aa r X i v : . [ m a t h . N T ] D ec Congruences involvingalternating multiple harmonic sum
Roberto Tauraso
Dipartimento di MatematicaUniversit`a di Roma “Tor Vergata”, Italy ∼ tauraso Abstract
We show that for any prime prime p = 2 p − X k =1 ( − k k (cid:18) − k (cid:19) ≡ − ( p − / X k =1 k (mod p )by expressing the l.h.s. as a combination of alternating multiple harmonic sums. In [8] Van Hamme presented several results and conjectures concerning a curious analogybetween the values of certain hypergeometric series and the congruences of some of theirpartial sums modulo power of prime. In this paper we would like to discuss a new exampleof this analogy. Let us consider ∞ X k =1 ( − k k (cid:18) − k (cid:19) = (cid:18) (cid:19) + 12 (cid:18) · · (cid:19) + 13 (cid:18) · · · · (cid:19) + 14 (cid:18) · · · · · · (cid:19) + · · · = Z − x (cid:18) √ x − (cid:19) dx = − (cid:20) log (cid:18) √ x (cid:19)(cid:21) − = 2 log 2 . Let p be a prime number, what’s the p -adic analogue of the above result?The real case suggests to replace the logarithm with some p -adic function which behaves ina similar way. It turns out that the right choice is the Fermat quotient q p ( x ) = x p − − p (which is fine since q p ( x · y ) ≡ q p ( x ) + q p ( y ) (mod p )), and, as shown in [7], the followingcongruence holds for any prime p = 2 p − X k =1 ( − k k (cid:18) − k (cid:19) ≡ q p (2) (mod p ) . Here we improve this result to the following statement.1 heorem 1.1.
For any prime p > p − X k =1 ( − k k (cid:18) − k (cid:19) ≡ q p (2) − pq p (2) + 23 p q p (2) + 712 p B p − ≡ − ( p − / X k =1 k (mod p ) where B n is the n -th Bernoulli number. In the proof we will employ some new congruences for alternating multiple harmonic sumswhich are interesting in themselves such as H ( − , − p −
1) = X
1) = X a , a , . . . , a r ) ∈ ( Z ∗ ) r . For any n ≥ r , we define the alternating multipleharmonic sum as H ( a , a , . . . , a r ; n ) = X ≤ k
Let a, b > then for any prime p = 2 H ( − a ; p −
1) = − H ( a ; p −
1) + 12 a − H (cid:18) a ; p − (cid:19) , H ( − a, − a ; p −
1) = H ( − a ; p − − H (2 a ; p − , and H ( − a, − b ; p − ≡ − (cid:18) − a + b − (cid:19) H ( a, b ; p − − ( − b a + b − H (cid:18) a ; p − (cid:19) H (cid:18) b ; p − (cid:19) (mod p ) . Proof.
The shuffling relation given by H ( − a ; p − yields the second equation. As regardsthe first equation we simply observe that ( − i /i a is positive if and only if i is even. Weuse a similar argument for the congruence: since ( − i + j / ( i a j b ) is positive if and only if i and j are both even or if ( p − i ) and ( p − j ) are both even then H ( − a, − b ; p − ≡ − H ( a, b ; p −
1) + 22 a + b (cid:18) H (cid:18) a, b ; p − (cid:19) + ( − a + b H (cid:18) b, a ; p − (cid:19)(cid:19) . Moreover, by decomposing the sum H ( a, b ; p −
1) we obtain H ( a, b ; p − ≡ H (cid:18) a, b ; p − (cid:19) + H (cid:18) a ; p − (cid:19) ( − b H (cid:18) b ; p − (cid:19) +( − a + b H (cid:18) b, a ; p − (cid:19) . that is H (cid:18) a, b ; p − (cid:19) +( − a + b H (cid:18) b, a ; p − (cid:19) ≡ H ( a, b ; p − − H (cid:18) a ; p − (cid:19) ( − b H (cid:18) b ; p − (cid:19) . and the congruence follows immediately. 3 orollary 2.2. For any prime p > H ( − p − ≡ − q p (2) + pq p (2) − p q p (2) − p B p − (mod p ) ,H ( − , − p − ≡ q p (2) − pq p (2) − p B p − (mod p ) . Moreover for a > and for any prime p > a + 1 H ( − a ; p − ≡ − a − a a − B p − a (mod p ) . Proof.
The proof is straightforward: apply Theorem 2.1, (i), (ii), and (iii).The following theorem is a variation of a result presented in [9].
Theorem 2.3.
Let r > then for any prime p > r + 1 H ( { } r − , − p − ≡ ( − r − p − X k =1 k k r (mod p ) . Proof.
For r ≥
1, let F r ( x ) = X
1, then the formal derivative yields ddx F r ( x ) = X The following congruences mod p hold for any prime p > H (1 , − p − ≡ − H ( − , p − ≡ q p (2) ,H ( − , p − ≡ H (1 , − p − ≡ H (2 , − p − ≡ H ( − , p − ≡ B p − ,H ( − , − p − ≡ − H ( − , − p − ≡ − B p − . Proof. By Theorem 2.3 and by [2] H (1 , − p − ≡ − p − X k =1 k k ≡ q p (2) (mod p ) . By (i) and by the shuffling relation given by the product H ( − p − H (2; p − 1) we get H ( − , p − 1) = 12 H ( − p − H (2; p − − H ( − p − ≡ B p − (mod p ) . By (ii) and by Theorem 2.1 H ( − , − p − ≡ − H (1 , p − − H (cid:18) p − (cid:19) H (cid:18) p − (cid:19) ≡ − B p − (mod p ) . The remaining congruences follow by applying the reversal relation of depth 2. Corollary 2.5. The following congruences mod p hold for any prime p > H ( − , , − p − ≡ ,H (1 , , − p − ≡ H ( − , , p − ≡ − q p (2) − B p − ,H ( − , − , p − ≡ − H (1 , − , − p − ≡ q p (2) + 78 B p − ,H (1 , − , p − ≡ q p (2) + 112 B p − ,H ( − , − , − p − ≡ − q p (2) − B p − . roof. By the reversal relation of depth 3, H ( − , , − p − ≡ − H ( − , , − p − ≡ H (1 , , − p − ≡ p − X k =1 k k ≡ − q p (2) + 712 H ( − , p − ≡ − q p (2) − B p − (mod p ) . By the shuffling relations given by the products H (1 , − p − H ( − p − , H (1 , − p − H (1; p − , and H ( − , − p − H ( − p − H (1 , − , − p − ≡ H (1 , − p − H ( − p − − H (1 , p − − H ( − , − p − ,H (1 , − , p − ≡ − H (1 , , − , p − − H (2 , − p − , H ( − , − , − p − ≡ H ( − , − p − H ( − p − − H (2 , − p − . The remaining congruences follow by applying the reversal relation of depth 3. The following useful identity appears in [7]. Here we give an alternate proof by usingRiordan’s array method (see [5] for more examples of this technique). Theorem 3.1. Let n ≥ d > d n X k =1 (cid:18) kk + d (cid:19) x n − k k = n − d X k =0 (cid:18) nn + d + k (cid:19) v k − (cid:18) nn + d (cid:19) where v = 2 , v = x − and v k +1 = ( x − v k − v k − for k ≥ .Proof. We first note that (cid:18) kk + d (cid:19) = (cid:18) kk − d (cid:19) = ( − k − d (cid:18) − k − d − k − d (cid:19) = [ z k − d ] 1(1 − z ) k + d +1 = [ z − ] z d − (1 − z ) d +1 · (cid:18) z (1 − z ) (cid:19) k . Since the residue of a derivative is zero then d n X k =1 (cid:18) kk + d (cid:19) x n − k k = [ z − ] x n dz d − (1 − z ) d +1 G (cid:18) xz (1 − z ) (cid:19) = − [ z − ] x n z d (1 − z ) d G ′ (cid:18) xz (1 − z ) (cid:19) · (cid:18) xz (1 − z ) (cid:19) ′ = [ z − ] z d − n − (1 − z ) n + d +1 − x n z n (1 − z ) n − xz + xz · (1 − z )= [ z − ] z d − n − (1 − z ) n + d +1 − z − xz + xz . G ( z ) = P nk =1 z k k and G ′ ( z ) = P nk =1 z k − = − z n − z . Moreover (cid:18) nn + d + k (cid:19) = (cid:18) nn − d − k (cid:19) = ( − n − d − k (cid:18) − n − d − k − n − d − k (cid:19) = [ z n − d − k ] 1(1 − z ) n + d + k +1 = [ z − ] z d − n − (1 − z ) n + d +1 · (cid:18) z − z (cid:19) k Letting F ( z ) = P ∞ k =0 v k z k = − ( x − z − ( x − z + z then n − d X k =0 (cid:18) nn + d + k (cid:19) v k − (cid:18) nn + d (cid:19) = [ z − ] z d − n − (1 − z ) n + d +1 · F (cid:18) z − z (cid:19) − [ z − ] z d − n − (1 − z ) n + d +1 = [ z − ] z d − n − (1 − z ) n + d +1 (cid:18) (2 − xz )(1 − z )1 − xz + xz − (cid:19) = [ z − ] z d − n − (1 − z ) n + d +1 − z − xz + xz . Corollary 3.2. For any n > n n X k =1 (cid:18) − k (cid:19) ( − k k = − − n n − X d =0 ( − d n − d d − X j =0 (cid:18) nj (cid:19) − − n n − X d =0 ( − d n − d (cid:18) nd (cid:19) . Proof. Since 0 = k X d = − k ( − d (cid:18) kk + d (cid:19) = (cid:18) kk (cid:19) + 2 k X d =1 ( − d (cid:18) kk + d (cid:19) then for any n ≥ k ( − k (cid:18) − k (cid:19) = 4 − k (cid:18) kk (cid:19) = − · − k n X d =1 ( − d (cid:18) kk + d (cid:19) . For x = 4 then v k = 2 for all k ≥ n n X k =1 ( − k k (cid:18) − k (cid:19) = − n X k =1 n − k k n X d =1 ( − d (cid:18) kk + d (cid:19) = − n X d =1 ( − d n X k =1 n − k k (cid:18) kk + d (cid:19) = − n X d =1 ( − d d n − d X k =0 (cid:18) nn + d + k (cid:19) + 2 n X d =1 ( − d d (cid:18) nn + d (cid:19) = − n X d =1 ( − d d n − d X k =1 (cid:18) nn − d − k (cid:19) − n X d =1 ( − d d (cid:18) nn − d (cid:19) = − − n n − X d =0 ( − d n − d d − X j =0 (cid:18) nj (cid:19) − − n n − X d =0 ( − d n − d (cid:18) nd (cid:19) . We will make use of the following lemma. 7 emma 3.3. For any prime p = 2 and for < j < p (cid:18) pj (cid:19) ≡ − p ( − j j + 4 p ( − j j H (1; j − 1) (mod p ) and (cid:18) pp (cid:19) ≡ − p B p − (mod p ) . Proof. It suffices to expand the binomial coefficient in this way (cid:18) pj (cid:19) = − p ( − j j j − Y k =1 (cid:18) − pk (cid:19) = ( − j j j − X k =1 ( − p ) k H ( { } k − ; j − . and apply (i). Proof of Theorem 1.1. Letting n = p in the identity given by Corollary 3.2 we obtain4 p p X k =1 ( − k k (cid:18) − k (cid:19) = 4 X ≤ j 1) = − q p (2) + pq p (2) − p q p (2) − p B p − (mod p ) . Since for 0 < d < p p − d = − d (1 − pd ) ≡ − d − pd (mod p )then by Lemma 3.3, (i), and (iii) we have that X 1) + 2 p H (3; p − − p H (1 , p − ≡ − p B p − . (mod p ) . In a similar way, by Lemma 3.3 and Corollaries 2.4 and 2.5 we get X 1) + 2 p H ( − , − p − − p H (1 , − , − p − ≡ pq p (2) + 43 p B p − (mod p ) . p − p − X k =1 ( − k k (cid:18) − k (cid:19) = 2 q p (2) + 3 pq p (2) + 23 p q p (2) + 712 p B p − (mod p ) . Since 4 p − = ( q p (2) p + 1) = 1 + 2 q p (2) p + q p (2) p then4 − ( p − = (1 + 2 q p (2) p + q p (2) p ) − ≡ − q p (2) p + 3 q p (2) p (mod p ) . 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