sp−1 (1, {2}4 ) ≡ 48 On the other hand, using [19, Theorem 6.4] we can rewrite the length 5 sum Sp−1 (1, {2}4 ) in terms of length 4 sums and Sp−1 (1, 8) as 2Sp−1 (1, {2}4 ) ≡Sp−1 (1, 2, 2, 4) + Sp−1 (1, 2, 4, 2) + Sp−1 (1, 4, 2, 2) 255 Sp−1 (1, 8) (mod p). + Sp−1 (3, 2, 2, 2) − 18 Now applying [19, Theorem 7.5] and expressing all length 4 sums in the congruence above in terms of the three quantities Sp−1 (1, 1, 1, 6), Sp−1 (1, 8) ≡ Bp−9 , and Sp−1 (1, 2)Sp−1 (1, 1, 4) ≡ 3 − 16 Bp−3 (mod p) we obtain 281 1 Sp−1 (1, 8) (mod p). 2Sp−1 (1,{2}4 ) ≡ 3Sp−1 (1, 1, 1, 6) + Sp−1 (1, 2)Sp−1 (1, 1, 4) − 3 54
(42)
18
KH. HESSAMI PILEHROOD, T. HESSAMI PILEHROOD, AND R. TAURASO
Comparing congruences (41) and (42) we conclude that 1 1889 Sp−1 (1, 1, 1, 6) ≡ − Sp−1 (1, 2)Sp−1 (1, 1, 4) + Sp−1 (1, 8) 9 648 1 3 1889 ≡ Bp−3 + Bp−9 (mod p). 54 648 From Corollary 4.1, [19, Theorems 7.3, 7.5], and [36, Section 2] we obtain the following description of multiple harmonic sums of weight 7 and 9. Corollary 4.2. Let p be a prime greater than 7. Then (i) all multiple harmonic sums Sp−1 (s), Hp−1 (s), of weight |s| = 7 belong to Q(2)Bp−7 modulo p; (ii) all multiple harmonic sums Sp−1 (s), Hp−1 (s) of weight |s| = 9 can be written modulo p 3 with coefficients in Q(3). as linear combinations of the two quantities Bp−9 and Bp−3 In [35, Theorem 3.16], Zhao proved that Sp−1 ({1}a , 2, {1}b ) ≡ Hp−1 ({1}a , 2, {1}b ) ≡ 0
(mod p)
if a + b is even and a prime p > a + b + 3. In the next theorem we prove a deeper result on the above sums. Theorem 4.3. For non-negative integers a, b and a prime p > a + b + 3, (−1)b a+b+2 a+b+2 (mod p) if a + b is odd, a+1 Bp−a−b−2 Sp−1 ({1}a , 2, {1}b ) ≡ pBp−a−b−3 1 + (−1)a a+b+3 (mod p2 ) if a + b is even; a+2 2(a+b+3) Hp−1 ({1}a , 2, {1}b ) ≡
(−1)b a+b+2
a+b+2 a+1 Bp−a−b−2
pBp−a−b−3 2(a+b+3)
Proof. From Theorem 2.2 for c = 2 we have a
b
Sn ({1} , 2, {1} ) =
n X (−1)k−1 k=1
n k
ka+b+2
+
(mod p)
a+b+3
1 + (−1)a
b+2
if a + b is odd,
(mod p2 ) if a + b is even.
n X Hk−1 (a + 1)(−1)k−1 k=1
kb+1
Setting n = p − 1 and observing that p−1 (p − 1)(p − 2) · · · (p − k) ≡ (−1)k (1 − pHk (1)) = k! k
n k
.
(mod p2 )
we have a
b
Sp−1 ({1} , 2, {1} ) ≡ −
p−1 X 1 − pHk (1) k=1
ka+b+2
−
p−1 X Hk−1 (a + 1)(1 − pHk (1)) k=1
kb+1
= −Hp−1 (a + b + 2) + pHp−1 (1, a + b + 2) + pHp−1 (a + b + 3)
− Hp−1 (a + 1, b + 1) + pHp−1 (a + 1, b + 2) + pHp−1 (a + 1, 1, b + 1) + pHp−1 (1, a + 1, b + 1) + pHp−1 (a + 2, b + 1)
(mod p2 ).
(43)
NEW PROPERTIES OF MULTIPLE HARMONIC SUMS
19
If a + b is odd, then by (ii) we obtain (−1)b a+b+2 Sp−1 ({1} , 2, {1} ) ≡ −Hp−1 (a + 1, b + 1) ≡ Bp−a−b−2 (mod p). a+b+2 a+1 a
b
If a + b is even, then from (43), (i)–(iii) after simplifying we get the required congruence. Considering equality (35) with n = p − 1, s = ({1}a , 2, {1}b ) modulo p we find (−1)a+b+1 Sp−1 ({1}b , 2, {1}a ) ≡ −Hp−1 ({1}a , 2, {1}b )
(mod p)
and therefore a
b
Hp−1 ({1} , 2, {1} ) ≡
( −Sp−1 ({1}b , 2, {1}a ) (mod p) if a + b is odd, 0
(mod p)
if a + b is even.
(44)
Similarly, from identity (35) considered modulo p2 we obtain (−1)a+b Sp−1 ({1}b , 2, {1}a ) ≡ Hp−1 ({1}a , 2, {1}b ) −
Y F
Hp−1 (s1 )Hp−1 (s2 ) (mod p2 ).
(45)
s1 s2 = ({1}a ,2,{1}b )
It is clear that the product Hp−1 (s1 )Hp−1 (s2 ) has one of the forms Hp−1 ({1}m )Hp−1 ({1}a−m , 2, {1}b ),
1 ≤ m ≤ a,
or Hp−1 ({1}a , 2, {1}l )Hp−1 ({1}b−l ),
0 ≤ l ≤ b.
If a + b is even, then by (i) and (44), it easily follows that both products are congruent to zero modulo p2 . Thus in this case from (45) we obtain Sp−1 ({1}b , 2, {1}a ) ≡ Hp−1 ({1}a , 2, {1}b )
(mod p2 )
and the proof is complete.
Remark 4.3.1. It is worth mentioning that there is a notion of duality for multiple harmonic sums: given s = (s1 , s2 , . . . , sr ) we define the power set P (s) to be the partial sum sequence (s1 , s1 + s2 , . . . , s1 + · · · + sr−1 ) as a subset of {1, 2, . . . , |s| − 1}. Then s∗ is the composition of weight |s| corresponding to the complement subset of P (s) in {1, 2, . . . , |s| − 1}, namely, s∗ = P −1 ({1, 2, . . . , |s| − 1} \ P (s)). It is easy to verify that (s∗ )∗ = s. Moreover, by [18, Theorem 6.7] we have that Sp−1 (s) ≡ −Sp−1 (s∗ ) (mod p) for any prime p. In particular, for any positive integers a, b it follows that Sp−1 ({2}a , 1, {2}b ) ≡ −Sp−1 (1, {2}a−1 , 3, {2}b−1 , 1) Sp−1 ({2}a , 3, {2}b ) ≡ −Sp−1 (1, {2}a , 1, {2}b , 1) a
b
Sp−1 ({1} , 2, {1} ) ≡ −Sp−1 (a + 1, b + 1)
(mod p),
(mod p),
(mod p),
and by Theorem 4.1 and Theorem 4.2 we are able to find Sp−1 (1, {2}a , 1, {2}b , 1) and Sp−1 (1, {2}a−1 , 3, {2}b−1 modulo a prime p in terms of Bernoulli numbers.
20
KH. HESSAMI PILEHROOD, T. HESSAMI PILEHROOD, AND R. TAURASO
5. Congruences for “odd” multiple harmonic sums In this section we prove some congruences involving “odd” multiple harmonic sums. These results will be needed in the next sections. Lemma 5.1. Let s ∈ Nr . Then for any prime p > 3, H p−1 (s) ≡ 2
(−1)|s| H p−1 (s) (mod p). 2 2|s|
Proof. Changing the order of summation in the definition of H n (s), with n = X 1 H n (s1 , s2 , . . . , sr ) = s 1 (2k1 + 1) · · · (2kr + 1)sr
p−1 2 ,
we obtain
0≤k1 kr ≥0
≡
X
1 (2(n − 1 − k1 ) + 1)s1 · · · (2(n − 1 − kr ) + 1)sr
0≤kr 1, then by (48) and induction hypothesis, we have mH n ({2}m ) ≡ (−1)m−1 H n (2m) (mod p2 ), and the theorem is proved. Theorem 5.2. Let a, b be non-negative integers and a prime p > 2a + 2b + 3. Then Bp−2a−2b−3 2a + 2b + 2 a b S p−1 ({2} , 3, {2} ) ≡ − (mod p), 2 b+1 2a + 1 a b a+b+1 Bp−2a−2b−3 2a + 2b + 2 (mod p). H p−1 ({2} , 3, {2} ) ≡ (−1) 2 a+1 2a + 1
NEW PROPERTIES OF MULTIPLE HARMONIC SUMS
Proof. Setting n = n
p−1 2
in identity (23) and observing that p 3 p p 2k−1 1 − − · · · − k 2 2 k = (−1)k 12 p2 23 p2 p ≡ (−1) n+k 2k−1 + + · · · + 2 2 2 2 2 2 k
we obtain
21
(mod p)
(49)
S p−1 ({2}a , 3, {2}b ) ≡ −2H p−1 (2a + 2b + 3) − 4H p−1 (2a + 1, 2b + 2) (mod p). 2
2
2
Now the required congruence easily follows by (iv) and (v). From Lemma 5.1 and Theorem 5.1 it follows that H p−1 ({2}m ) ≡ 22m H p−1 ({2}m ) ≡ 0 (mod p). 2
2
Therefore applying identity (35) with n =
p−1 2
and s = ({2}a , 3, {2}b ) we obtain
(−1)a+b+1 S p−1 ({2}b , 3, {2}a ) ≡ −H p−1 ({2}a , 3, {2}b ) (mod p) 2
2
and the theorem is proved.
Theorem 5.3. Let a, b be non-negative integers, not both zero, and a prime p > 2a + 2b + 1. Then 21−2a−2b − 2 2a + 2b a b Bp−2a−2b−1 (mod p), S p−1 ({2} , 1, {2} ) ≡ 2 2b + 1 2a (−1)a+b (21−2a−2b − 2) 2a + 2b a b H p−1 ({2} , 1, {2} ) ≡ Bp−2a−2b−1 (mod p). 2 2a + 1 2a Proof. If a ≥ 1, then from identity (26) with n = (p − 1)/2 and congruence (49) we find S p−1 ({2}a , 1, {2}b ) ≡ −2H p−1 (2a + 2b + 1) − 4H p−1 (−2a, −2b − 1) 2
2
(mod p).
2
Now the required result easily follows by (iv) and Lemma 3.2. If a = 0, then from (25) we have S p−1 (1, {2}b ) ≡ 2H p−1 (−2b − 1) (mod p). (50) 2
2
Moreover (p−1)/2
X (−1)k = H p−1 (−2b − 1) = 2 k2b+1 k=1
≡
p−1 X
p−1 X
k=(p+1)/2
k=(p+1)/2
(−1)p−k (p − k)2b+1
(−1)k = Hp−1 (−2b − 1) − H p−1 (−2b − 1) 2 k2b+1
(mod p).
Therefore, since b > 0, by (vi) it follows that H p−1 (−2b − 1) ≡ 2
1 2−2b − 1 Hp−1 (−2b − 1) ≡ Bp−2b−1 2 2b + 1
(mod p).
(51)
From (50) and (51) we conclude the truth of the first congruence for a = 0. From identity (35) we easily obtain (−1)a+b S p−1 ({2}b , 1, {2}a ) ≡ H p−1 ({2}a , 1, {2}b ) (mod p) 2
and the proof is complete.
2
22
KH. HESSAMI PILEHROOD, T. HESSAMI PILEHROOD, AND R. TAURASO
Lemma 5.2. Let r, a be positive integers. Then for any prime p > r + 2, we have a X r−1+k r H p−1 (r) ≡ H⌊ p ⌋ (r) + (−2) H p−1 (r + k)pk 4 2 2 k k=0 a X r − 1 + k H⌊ p4 ⌋ (r + k) k − (−1)r p (mod pa+1 ). k 2k
(52)
k=0
If r is a non-negative integer, then for any prime p > 2r + 3, we have p+1
(−1) 2 H p−1 (2r + 1) (mod p2 ). 2 2 42r+1 Proof. Let n = (p−1)/2, m = ⌊p/4⌋. Then the first congruence follows easily from the equality n m X X 1 1 Hn (r) = Hm (r) + 2r . − (p − k)r (p − 2k)r H p−1 (−2r − 1) ≡
k=1
k=1
To prove the second one, we have (−1)n
n−1 X k=0
n
X (−1)k (−1)k = (2k + 1)2r+1 (p − 2k)2r+1 k=1
=2
m X k=1
m X
n
X 1 1 − 2r+1 (p − 4k) (p − 2k)2r+1 k=1
n
X 1 1 − 2r+1 2r+1 2r+1 (4k) (p/(4k) − 1) (2k) (p/(2k) − 1)2r+1 k=1 k=1 1 (2r + 1)p ≡ − 4r+1 Hm (2r + 1) + Hm (2r + 2) 2 4 1 + 2r+1 Hn (2r + 1) (mod p2 ). 2 From (52) with a = 1 and r replaced by 2r + 1, we get =2
(2r + 1)p Hm (2r + 2) ≡ (1 + 22r+1 )Hn (2r + 1) (mod p2 ), 2 which after substitution into the right-hand side of the above congruence implies the result. 2Hm (2r + 1) +
6. Results on p-analogues of Leshchiner’s series In this section, we consider finite p-analogues of zeta and beta values arising from the truncation of Leshchiner’s series (5)–(8) and defined by the finite sums p−1
m p−1
k=1
j=1 k=1
X X (−1)m−j Hk−1 ({2}m−j ) 3 X (−1)m Hk−1 ({2}m ) + 2 , ζp (2m + 2) := 2 k2 2k k2j+2 2k k k p−1
5 X (−1)k+m−1 Hk−1 ({2}m ) ζp (2m + 3) := 2 k3 2k k +2
k=1 p−1 m XX
(−1)k+m−j−1 Hk−1 ({2}m−j ) , k2j+3 2k k j=1 k=1
NEW PROPERTIES OF MULTIPLE HARMONIC SUMS
23
(p−3)/2 2k 5 X (−1)k+m k H k ({2}m ) ζ p (2m + 2) := 4 16k (2k + 1)2 k=0 m (p−3)/2 X X (−1)k+m−j 2k H k ({2}m−j ) k , + 16k (2k + 1)2j+2 j=1 k=0 (p−3)/2 2k 3 X (−1)m k H k ({2}m ) βp (2m + 1) := 4 16k (2k + 1) k=0 m (p−3)/2 X X (−1)m−j 2k H k ({2}m−j ) k . + 16k (2k + 1)2j+1 j=1
k=0
Applications of our results on congruences obtained in Sections 4 and 5 allowed us to establish the following congruences for the finite p-analogues of Leshchiner’s series. Theorem 6.1. Let m be a non-negative integer and p be a prime such that p > 2m + 3. Then 4m(1 − 2−2m ) Bp−2m−1 (mod p), 2m + 1 (m + 1)(2m + 1) ζp (2m + 3) ≡ − Bp−2m−3 (mod p). 2m + 3 Note that for ζp (2), ζp (3), ζp (4), stronger congruences were proved by the authors in [26, 14]. For ζp (5), we recover the congruence ζp (5) ≡ −6/5Bp−5 (mod p) proved in [14]. 2p ζp (2m + 2) ≡ −
Theorem 6.2. Let m be a non-negative integer and p be a prime such that p > 2m + 3. Then (−1)(p+1)/2 m(4m − 1) Bp−2m−1 (mod p), 16m (2m + 1) 2m2 + 3m + 2 p Bp−2m−3 (mod p2 ). ζ p (2m + 2) ≡ − 2m+3 2 (2m + 3)
βp (2m + 1) ≡
Note that for βp (1) and ζ p (2), sharper congruences were proved in [15]. 7. Finite identities and their formal proofs By revisiting the combinatorial proof (due to D. Zagier) presented in Section 5 in [22], it is possible to obtain finite versions of identities (5)–(8). Here we derive these identities in a formal unified way based on applications of appropriate WZ pairs. Lemma 7.1. Let m be a non-negative integer. Then for every positive integer n, we have n n X (−1)k−1 1 X (−1)n+k+m Hk−1 ({2}m ) − k2m+2 2 k2 nk n+k k k=1 k=1 (53) n m X n X (−1)m−j Hk−1 ({2}m−j ) 3 X (−1)m Hk−1 ({2}m ) +2 , = 2 k2 2k k2j+2 2k k k j=1 k=1
k=1
n X k=1
1
k2m+3
5 = 2
n
−
1 X (−1)k+m Hk−1 ({2}m ) 2 k3 n n+k k=1
k
k
n m X X (−1)k+m−j−1 Hk−1 ({2}m−j ) (−1)k+m−1 Hk−1 ({2}m ) + 2 . k3 2k k2j+3 2k k k j=1 k=1 k=1
n X
(54)
24
KH. HESSAMI PILEHROOD, T. HESSAMI PILEHROOD, AND R. TAURASO
Proof. For non-negative integers n, k, we consider a pair of functions F1 (n, k) =
(−1)n+k (n − k − 1)!(1 + a)k (1 − a)k , 2(n + k + 1)!
G1 (n, k) =
(−1)n+k (n − k)!(n + 1)(1 + a)k (1 − a)k , ((n + 1)2 − a2 )(n + k + 1)!
n ≥ k + 1, n ≥ k,
where (c)k is the shifted factorial defined by (c)k = c(c + 1) · · · (c + k − 1), k ≥ 1, and (c)0 = 1. It is easy to check that (F1 , G1 ) is a WZ pair, i.e., it satisfies the relation F1 (n + 1, k) − F1 (n, k) = G1 (n, k + 1) − G1 (n, k). Summing the above equality first over k and then over n, we get the following summation formula (see [14, Section 4]): N X
G1 (n − 1, 0) =
n=1
N X
(G1 (n − 1, n − 1) + F1 (n, n − 1)) −
n=1
F1 (N, k − 1),
N ∈ N. (55)
k=1
Substituting (F1 , G1 ) in (55) we obtain N X (−1)n−1
n=1
N X
n 2 − a2
n−1 N a2 1 X 1 3n2 + a2 Y 1 − 2 n 2 − a2 m2 n2 2n n
=
m=1
n=1
k−1 N 1 X (−1)N +k Y a2 + 1 − . 2 m2 k2 Nk N k+k k=1
(56)
m=1
Expanding both sides of (56) in powers of a2 and comparing coefficients of a2m we derive the finite identity (53). Similarly, considering the WZ pair (−1)n (−1)n F1 (n, k), G2 (n, k) = G1 (n, k) k+1 n+1 and applying the same argument as above we obtain the second identity. F2 (n, k) =
Lemma 7.2. For any non-negative integers m, n we have n n 2k X 1 X (−1)n+k+m k H k ({2}m ) (−1)k − (2k + 1)2m+1 4 16k (2k + 1) n+k+1 3 = 4
n X k=0
2k+1
k=0
k=0
n X k=0
2k
(−1)m k H k ({2}m ) + 16k (2k + 1)
n m X X j=1 k=0
m−j ) (−1)m−j 2k k H k ({2} , 16k (2k + 1)2j+1
n 2k 1 1 X (−1)k+m k H k ({2}m ) + (2k + 1)2m+2 4 16k (2k + 1)2 n+k+1 2k+1 k=0
m n n 2k 5 X (−1)k+m k H k ({2}m ) X X (−1)k+m−j k H k ({2}m−j ) = + . 4 16k (2k + 1)2 16k (2k + 1)2j+2 k=0
2k
j=1 k=0
Proof. As in the proof of Lemma 7.1, applying summation formula (55) to the WZ pairs (−1)n+k (n − k − 1)! 1 + a 1 − a , n ≥ k + 1, F1 (n, k) = 4(n + k)! 2 2 k k
(57)
(58)
NEW PROPERTIES OF MULTIPLE HARMONIC SUMS
G1 (n, k) = and
(−1)n+k (n − k)!(2n + 1) 1 + a 1 − a , (n + k)!((2n + 1)2 − a2 ) 2 2 k k
(−1)n F1 (n, k), 2k + 1 we get the required identities. F2 (n, k) =
G2 (n, k) =
25
n ≥ k,
(−1)n G1 (n, k), 2n + 1
8. Proofs of Theorems 6.1 and 6.2 Proof of Theorem 6.1 Setting n = p − 1 in (53), we get p−1
ζp (2m + 2) = Hp−1 (−2m − 2) −
1 X (−1)k+m Hk−1 ({2}m ) . 2 k2 p−1 p−1+k k
k=1
Since
k
k−1 k (−1)k p −1 1 Y p −1 Y 1 + = 1 − p−1+k pk m=1 m m k2 p−1 k k m=1
≡
1 1 pHk (2) + + pk k2 k
(59)
(mod p2 ),
we obtain ζp (2m + 2) ≡ −Hp−1 (−2m − 2) −
(−1)m Hp−1 ({2}m , 1) 2p
(−1)m Hp−1 ({2}m+1 ) (mod p), 2 which by (i), (vi), and Theorem 4.2 implies the required congruence. Similarly, setting n = p − 1 in (54) and applying (59), (i) and Theorem 4.1 we get −
p−1
1 X (−1)k+m Hk−1 ({2}m ) p−1+k 2 k3 p−1 k k k=1 m−1 (−1) 1 ≡ Hp−1 (2m + 3) + Hp−1 ({2}m+1 ) + Hp−1 ({2}m , 3) 2 p m−1 Bp−2m−3 (−1) + Hp−1 ({2}m , 3) ≡− 2m + 3 2 (m + 1)(2m + 1) Bp−2m−3 (mod p). ≡− 2m + 3 Proof of Theorem 6.2 Setting n = (p − 3)/2 in (57), we get (p−3)/2 2k (−1)(p−1)/2+m X (−1)k k H k ({2}m ) βp (2m + 1) = H p−1 (−2m − 1) + . 2 4 16k (2k + 1) (p−1)/2+k ζp (2m + 3) = Hp−1 (2m + 3) −
k=0
Taking into account that 2k k (−16)k (p−1)/2+k 2k+1
k−1 −1 2(2k + 1) Y p2 2p = 1− ≡ −2 − 2 p − 1 − 2k (2j + 1) 2k + 1 j=0
2k+1
(mod p2 ),
(60)
26
KH. HESSAMI PILEHROOD, T. HESSAMI PILEHROOD, AND R. TAURASO
we obtain by Theorem 5.1, βp (2m + 1) ≡ H p−1 (−2m − 1) + 2
(−1)(p+1)/2+m 2 (−1)(p+1)/2+m
(p−3)/2
X H k ({2}m ) p 1+ 2k + 1 2k + 1 k=0
H p−1 ({2}m , 1) (mod p2 ), 2 2 which implies the first congruence of the theorem by Lemmas 5.2 and 5.1, and Theorem 5.3. Similarly, from (58) by (60), Theorem 5.1, Lemma 5.1 and Theorem 5.2, we have (−1)m−1 ζ p (2m + 2) ≡ H p−1 (2m + 2) + H p−1 ({2}m+1 ) + pH p−1 ({2}m , 3) 2 2 2 2 m−1 2m + 1 (−1) ≡ 2m+3 p Bp−2m−3 + p H p−1 ({2}m , 3) 2 2 (2m + 3) 2 2m2 + 3m + 2 ≡ − 2m+3 pBp−2m−3 (mod p2 ). 2 (2m + 3) ≡ H p−1 (−2m − 1) + 2
Acknowledgements. The authors wish to thank the referees of the journal whose valuable
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KH. HESSAMI PILEHROOD, T. HESSAMI PILEHROOD, AND R. TAURASO
Department of Mathematics and Statistics, Dalhousie University, Halifax, Nova Scotia, B3H 3J5, Canada E-mail address:
[email protected] Department of Mathematics and Statistics, Dalhousie University, Halifax, Nova Scotia, B3H 3J5, Canada E-mail address:
[email protected] ` di Roma “Tor Vergata”, 00133 Roma, Italy Dipartimento di Matematica, Universita E-mail address:
[email protected]
