#A45 INTEGERS 12 (2012) SUPERCONGRUENCES FOR A TRUNCATED HYPERGEOMETRIC SERIES

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1 #A45 INTEGERS 2 (202) SUPERCONGRUENCES FOR A TRUNCATED HYPERGEOMETRIC SERIES Roberto Tauraso Diartimento di Matematica, Università di Roma Tor Vergata, Italy tauraso@mat.uniroma2.it Received: /7/, Acceted: 6/30/2, Published: 8/8/2 Abstract The urose of this note is to obtain some congruences modulo a ower of a rime involving the truncated hyergeometric series (x) ( x) () 2 a for a and a 2. In the last section, the secial case x /2 is considered.. Introduction In [6], E. Mortenson develoed a general framewor for studying congruences modulo 2 for the truncated hyergeometric series 2F x, x ; tr() (x) ( x) () 2. Here we would lie to resent our investigations about a similar class of finite sums, namely and x( x) 4 F 3 2 x, + x,, 2, 2, 2 x( x) 5 F 4 2 x, + x,,, 2, 2, 2, 2 (x) ( x) ; tr( ) () 2, (x) ( x) ; tr( ) () Mathematics Subject Classification: 33C20, B65, (Primary) 05A0, 05A9 (Secondary)

2 INTEGERS: 2 (202) 2 In order to state our main result we need to introduce the definition of what we call Pochhammer quotient Q (x) (x) ( x) m 2 () 2 [r/] m [ r/] m where is an odd rime, x r/m, 0 < r < m are integers with m rime to, (x) n x(x + ) (x + n ) denotes the Pochhammer symbol, and [a] m is the unique reresentative of a modulo m in {0,,..., m }. Note that the Pochhammer quotient is really a -integral because by the artial-fraction decomosition (x + )() () lim x (x) ( ) (x) x + x + x + m [r/] m (mod ). It is easy to verify that for some articular values of x the quotient Q (x) is connected with the usual Fermat quotient q (x) (x )/ : Q ( 2 ) q ( [/] 2 [ /] 2 6 ) (mod 2 ), Q ( 3 ) Q ( 4 ) Q ( 6 ) q ( [/] 3 [ /] 3 27 ) (mod 2 ), 2 6 q ( [/] 4 [ /] 4 64 ) (mod 2 ), q (mod 2 ). 2 2 [/] 6 [ /] (6 27) where we used one of the equivalent statement of Wolstenholme s theorem, that is a b a b modulo 3 for any rime > 3. Our main goal in this aer is to show the following: if > 3 is a rime then (x) ( x) () 2 (x) ( x) () 2 Q (x) + 2 Q (x) 2 (mod 2 ), 2 2 Q (x) 2 (mod ). The next section contains some reliminary results about certain hyergeometric identities and congruences. We state and rove our main result in the third section.

3 INTEGERS: 2 (202) 3 We conclude the aer by considering the secial case x /2 and by roving that (/2) 2 () 2 2H ( )/2() (mod 3 ) where H n (r) n r is the finite harmonic sum of order n and weight r. 2. Preliminaries The hyergeometric identity resented in the next theorem is nown (see, for examle, equation (2), Ch. 5.2 in [5]). Here we give a simle roof by using the Wilf-Zeilberger (WZ) method. Theorem. For 0 < x < and for any ositive integer n, n n (x) ( x) () 2 Proof. For 0..., n, let n (x) n( x) n () 2 n F (n, ) Then WZ method yields the certificate n x + + () 2 n (x) ( x) (x) n ( x) n () 2 n. 2 (n ) R(n, ) (n + )(x + n)( x + n) with G(n, ) R(n, )F (n, ), such that It is easy to verify by induction that F (n +, ) F (n, ) G(n, + ) G(n, ). n n F (n, ) + G(j, 0) n (F ( +, ) + G(, )). Thus the desired identity follows by noting that G(j, 0) 0 and F (+, )+G(, ). () x + ( + ) 2 (x + )( x + ) 2 (x + )( x + ) x + + x +.

4 INTEGERS: 2 (202) 4 Let B n (x) be the Bernoulli olynomial in x of degree n 0, given by B n (x) n n B x n where B are rational numbers called Bernoulli numbers which are defined recursively as follows: B 0, and n n B 0 for n 2. For the main roerties of B n (x) we will refer to Chater 5 in [3] and to the nice introductory article []. Lemma 2. If > 3 is a rime and x r/m, where 0 < r < m are integers with m rime to, then for any ositive integer a a x + + m x + a [r/] m + jm + [ r/] m + jm 2 3 (a)2 B 3 (x) (mod 3 ), (2) (x) a ( x) a () 2 a (x) (a )( x) (a ) () 2 (a ) a 2 + (x) ( x) () 2 a(a )m2 [r/] m [ r/] m (mod 3 ). (3) Proof. We first show (2). By well-nown roerties of the Bernoulli olynomials a (x + ) ϕ(3 ) B ϕ( 3 )(a + x) B ϕ(3 )(x) ϕ( 3 ) ϕ( 3 ) ϕ( 3 ) ϕ( 3 ) B ϕ(3 ) (x)(a) where ϕ() is the Euler s totient function. Since m n B n (x) B n Z (see [] for a short roof), it follows from the Clausen-von Staudt congruence (. 233 in [3]) that B n (x) B n m n if ( ) divides n and n > 0 otherwise (mod ). Moreover, by Kummer s congruences (. 239 in [3]), if q 0 and 2 r then B q( )+r (x) q( ) + r B r(x) r (mod ).

5 INTEGERS: 2 (202) 5 Because divides the numerator of the fraction x + for 0,..., if and only if x + [r/] m /m, we have that a x + m a a [r/] m + jm (x + ) ϕ(3 ) a B ϕ(3 ) (x) 2 (a)2 B ϕ(3 ) 2(x) a B 2 ( ) (x) 3 (a)2 B 3 (x) (mod 3 ). Finally, since by the symmetry relation B n (x) ( ) n B n ( x), we have that (2) holds. As regards (3), let 0 [r/] m /m x, then (x) a ( x) a (x) (a ) ( x) (a ) (x) ( x) (a ) + x + a(a ) 2 (x + )( (x + )) a(a ) 2 (x + 0 )( (x + 0 )) a(a )m2 [r/] m [ r/] m (mod 3 ). (a ) (x + ) Moreover, by an equivalent statement of Wolstenholme s theorem (see for examle [4]), () a a a (mod 3 ) () (a ) () and the roof of (3) is comlete. The next lemma rovides a owerful tool which will be used in the next section in the roof of the main theorem. Lemma 3. If > 3 is a rime and x r/m, where 0 < r < m are integers with m rime to, then a+ a+ (x) ( x) () 2 (x) a( x) a (x) ( x) () 2 a () 2 a + (mod 2 ). (4) Proof. Since (x) a+ (x + j + a) (x) + a (x) a (mod 2 ) x + j

6 INTEGERS: 2 (202) 6 we have (x) a+ ( x) a+ () 2 a (x) a ( x) a () 2 a+ (x) ( x) () 2 + a x + j + x + j 2 (mod 2 ). + j Hence it suffices to rove that a that is, (x) ( x) () 2 a + x + j + x + j 2 0 (mod 2 ), + j (x) ( x) () 2 By (), the left-hand side becomes (x) j ( x) j () 2 j 2 (x) j ( x) j () 2 j 2 H (2) + j 2 H (2) + 2 j j (x) j ( x) j () 2 j x + j + x + j j j+ (x) j ( x) j () 2 j (x) j ( x) j () 2 j 2H j() j ( j) j j+ (x) ( x) () 2 2H () (mod ). j j (H j() H () + H j ()) j (x) j ( x) j () 2 j 2H j() j (mod ), because H j () H j () (mod ) and H () H (2) 0 (mod ). The roof is now comlete.

7 INTEGERS: 2 (202) 7 3. The Main Theorem Theorem 4. If > 3 is a rime and x r/m, where 0 < r < m are integers with m rime to, then (x) ( x) () 2 (x) ( x) () 2 Q (x) + 2 Q (x) 2 (mod 2 ), (5) 2 2 Q (x) 2 (mod ). (6) Proof. Let S a (b) a+ a+ By () with n and by (2), we obtain where S 0 () S 0 (2) + (x) ( x) () 2 βtm (x) ( x) () 2 β (x) ( x) () 2, and t + [r/] m Moreover by (3) b. x + + x + (mod 2 ). (7) [ r/] m m [r/] m [ r/] m. (x) 2 ( x) 2 () 2 2 β2 4 ( + 2mt) (mod 3 ). and [r/] m + m + [ r/] m + m 3t + 2mt. Hence, by () with n 2 and by (2), we get S 0 () 2 S 0 (2) S () 2 S (2) β 2 + (x) 2( x) 2 () 2 2 β 2 + β2 4 ( + 2mt) m 2 t + 2 x + + 3t + 2mt x + (mod 2 ) (8)

8 INTEGERS: 2 (202) 8 According to Lemma 3, S () β(s 0 () S 0 (2)) (mod 2 ), S (2) β S 0 (2) (mod ), and (7) and (8) yield the linear system S 0 () S 0 (2) βtm (mod 2 ), S 0 () 2 S 0 (2) β S 0 () β S 0 (2) β 2 + β2 4 ( + 2mt)m t + 3t + 2mt (mod 2 ). By substituting the first congruence in the second one, we obtain S 0 (2) + βtm + β βtm β 2 + β2 4 ( + 2mt)m 3t t + + 2mt (mod 2 ), which yields Finally, S 0 (2) 2 βtm 2 2 Q (x) 2 (mod ). S 0 () Q (x) + 2 Q (x) 2 (mod 2 ). We conclude this section by osing a conjecture which extends (6). Conjecture 5. If > 3 is a rime and x r/m, where 0 < r < m are integers with m rime to, then (x) ( x) () Q (x) 2 2 Q (x) 3 (mod 2 ). More conjectures of the same flavor can be found in Section A30 of [0]. 4. The Secial Case x 2 As we already noted, if x 2 then (x) ( x) () 2 (/2)2 ()

9 INTEGERS: 2 (202) 9 Let be an odd rime. Then for 0 n ( )/2 we have that n + j (2 (2j ) 2 ) j (2j )2 2 ( ) 2 4 (2)! ( 4) (2)! 6 (mod 2 ), which means that n n + 2 ( ) ( ) n (mod 2 ). Since divides 2 for n < <, it follows that for any -adic integers a0, a,..., a we have 2 n 2 a 6 n n + ( ) a (mod 2 ). This remar is interesting because the sum on the right-hand side could be easier to study modulo 2. With this urose in mind, we consider Identity 2. in [7]: n ( ) n n + ( z)n. (9) z + z ( + z) n As a first examle, we can give a short roof of (.) in [6]. By multilying (9) by z, and by letting z, we obtain that n n n + ( ) ( ) n which imlies that for any rime > ( ) 2 6 (mod 2 ). For more alications of the above remar see [9], [2] and [4]. Let s (s, s 2,..., s d ) be a vector whose entries are ositive integers then we define the multile harmonic sum for n 0 as H n (s) < 2<< d n s s2 2. s d We call l(s) d and s d i s i its deth and its weight resectively. Theorem 6. For n, r n ( ) n n + r 2 l(s) H n (s). (0) s r d

10 INTEGERS: 2 (202) 0 Proof. Note that (+z) n n! + d times n H n (,,..., )z d, d and n d dz (+z) n (+z) n z +. Then the left-hand side of (0) can be obtained by differentiating (r ) times the identity (9) with resect to z and then by letting z 0: n ( ) n n + r ( )r d r ( z)n (r )! dz r z ( + z) n ( )r d r ( z)n r! dz r ( + z) n. z0 Taing the derivatives we get a formula which involves roducts of multile harmonic sums. This formula can be simlified to the right-hand side of (0) by using the so-called stuffle roduct (see for examle [3]). In the next theorem, we rove two generalizations of (5) and (6) for x /2. Note that the first of these congruences can be considered as a variation of another congruence roved by the author in [2]: for any rime > 3 Theorem 7. For any rime > 3 (/2) () H ( )/2() (mod 3 ). z0 (/2) 2 () 2 (/2) 2 () 2 2H ( )/2() (mod 3 ), 2 2H ( )/2() 2 (mod 2 ). Proof. We will use the same notations as in Theorem 4. For x /2 we have that m t 2. By () with n and by (2), we obtain S 0 () S 0 (2) 2 S 0 (3) + β βtm x + + x + 2β 3 2 B 3 (/2) (mod 3 ). ()

11 INTEGERS: 2 (202) On the other hand, by [4] β B 3 4( + q (2)) B 3 (mod 4 ), and by Raabe s multilication formula B 3 (/2) 2 4 B 3 7 B 3 (mod ). Moreover, by letting n ( )/2 in (0) for r 2, 3, we have that S 0 (2) S 0 (3) n ( ) n n + 2 2H n (2) 4H n (, ) 2 H n () 2 (mod 2 ), n ( ) n n + 3 2H n (3) 4H n (2, ) 4H n (, 2) 8H n (,, ) 4 3 H n() H n(3) (mod 2 ). Finally by [8] H n () 2q (2) + q (2) q (2) B 3 (mod 3 ), H n (3) 2B 3 (mod ). By lugging all of these values in (), after a little maniulation, we easily verify the desired congruence for S 0 (). References [] T. M. Aostol, A rimer on Bernoulli numbers and olynomials, Math. Mag. 8 (2008), [2] H. H. Chan, L. Long and W. Zudilin, A suercongruence motivated by the Legendre family of ellitic curves, Math. Notes 88 (200), [3] K. Ireland and M. Rosen, A Classical Introduction to Modern Number Theory, Sringer, New Yor, 990. [4] L. Long, Hyergeometric evaluation identities and suercongruences, Pacific J. Math. 249 (20), [5] Y. L. Lue, Mathematical Functions and Their Aroximations, Academic Press, New Yor, 975.

12 INTEGERS: 2 (202) 2 [6] E. Mortenson, Suercongruences between truncated 2 F hyergeometric functions and their Gaussian analogs, Trans. Amer. Math. Soc. 355 (2003), [7] H. Prodinger, Human roofs of identities by Osburn and Schneider, Integers 8 (2008), A0, 8. [8] Z. H. Sun, Congruences concerning Bernoulli numbers and Bernoulli olynomials, Discrete Al. Math. 05 (2000), [9] Z. H. Sun, Congruences concerning Legendre olynomials, Proc. Amer. Math. Soc. 39 (20), [0] Z. W. Sun, Oen conjectures on congruences, rerint arxiv:math.nt/ v59 (20), 88. [] B. Sury, The value of Bernoulli olynomials at rational numbers, Bull. London Math. Soc. 25 (993), [2] R. Tauraso, Congruences involving alternating multile harmonic sum, Electron. J. Combin. 7 (200), R6,. [3] R. Tauraso, New harmonic number identities with alications, Sém. Lothar. Combin., 63 (200), B63g, 8. [4] J. Zhao, Bernoulli numbers, Wolstenholme s theorem, and 5 variations of Lucas theorem, J. Number Theory 23 (2007), 8 26.