Almost all hyperharmonic numbers are not integers

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1 Joural of Number Theory 7 (207) Cotets lists available at ScieceDirect Joural of Number Theory Almost all hyerharmoic umbers are ot itegers Haydar Göral a, Doğa Ca Sertbaş b, a Deartmet of Mathematics, Koç Uiversity, Rumelifeeri Yolu, 34450, Sarıyer, Istabul, Turkey b Deartmet of Mathematics, Faculty of Scieces, Cumhuriyet Uiversity, 5840, Sivas, Turkey a r t i c l e i f o a b s t r a c t Article history: Received 25 Aril 206 Acceted 3 July 206 Available olie 6 Setember 206 Commuicated by D. Goss MSC: B83 5A0 B75 Keywords: Hyerharmoic umbers Harmoic umbers Prime umber theory It is a oe questio asked by Mezö that there is o hyerharmoic iteger excet. So far it has bee roved that all hyerharmoic umbers are ot itegers u to order r = 25. I this aer, we exted the curret results for large orders. Our method will be based o three differet aroaches, amely aalytic, combiatorial ad algebraic. From aalytic oit of view, by exloitig rimes i short itervals we rove that almost all hyerharmoic umbers are ot itegers. The usig combiatorial techiques, we show that if is eve or a rime ower, or r is odd the the corresodig hyerharmoic umber is ot iteger. Fially as algebraic methods, we relate the itegeress roerty of hyerharmoic umbers with solutios of some olyomials i fiite fields. 206 Elsevier Ic. All rights reserved.. Itroductio The goal of this aer is to aalyse the itegeress roerty of hyerharmoic umbers. We aswer Mezö s questio [8] i almost all cases, which states that all hy- * Corresodig author. addresses: hgoral@gmail.com (H. Göral), dogaca.sertbas@gmail.com (D.C. Sertbaş). htt://dx.doi.org/0.06/j.jt X/ 206 Elsevier Ic. All rights reserved.

2 496 H. Göral, D.C. Sertbaş / Joural of Number Theory 7 (207) erharmoic umbers are ot itegers excet. I the same aer, he also roved that there are oe of them of order r =2, 3. I [5,6], this was exteded to r 25 ad they gave a set of iteger airs (, r) such that h (r) is ot a iteger. I the curret aer, we exted all these results i the literature alyig aalytic, combiatorial ad algebraic tools. Harmoic umbers are defied as the sequece of artial sums of the harmoic series h = k= for. These umbers have bee studied extesively ad they are equied with a lot of combiatorial ad aalytic roerties. As a combiatorial oe, it was roved that there is o harmoic umber which is a iteger excet [2]. This ca also be see by Bertrad s ostulate, which was reroved by Erdös [4] from combiatorial oit of view. Although almost all harmoic umbers are ot itegers, Wolsteholme [22] roved that h 0 mod 2 for all rimes 5. More recetly this result was exteded by Alka i []. O the aalytical side, Euler s harmoic zeta fuctio which is defied by ζ h (s) = for R(s) > satisfies the well-kow strikig relatio (see [0,. 252]) = k h s m 2 2ζ h (m) =(m +2)ζ(m +) ζ(m k)ζ(k + ) (.) for all m 2, where the sum is zero whe m = 2ad ζ(s) is the Riema zeta fuctio defied by ζ(s) = = k= for R(s) >. Euler s formula yields some imortat evaluatios s = h 2 =2ζ(3), = h 3 = 5 π4 ζ(4) = Recetly, by makig use of coectios to log-sie itegrals which have remarkable alicatios to hysical roblems via otetial eergy of charged article systems o the uit circle, Alka (see [2] ad [3]) showed that real umbers ca be strogly aroximated by combiatios of values of ζ h (s) ad ζ(s), which resembles the classical result of Liouville s theorem o Diohatie aroximatio.

3 H. Göral, D.C. Sertbaş / Joural of Number Theory 7 (207) Hyerharmoic umbers were first itroduced by Coway ad Guy []. They are ideed a geeralizatio of harmoic umbers. More recisely, the -th hyerharmoic umber of order r is defied recursively by h (r) := k= h (r ) k, where h () = h. Hyerharmoic umbers also ejoy a lot of combiatorial roerties. For istace a combiatorial aroach to h (r) was give i [9] where they gave ew exressios for h (r). A coectio betwee hyerharmoic umbers ad r-stirlig umbers, ad ew idetities were obtaied i [9]. It is kow by [] that h (r) ca be exressed i terms of biomial coefficiets ad harmoic umbers with the formula h (r) = ( + ) (h +r h r ). (.2) May roofs related to the distributio of rimes are based o the coectio betwee combiatorial ad aalytic methods at the same time, such as sieve methods. I this aer, we also aly this iteractio ad we give a huge class where the corresodig hyerharmoic umber is ot a iteger. Our first result is based o distributio of rimes ad it is a desity estimate for o-iteger hyerharmoic umbers. To do so, we tur our attetio to the distributio of rime umbers. Let π(x) be the umber of rime umbers that are less tha or equal to x. The rime umber theorem, which was first roved i 896 ideedetly by Hadamard ad de la Valée Poussi, states that π(x) x log x. Usig rime umber theorem, oe ca rove that for ay ositive ε > 0 there exists a real umber x 0 (ε) deedig o ε such that if x x 0 the the iterval [( ε)x, x] cotais a rime umber. The geeral questio about rimes i short itervals is the followig: for which fuctio Φ(x) = o (x) the iterval [x Φ(x), x] cotais a rime umber for large x deedig o the fuctio Φ? The curret ucoditioal result is Φ(x) = x which was roved i [8]. Note that the Riema hyothesis is equivalet to where π(x) =Li(x)+O ε ( x 2 +ε) Li(x) = x 2 log t

4 498 H. Göral, D.C. Sertbaş / Joural of Number Theory 7 (207) is the logarithmic itegral. Coditioally, assumig the Riema hyothesis oe ca choose Φ(x) of size x 2 +ε for ay ositive ε. There is a far-reachig cojecture by Cramér, which is called the Cramér s model ad it claims that Φ(x) = c log 2 x, for some c > 0. For more o Cramér s model ad its aalog i algebraic umber theory, we direct the reader to [4]. Now we will state our first theorem ad it states that almost all hyerharmoic umbers are ot itegers. More recisely, we will give a quatitative estimate for the umber of airs (, r) lyig i a rectagle where the corresodig hyerharmoic umber h (r) is ot a iteger. To achieve this, we eed to wait for large values of i order to catch rimes i short itervals. For the followig theorem, see Proositio 9, Theorem 0 ad Theorem : Theorem. { () Let S(x) = (, r) [0,x] [0,x] h (r). / Z} I other words, S(x) couts the umber of airs (, r) i the rectagle [0, x] [0, x] where the corresodig ( hyerharmoic umber h (r) is ot a iteger. The we have S(x) = x 2 + O x , ) which meas that o-iteger hyerharmoics have the full asymtotic i the first quadrule. Coditioally, assumig the ( Riema ) hyothesis ( or Cramér s ) cojecture, the error term ca be take of size O ε x 5 3 +ε ad O x 3 2 log x resectively. (2) For a give atural umber d, there exists a atural umber d such that if d ad r d the the corresodig hyerharmoic umber h (r) is ot a iteger. Moreover there exists a atural umber 0 such that if 0 ad r the the corresodig hyerharmoic umber h (r) is ot a iteger. I articular for a fixed r, there are oly fiitely may ossible values of where h (r) may be a iteger. (3) For a fixed >, there are ifiitely may values of r such that h (r) / N. Our ext theorem is of combiatorial flavour ad it is based o the -ary reresetatios of ad r for a certai rime. Also the roof uses the coectio betwee the -adic valuatio of biomial coefficiets ad coutig carries. Let I be a set ad be a rime. We defie μ (I) := max {ν (s) s I } as the maximal ower of i I. We also ut I(, r) := {r,..., +r }. For the followig result, see Theorem 8 ad Theorem 2: Theorem 2. If oe of the followig coditios () (3) o ad r is satisfied, the the corresodig hyerharmoic umber h (r) is ot a iteger: () is eve. (2) r is odd. (3) is a rime ower.

5 H. Göral, D.C. Sertbaş / Joural of Number Theory 7 (207) Now let a be oegative ad m, k, b be ositive itegers. Let be a rime umber such that a + b. Put = m + a =( α,..., 0 ) r = k + b =(r β,...,r 0 ), where ( α,..., 0 ) ad (r β,..., r 0 ) are the -ary reresetatios of ad r, resectively. If there exist c N ad κ > αsuch that r (c κ, c κ ], the the corresodig hyerharmoic umber h (r) is ot a iteger. Otherwise the o-itegeress roerty still holds, if is sufficietly large deedig o m ad k. Our third theorem is o the algebraic side ad it coects the o-itegeress roerty of hyerharmoic umbers with solutios of certai olyomials i fiite fields (see Theorem 23). Theorem 3. Let = k α be a odd iteger where is a rime, α ad r is give. Defie a = k 2, c = r ad α F k (x) := a i= a a (x j) x i. j= a Assume that ν (F k (c + a)) ν (k!). The the corresodig hyerharmoic umber h (r) is ot a iteger. I articular if c + a is ot a root of F k (x) i modulo, the h (r) / N. Our fial theorem is a comutatioal asect of our methods ad it is a alicatio of the revious three theorems (see Corollary 24 ad Corollary 26): Theorem 4. If oe of the followig coditios o ad r is satisfied, the the corresodig hyerharmoic umber h (r) is ot a iteger: () r 6. (2) < 32. (3) r (4) = 3 α where α, is a rime ad ±5 mod 2. (5) = 5 α where α, is a rime ad 7,, 3, 4, 9, 2, 22, 23, 26, 28, 3, 33, 38, 39, 4, 42, 44, 46, 52, 53, 56, mod , 6, 62, 63, 66, 67, 69, 76, 78, 79, 82, 83, 84, 88, 89, 92, 93, 99, 0, 03, 04, 06, 07, 2, 4, 7, 9, 22, 23, 24, 26, 3, 32, 34, 38

6 500 H. Göral, D.C. Sertbaş / Joural of Number Theory 7 (207) Throughout the aer, the umbers, r will be always atural umbers. The set P deotes the set of rime umbers. We deote as a rime i the etire aer. Also we use the -adic valuatio of a atural umber i the usual way, i.e. for a give a Z we deote ν (a) := { m if m a if a =0 as the -adic valuatio (or the -adic order) of a. Here m a meas m a but m+ a. We exted this otatio to a ratioal umber q = a/b Q by settig ν (q) = ν (a) ν (b) where a, b Z. Now we will make a crucial observatio which we will use subsequetly i the whole aer for the o-itegeress roerty. I order to show the o-itegeress roerty of hyerharmoic umbers, it is eough to check the followig: from (.2), we have where h (r) Per( +,r ) =! = P! = +r i=r +r i=r! P P i P i P=Per( +,r ) = ( r + ) r , (.3) ( + )! ()! = r (r +) ( + ) ad P i = P i for i {r,...+ }. This imlies that h (r) exists a rime such that ν (h (r) ) < 0. / N if ad oly if there 2. Aalytic methods I this sectio we will rove Theorem. For this we eed the followig corollary of the rime umber theorem. For the details of the rime umber theory, we refer the reader to [7] or [2]. Fact 5. For all ɛ R >0, there exists x 0 = x 0 (ɛ) R deedig o ɛ such that for all x x 0, there exists a rime i the iterval (( ɛ)x, x]. Recall that I(, r) = {r,..., +r }. We defie I (, r) as the set of iteger multiles of which lie i I(, r) = {r,..., + }. The followig roositio is the first ste toward Theorem.

7 H. Göral, D.C. Sertbaş / Joural of Number Theory 7 (207) Proositio 6. Let be a rime umber such that. If I (, r) =, the h (r) / N. Moreover for a fixed iteger r, there exists 0 = 0 (r) N deedig o r such that for all 0, we have h (r) / N. Proof. First observe that whe = 2the hyerharmoic umber h (r) = 2r+ 2 is ot a iteger. So we may assume that 3. Also we may suose that r 2as harmoic umbers are ot itegers excet. By (.3), observe that h (r) / N if ν (!) +r i=r P i. (2.) Note that P i if i, ad P i if i for i I(, r). Sice I (, r) =, there is oly oe such i I(, r) that is divisible by ad hece +r i=r P i. This imlies that h (r) / N by (2.). Now we rove the secod art of the roositio. Observe that if there exists a rime ( +r 2, ] for r 2, the we have + < 2 ad so <. These iequalities yield that that I (, r) =. By the first art of the roositio, we deduce that h (r) / N. Therefore it is eough to show that there exists a rime ( +r 2, ]. By Fact 5, we kow that there exists 0 such that for all 0, there is a rime i the iterval ( 2 3, ]; i fact we ca take that 0 =2, sice N. Note that whe 3r 3we have ( ] ( ] 2 + 3,,. 2 Thus we see that give r 2if 3r 3 2the there is always a rime i the iterval ( +r 2, ] with > 2 3 2r 2 r. Hece we coclude that if r 2ad 3r 3the the corresodig hyerharmoic umber h (r) is ot a iteger. Examle 7. For istace, oe ca choose =2 ad r = k + for some P ad k N. I that case I(, r) = {k +,...,(k +2) }, ad observe that the oly multile of i I(, r)is (k+). Sice P ( 2, 2 ], we coclude that h (r) / N. Util ow, we have oly dealt with the case where P ad r < + r 2 as i the revious roositio. However oe ca otice that it is eough to have c I(, r), whereas (c ), (c + ) / I(, r) for some c N. Therefore oe ca deduce the followig corollary, as a cosequece of Proositio 6: Corollary 8. For a give, r N, suose that there exist itegers c, d ad, q P such that either + r (c ) r<c, << or, (2.2) c + d 2 <r d, + r d +2 <q< r holds. (2.3) d The i ay two cases, we have h (r) / N.

8 502 H. Göral, D.C. Sertbaş / Joural of Number Theory 7 (207) Proof. For the first art of the theorem, observe that 2 +r c+ ad r c < +r c+ from (2.2). This imlies that P ( 2,) ad r<c. The iequality (c ) < (c ) r ca also be show by usig (2.2). Note that c < + r, because if we have c + r the c > + r i.e. (c ) > (c ) > r, which is a cotradictio. Therefore (c ) <r<c < + r<(c + ) holds. So there exists a rime umber such that the coditios ( 2,] ad I (, r) = are both satisfied. By Proositio 6, we deduce that h (r) / N. For the other case, similarly we have 2 < +r d+2 ad r d. Therefore the rime q lies i ( 2,). Moreover sice dq < r, + r<(d +2)q ad q <, we obtai that (d + )q < q + r < + r ad (d + )q > + r q > r. Agai by Proositio 6, we have h (r) / N. As a examle of Corollary 8, take r =3. I that case we ca take c = 4or d = 3, ad the we try to fid the lowest 0 value such that there exists a rime betwee 4 5 ad, for all 0. We ca also use the iequalities give i (2.3) with aother value of d, for istace d = 5; ad try to fid a rime i the iterval ( 4 7, ) 3 5. However as oe ca see that this iterval is much restricted tha the revious oe. So it is better to use smaller values of c ad d to obtai more geeral results. The followig result is the third art of our Theorem. Proositio 9. For ay fixed 2, there are ifiitely may values of r such that / N. More recisely, deote h (r) { R (m) := r m h (r) / N}. If 2 is a fixed iteger, the R (m) m as m teds to ifiity. Moreover if m = m() is a fuctio of where = o(m) the R (m) m, as teds to ifiity. Proof. Let 2be a fixed iteger ad () P be the greatest rime that is less tha or equal to, so it is also fixed. By the well-kow Bertrad s Postulate, we see that () ( 2,]. Also by Proositio 6, we kow that if I ()(, r) =the h (r) / N. Defie r c =(c +) (). Our claim is that if r ( ] c= (c ) () (r),r c the h / N. We ut I c = ( ] (c ) (),r c. So suose that r N is give ad there exists c N such that r I c = ( (c ) (),r c ]. I that case, (c ) () <r r c =(c +) () = c () ( ) c () holds. Also we have + (c ) () < + r (c + ) (). By usig both of these iequalities, we obtai the coditio that I ()(, r) =as oly c () I(, r) ad thus h (r) / N.

9 H. Göral, D.C. Sertbaş / Joural of Number Theory 7 (207) Note that N ((c ) (),r c ] =(c +) () (c ) () =2 (), due to the Bertrad s Postulate. Now let x be a ositive real umber. Observe that if c ad c are distict the I c I c is emty. Therefore we have N ] ((c ) (),r c c x = ] N ((c ) (),r c x. c x c x As a cosequece, we deduce that there are ifiitely may values of r which satisfies the coditio h (r) / N, for a fixed N 2. Now let 2be fixed. Also take m N. Defie s = max{c r c m}. We kow by the first art of the roositio that if r N ( ] c= (c ) () (r),r c the h / N. Also by the defiitio of R (m), we ca say that { R (m) = r m h (r) / N} s = c= ( ( ]) N (c ) (),r c s (r c (c ) () ). c= From the defiitio of r c, we obtai that R (m) = s ((c +) () (c ) () ) c= s (2 () ) =s (2 () ). (2.4) c= Furthermore for all c, we have r c =(c +) () (c +) = c. Hece Therefore we get that r s m<r s+ (s +). s> m. (2.5) Combiig this fact with (2.4) ad Bertrad s ostulate, we obtai that R (m) > m (2.6) which leads to the desired iequality. For the last art of the roositio, we will aly [8]. Let be a sufficietly large fixed iteger so that P ( 0.525, ] is ot emty. Sice

10 504 H. Göral, D.C. Sertbaş / Joural of Number Theory 7 (207) () ( 0.525, ], we have 2 () > Now suose that m = m() is a fuctio of where = o(m). Thus (2.4) ad (2.5) yield that R (m) > ( m ) ( ( )=(m ) 2 ) (2.7) From (2.7), we see that R (m) = m + O ( ) + m ad this gives the asymtotic R (m) m as goes to ifiity. Although the first result give i Proositio 9 ca be obtaied by the first ad the secod art of Theorem 2, we eed to use the ideas give i the roof of it; so that we ca deduce a more geeral result. I fact, we ca exted the result give i Corollary 8 by usig Proositio 9 ad this is the secod art of our Theorem. Theorem 0. For a give atural umber d, there exists a atural umber d such that if d ad r d the the corresodig hyerharmoic umber h (r) is ot a iteger. Moreover there exists a atural umber 0 such that if 0 ad r the the corresodig hyerharmoic umber h (r) is ot a iteger. I articular, for a fixed r, there are oly fiitely may ossible values of where h (r) may be a iteger. Proof. By the Prime Number Theorem we kow that for a give d (( there exists ) a real ] umber x d such that for all x x d there is a rime i the iterval 2d+7 x, x. Now we rove the followig claim by iductio: Claim. For a give d ad for all 9 7 x d, if r d the the corresodig hyerharmoic umber h (r) is ot a iteger. So first we deal with the case d =. Let x =53ad () be the greatest rime that is less tha or equal to. Assume that x. The we have that () ( 8 9,]. Observe by Proositio 6 ad Proositio 9 that if r Z \(2 (), () ]the ν ()(h (r) ) =. So it is eough to check the o-itegeress roerty for r (2 (), () ]. By (2.3) i Corollary 8, we ca say that if r ( 2,] ad there is a rime i the iterval ( +r 3,r) the the corresodig hyerharmoic umber is ot a iteger. Sice r (2 (), () ] ad () ( 8 9,], we get that () r>2 () > 6 9 = 7 9. I articular we have r ( 2,]. Moreover we obtai that +r Thus I := ( 2 3, ] ( 7 9 +r 3,r). Note that if 9 7 x the there is a rime i the iterval ( 2 3, ] 7 9. Thus we deduce that for all x ad r, we have h (r) / N. (( Assume ) that ] there exists a real umber x d such that there is a rime i 2d+5 x, x ad the claim holds for d. Also suose that for all x x d,

11 H. Göral, D.C. Sertbaş / Joural of Number Theory 7 (207) (( ) ] the set P 2d+7 x, x is o-emty. We ca assume that x d x d. As we did i the case d =, let 9 7 x d. The agai we have () (( ) ],. (2.8) 2d +7 By Proositio 9, we kow that the ossible values of r that satisfy the roerty ν ()(h (r) ) 0belog to the set ( (j +) (), j ()] where j {,...,d}. Sice (( ) ] () 2d+7,, we see that for all j {,...,d} we have j () j < (j +) (). This imlies that ( (j +) (), j ()] ((j ), j]. So it is eough to check whe r ( (d +) (), d ()] (2.9) by the iductio hyothesis. We agai use (2.3) i Corollary 8 to rove the claim for d. Note that as r >(d ) ad d 2, we see that r > d 2. Also r d () d (2.0) holds. I order to show the claim, we should see that the iterval rime umber. By (2.8), (2.9) ad (2.0) observe that I d := ( ] ( d + d +2, 2d2 +6d + r 2d 2 +7d d +2, r ). d ( ) +r d+2, r d cotais a So fidig a rime i the iterval I d leads to the desired result. Note that sice the iterval I d 2d 2 +6d 2d 2 +7d 2d 2 +6d 2d 2 +7d < does ot itersect the iterval, the the lower boud of I d becomes d + d +2 2d 2 +7d 2d 2 +6d = 2d +6 2d +7, (( ) ] 2d+7,. If we deote = 2d 3 +9d 2 +7d 2d 3 +0d 2 +d 2. Comutatio ad simlificatio idicate that the iequality 2d 3 +9d 2 +7d 2d +6 2d 3 +0d 2 < +d 2 2d +7 holds as we have 5d 2 +3d 2 > 0for all d. So if x d, the there is a rime i I d. This ca oly be achieved whe 2d2 +7d 2d 2 +6d x d. As

12 506 H. Göral, D.C. Sertbaş / Joural of Number Theory 7 (207) x d > 2d2 +7d 2d 2 +6d x d, (( ) ] we have I d P is o-emty by the fact that the iterval 2d+7 x, x cotais a rime umber. By usig this fact, we deduce that h (r) is ot a iteger for 9 7 x d ad r ((d ), d]. Thus we have the claim. O the other had, the above roof yields that we ca choose d = d() as a fuctio of. We kow that by [8], there exists x 0 such that if x 0 the the iterval ( 0.525,]= ( ( ) ] 0.475, always cotais a rime (actually the umber of rimes i this iterval teds to ifiity as teds to ifiity). Thus we ca take 2d + 7 = 2d() + 7 = 0.475, i other words d = d() = Hece if 0 = 9 7 x 0 ad r d = , the the corresodig hyerharmoic umber h (r) is ot a iteger. Thus for a fixed r there are oly fiitely may ossible values of where h (r) is a iteger. Note that the fial result ca also be obtaied by Proositio 6. Some remarks are as follows. As it was metioed before, we used (2.3) i Corollary 8 to show Theorem 0. Observe that the same result caot be show by usig (2.2). Moreover the lower boud for is ot the best ossible oe. For istace for d =, the lower boud for ca be decreased to 53 istead of 68. Oe ca also take larger iterval for (). For examle if d = 2the we ca take () i ( 9 0,] istead of ( 0,]. I that case, we obtai x 2 = 27 ad there is also a rime i the iterval ( 3 4, ] 7 20 for x2. However it is ecessary to fix the deomiator with resect to d to show the theorem by iductio. Also the deomiator 2d +7is otimal to rove such a theorem with full geerality as choosig the deomiator of the form d + k does ot work where k is a iteger. Fially we rove that almost all hyerharmoic umbers are o-itegers which is the first art of Theorem ad that also comletes the roof of Theorem. { Theorem. Let S(x) = (, r) [0,x] [0,x] h (r). / Z} I other words, S(x) couts the umber of airs (, r) i the rectagle [0, x] [0, x] where the corresodig ( ) hyerharmoic umber h (r) is ot a iteger. The we have S(x) = x 2 + O x , which meas that o-iteger hyerharmoics have the full asymtotic i the first quadrule. Coditioally, assumig ( the Riema ) ( hyothesis ) or Cramér s cojecture, the error term ca be take of size O ε x 5 3 +ε ad O x 3 2 log x resectively. Proof. We aly the revious Theorem 0. We kow by Theorem 0 if 0 ad r = O (.475)

13 H. Göral, D.C. Sertbaş / Joural of Number Theory 7 (207) r r =.475 x 0 x.475 x Fig.. The grah of r =.475. The shaded area idicates the lattice oits (, r) where the corresodig hyerharmoic umber h (r) is ot a iteger. the the corresodig hyerharmoic umber h (r) is ot a iteger. Now let x be a sufficietly large ositive real umber. Cosider the grah of the fuctio r =.475 (see Fig. ). It itersects the lie r = x whe = x.475 (.) Therefore we have S(x) x 2 0 x + x x.475 i other words S(x) = x 2 + O x ad i articular S(x) x 2. Assumig ( ) the Riema hyothesis, i the revious Theorem 0, we ca take d = O ε 2 ε ad ( so d = O ε 3 2 ). ε Followig the revious idea we have that S(x) x 2 ε (ε)x + x x /( 3 2 ε) = O ε (x 5 3 +ε). Fially if we assume Cramér s cojecture, we ca take d = d = whe c c 2. Similarly the grah of the fuctio r = c 2 log 2 log 2 c log 2 for some c > 0, ad itersects the lie r = x log = x, ad this i tur gives that = O ( x log x). This yields that S(x) = ( ) x 2 + O x 3 2 log x. Accordig to the calculatios, the ower i the error term does ot chage very much whe we use the result give i [8] istead of the Riema Hyothesis. I fact, the differece betwee the owers is aroximately Combiatorial methods I this sectio we will rove Theorem 2. Before begiig its roof recall that we defie the set I a (, r) := az [r, + r) i the revious sectio, where a is a give rime

14 508 H. Göral, D.C. Sertbaş / Joural of Number Theory 7 (207) umber. I this sectio we use the same otatio for a arbitrary ositive iteger a, istead of just a rime. Ad as a start for this sectio, we give a relatioshi betwee the umber of elemets of I a (, r) ad I a (, ), for ay a N >0. Lemma 2. Let a be a give ositive iteger ad defie k = I a (, ) = a. The we have Moreover if a, the I a (, r) = k. I a (, r) {k, k +}. Proof. Let b = I a (, r) ad c = r a. Sice < (k +)a ad r ca, we have + r< (k + + c)a; or i other words (c + b )a + < (c + k +)a. Hece b < k+ which is equivalet to b < k+2. Also sice ka ad (c )a < r, we have (c + k )a < + r. Therefore, (c + k )a + < (c + b)a which leads to the fact that k < b. Thus, b {k, k +}. Moreover, if a divides the = ka. Hece ( + ) = ka + (k + c)a ; ad this is equivalet to (b k )a. So b caot be equal to k +. Thus, I a (, r) = b = k = a. Defiitio 3 (Maximal ower of a rime i a set). Let I be a fiite set ad be a rime. We defie μ (I) = max{ν (s) s I } as the maximal ower of i I. The umber of itegers i ( μ (I) Z ) I is deoted by M (I). Note that for ay o-emty set I ad P, we have M (I). Also for ay I = I(, r), observe that M (I) < : if M (I) the there exists c N such that c θ, (c +) θ,...(c + ) θ I, where θ deotes the μ (I). By the defiitio of θ, we deduce that c + i for all i {0,..., }. However must divide either oe of the cosecutive itegers. This leads to a cotradictio. Examle 4. Take r =ad let P ad Z 2 be give where α < α+ holds for some α N. I that case, we will have where I(, r) = I(, ) = {, 2,...,}. μ (I(, )) = α = log

15 H. Göral, D.C. Sertbaş / Joural of Number Theory 7 (207) Lemma 5. Let 2 ad r be give. The for all rimes, we have log μ (I(, r)) log ( + ). Proof. Let θ = μ (I(, r)). By defiitio, there exists a I(, r) such that a = c θ for some c Z + where c. Therefore, θ a = c θ + which yields that θ log ( + ). As θ is i N, we have θ log ( + ). Let α < α+ be give. So to show the secod iequality, we assume that μ (I(, r)) = θ<α. That meas by defiitio, for all c N, c α / I(, r). I other words, there exists c 0 N such that c 0 α <rad + < (c 0 +) α. This leads to < α, which is a cotradictio. These bouds are shar ideed. For examle, if we take = 4, r =25ad = 2, the I(, r) = {25, 26, 27, 28} ad therefore μ (I(, r)) = 2 = log. I order to rove Theorem 2, we eed to use some combiatorial tools. I articular, we use the followig lemma which ca be foud i either [7,. 6] or [5]: (( )) a + b Lemma 6 (Kummer). Let a, b be give two atural umbers. The ν is b equal to the umber of carries that occur i the additio of a ad b i their -ary reresetatios. The followig roositio will be a key ste for Theorem 2. Proositio 7. Let be a give rime umber. Assume that = ( α, α,..., 0 ) N 2 ad = (r β, r β,..., r 0) N are the -ary reresetatios of ad, resectively. Also let δ =max{α, β} ad + = δ+ s i i =(s δ+, s δ,..., s 0 ) where s δ+ {0, }. The μ (I(, r)) = max {i {0,,...,δ+} s i >r i }. (3.) Moreover we have (( )) + ν μ (I(, r)). Proof. Let γ := max {i {0,,...,δ+} s i >r i } ad θ := μ (I(, r)). We wat to rove that θ = γ. Note that this γ exists because if ot, the for all i {0,,...,δ+} the iequality s i r i holds. This imlies that +r = δ+ s i i δ r i i = r, which is a cotradictio sice > 0. Furthermore we claim that for all j >γwe have s j = r j. To show this suose cotrary, i.e. take the largest j >γsuch that s j <r j, ad call it j 0. The cosider

16 50 H. Göral, D.C. Sertbaş / Joural of Number Theory 7 (207) >r ( + ) = δ δ+ r i i s i i = j 0 (r i s i ) i j 0 =(r j 0 s j0 ) j 0 + (r i s i ) i j 0 = j 0 ( ) j 0. A cotradictio, hece we have the claim. Now cosider δ+ b = s i i i=γ j 0 ( ) i where s δ+ is ossibly zero. We first have to show that b I(, r). To do this, we rereset as δ+ r i i where r β+ =...= r δ+ =0. It is easy to see that δ+ δ+ b = s i i s i i = +. i=γ Also by defiitio of b ad γ, we have s i = r i for all i {γ +,...,δ+} ad s γ >r γ; which imlies that b >r. If {γ +,...,δ+} is emty, the γ = δ + ad by defiitio of δ we have agai b > r ; as s δ+ >r δ+ =0. Thus b I(, r); ad hece oe ca easily see that γ = ν (b) μ (I(, r)) = θ. So it is eough to rove that θ γ. If γ = δ +, the γ = log ( + ) ad hece by Lemma 5 we have θ γ. For the secod case, suose that γ <δ+. That meas s δ+ = r δ+ ad sice β δ, we see that s δ+ =0. Also assume that θ >γ. Sice θ, γ Z, we have θ γ +. By defiitio of μ (I(, r)), there exists a I(, r) such that a = c a θ for some c a N ad c a. I other words, we have a = δ a i i I(, r) where a 0 = a =...= a θ =0 ad a θ 0. If δ = α, the sice α > 0we have s δ >r δ. This imlies that γ δ, but sice γ<δ+, this yields that γ = δ. However by Lemma 5 we have θ δ = log ( +r ), which leads to a cotradictio. Thus δ >αas δ =max{α, β}. So assume that δ = β, or more recisely β >α. Due to the defiitio of γ, we have s i = r i for all i {γ +,...,δ}. Also sice a I(, r), observe that a = δ i=θ a i i δ i=θ s i i. Because otherwise the iequality a δ i=θ s i i = c 0 θ > 0holds for some c 0 N. Therefore θ θ s i i ( ) i =( ) θ <θ c 0 θ = a δ s i i. This imlies that a > δ s i i = +, which is imossible. So by usig this fact ad the iequality θ γ +, we obtai that i=θ

17 H. Göral, D.C. Sertbaş / Joural of Number Theory 7 (207) a = δ a i i i=θ δ s i i i=θ δ i=γ+ s i i = δ i=γ+ r i i δ r i i = which is a cotradictio. Thus θ γ ad hece we have the equality. To rove the secod art, we use Lemma 6 ad aly the first art of the roositio. Defie i =0, for all i {α +,...,δ+}. Observe that there ca t be ay carry for the values of i = θ +,..., δ +, due to (3.). The situatio is also same for the case i = θ. Because if a carry occurs for that case, the s θ+ r θ+ + which cotradicts the fact that s i = r i for all i {θ +,...,δ+}. This i tur imlies that the oly ossible carries i the additio of ad i their -ary reresetatios are for the etries (( i =0,...,θ )). Thus, there are at most θ carries i the additio; ad hece + ν θ = μ (I(, r)). Due to Lemma 6 ad Proositio 7, we observe that it is eough to fid digits where a carry does ot occur i the additio (( of )) ad i their -ary reresetatios + i order to obtai μ (I(, r)) >ν. This fact will also hel us to fid airs (, r) where the corresodig hyerharmoic umber h (r) is ot a iteger. The followig theorem is a art of Theorem 2: Theorem 8. Let, r be give (( atural umbers. )) Suose that there exists a rime umber + such that μ (I(, r)) >ν ad M (I(, r)) =. The h (r) / N. I articular for all 2 ad r, if is eve or r is odd the h (r) / N. Moreover for all α ad rime, we have h (r) α / N. Proof. Let θ = μ (I(, r)). From (.2), we ca see that h (r) N P ν ( h (r) ) 0 P ν (( + )) + ν (h +r h r ) 0. (3.2) So take the rime umber which satisfies the give assumtios of the theorem. The cosider h +r h r = +r i=r i = +r c θ + i=r i c θ i, (3.3) where c. Sice M (I(, r)) =, we ca say that the maximal ower of i the set I(, r) \ { c θ} is at most θ. By the o-archimedia roerty of -adic valuatio, we have

18 52 H. Göral, D.C. Sertbaş / Joural of Number Theory 7 (207) ν +r i=r i c θ i > θ. (3.4) Agai (( by the same )) roerty, oe ca also see that ν (h +r h r ) = θ. Sice + θ>ν, we obtai that ν (h (r) ) < 0. By (3.2), we coclude that h (r) / N. For the secod art of the theorem assume that 2 α 2 < 2 α2+ holds, where α 2. The we have 2 < 2α 2. This imlies that for all r N we have I 2 α 2 (, r) {, 2}, due to Lemma 2. Defie θ 2 = μ 2 (I(, r)). Sice θ 2 α 2, we have I 2 θ 2 (, r) I 2 α 2 (, r). Therefore M 2 (I(, r)) = I 2 θ 2 (, r) {, 2}. We will show that I 2 θ 2 (, r) =. Suose ot. The by Defiitio 3 there exists c N such that c2 θ 2, (c +)2 θ 2 I(, r). However either c or c + is eve ad this imlies that μ 2 (I(, r)) θ 2 +, which cotradicts the defiitio of θ 2. Thus M 2 (I(, r)) =. Without loss of geerality, suose that is eve. If we write =( α,..., 0 ) 2, the we have 0 =0. This imlies that there does t exist ay carry i the first ste of (( the additio )) of ad. Due to Lemma 6 ad Proositio 7, we obtai that + ν 2 <θ 2 ; ad thus h (r) / N. The case that r is odd follows similarly. For the last art, we will follow the same lies that we did for the case = 2. Let θ := μ (I(, r)). Sice = α, observe that M (I(, r)) = I θ (, r) =. Also if =( α, α,..., 0 ), we have α =ad i =0, for all i {0,...,α }. This imlies that there exist α may etries such that a carry does ot occur i the additio of ad. Thus we have Ad hece h (r) / N. (( )) + ν θ α θ <θ. Remark 9. Note that for the case = 2, it is eough to obtai a digit where a carry does ot occur i the biary additio of ad. Let = ( α, α,..., 0 ) 2 N 2 ad =(r β, r β,..., r 0) 2 N be biary reresetatios of ad. Defie δ =max{α, β} ad i = r j =0, for all i {α +,...,δ} ad j {β +,...,δ}. (( )) + Observe that if there exists t δ such that t = r t =0the ν 2 <θ 2 eve if a carry occurs i the additio of the revious values. I that case for 2, the hyerharmoic umber h (r) is ot a iteger by Theorem 8. The case that we metioed i the revious aragrah ca be satisfied whe both t ad r t are eve simultaeously for some t {0,...,δ}. I fact by Theorem 8, we kow that ad should be odd if h (r) is a iteger. So eve this is the case, amely 0 = r 0 =, the = r = 0 imlies that h (r) / N. This situatio holds whe both mod 4. Cosequetly if mod 4 ad r 2 mod 4, we coclude that

19 H. Göral, D.C. Sertbaş / Joural of Number Theory 7 (207) / N. Note that this rocess ca be develoed to obtai similar equivalece classes i ay ower of 2. h (r) Remark 20. Whe we cosider the rimes other tha 2, we may ot obtai the same result. For istace, take = 3ad 3. The for (( ay r N)) we have M 3 (I(, r)) + {, 2}. By Theorem 8 if M 3 (I(, r)) =ad ν 3 <μ 3 (I(, r)), the we obtai that h (r) / (( N. However )) if M 3 (I(, r)) = 2, we might ot get the coditio ν 3 (h (r) + ) < 0eve if ν 3 <μ 3 (I(, r)). Because i that case, there exists c N such that c mod 3 ad c3 θ 3, (c + )3 θ 3 I(, r), where θ 3 = μ 3 (I(, r)). Note that c caot be equivalet to 0, 2 mod 3; because if so the either oe of c or c + is a multile of 3, ad hece M 3 (I(, r)) =. This imlies that h +r h r = +r c3 θ + 3 (c +)3 θ + 3 i=r 3 θ 3 i i = 2c + c(c +)3 θ 3 + q, where q deotes the sum of recirocals of the elemets of I(, r) which are ot divisible by 3 θ 3. Similar to (3.4) we ca say that ν 3 (q) θ 3, due to the o-archimedia roerty of the -adic valuatio. As c ( mod) 3, we see that 2c + 0 mod 3 ad 2c+ ν 3 (c) = ν 3 (c +) = 0. Thus we have ν 3 = ν 3 (2c +) θ 3 θ 3 ; ad hece c(c+)3 θ 3 { ( ) } 2c + ν 3 (h +r h r ) mi ν 3 c(c +)3 θ,ν 3 3 (q) θ 3. (( )) + So if ν 3 = θ 3 < μ 3 (I(, r)), the by (.2) we get (( )) + ν 3 (h (r) )=ν 3 + ν 3 (h +r h r ) 0. Cosequetly for some secific r N, the iequality ν 3 (h (r) ) 0ca be obtaied. Nevertheless Theorem 8 ca be used to cover most of the (, r)-tules where h (r) / N. More recisely, oe ca give some (( secific values )) of ad r which may deed o a rime + such that μ (I(, r)) >ν ad M (I(, r)) =. However, we do ot metio here each case searately. Istead we will give the followig theorem which may widely be used to cover most of the cases ad it is the other art of Theorem 2:

20 54 H. Göral, D.C. Sertbaş / Joural of Number Theory 7 (207) Theorem 2. Let a be oegative ad m, k, b be ositive itegers. Let be a rime umber such that a + b. Put = m + a =( α,..., 0 ) r = k + b =(r β,...,r 0 ), where ( α,..., 0 ) ad (r β,..., r 0 ) are the -ary reresetatios of ad r, resectively. If there exist c N ad κ > αsuch that r (c κ, c κ ], the the corresodig hyerharmoic umber h (r) is ot a iteger. Otherwise the o-itegeress roerty still holds, if is sufficietly large deedig o m ad k. Proof. Let θ = μ (I(, r)). We will show that the existece of c N ad κ > αwhere c κ < r c κ is equivalet to θ >α. Note that this coditio imlies that r c κ < + r. Hece θ ν (c) + κ > α. For the vice versa, if θ >αthe there exists c 0 N such that r c 0 θ < + r. This idicates that r (c 0 θ, c 0 θ ]. Takig κ = θ ad c = c 0 lead to the coditio o r give i the theorem. Observe that 0 = a ad r 0 = b > 0. Sice 0 < a + b, a carry does ot occur i the first ste of (( the additio )) of = ( α,...,, a) ad = (r β,..., r, b ). This + imlies that ν <θ. So for the first art of the theorem, assume that θ >α. The we see that < θ, sice = α i i α ( ) i = α+. By Lemma 2, we obtai that M (I(, r)) = I θ(, r) { } θ, θ + = {0, }. Sice (( I(, r) is )) o-emty, we deduce that M (I(, r)) =. By the iequality + ν <θad Theorem 8, we deduce that h (r) / N. For the secod art of the theorem, we may assume that θ = α by the first art of r the theorem ad by Lemma 5. Defie d = ad α s = M (I(, r)) = I α(, r). Note that d ad s deed o k ad m, resectively. Cosider h +r h r = ( ) +r α d d + s i. (3.5) i=r α i

21 H. Göral, D.C. Sertbaş / Joural of Number Theory 7 (207) I order to obtai the -adic valuatio of h +r h r, we have to calculate the -adic valuatio of each term o the right had side. So first of all, cosider the secod summad. Observe that the largest ower of that ca divide the deomiator of oe of these recirocals i this sum is α. By the o-archimedia roerty of the -adic valuatio, we fid that ν +r i=r α i For the first summad sice α α = α + i i α = α ad β r α = r i i α + i=α α. (3.6) i α r i i α = β α r α+i i +, we get d = (r α +) + β α i= r α+i i ad s { α, α +} by Lemma 2. Note that oe of j {d,...,d+ s } is a multile of. Because if ot the there is l N such that j = l ad j α = l α+ I(, r). I that case μ (I(, r)) = α+ > θ, which cotradicts the defiitio of θ. We set z = d+s j=d j = N z D z, where N z ad D z deote the umerator ad the deomiator of z, resectively. We also assume that gcd(n z, D z ) =. If is greater tha N z > 0, the ν (N z ) = 0. Moreover sice {d,...,d+ s } Z =, we see that d (d +) (d + s ). Sice D z d (d +) (d + s ) we deduce that D z, ad so ν (D z ) = 0. These facts imly that ν (z) = 0ad hece ( ( )) ( ) ν α d + + Nz = α + ν = α. d + s D z Combiig this fact with (3.6) leads to ν (( (h +r h )) r ) = α, due to the o + Archimedia roerty agai. Recall that ν <θ= α. Thus we coclude by (3.2) that h (r) / N.

22 56 H. Göral, D.C. Sertbaş / Joural of Number Theory 7 (207) This theorem simly elimiates most of the cases. For istace let P, = m + a for some a {0,..., }, α ad m {,..., α }. If κ > αad r = k κ i for some k N ad i {a, a +,...,m } {i =(i α,...,i 0 ) a i 0 <}, the we ca see that h (r) / N due to the fact that k κ I(, r). Therefore oe ca choose aroriately the values of m,, a, k, κ ad i to show that h (r) is ot a iteger for a huge class of airs (, r). Moreover for r = k + b as i Theorem 2, oe ca fix m, b, a ad k to check the o-itegeress roerty of h (k+b) m+a for differet values of. However we do ot discuss each case searately here. Nevertheless we would like to oit out that these classes of (, r) tules caot be obtaied from aalytic methods, esecially whe r is large. Examle 22. To give a examle for Theorem 2, we first take = 33ad = 3. I that case we have a = 2ad m =. Also take r = = Note that r ( , ]. This imlies that h (r) is ot a iteger for the give values of ad r. Observe that this result may ot be achieved by usig aalytic methods. Namely if we use Theorem 0, the oly way to achieve this boud may be to set d = = ad try to fid the corresodig d so that for all d, r d imlies that h (r) / N. However the corresodig d boud may be far beyod the value of = 33, because we kow by revious examles that the corresodig lower boud for d = is 53. For the secod art of the theorem, we will calculate the corresodig uer boud for r where = (3, 0,..., 0) =3 α ad α. Sice = 3 α ad s = α Z I(, r) we kow that s = 3. Moreover is of the form m + a where m = 3 α ad a = 0. Therefore we ca take r = k + b where b {,..., }. Note that we ca also take b = 0sice a carry does ot occur i the first ste of the additio of ad i their -ary reresetatios due to the fact that a = 0. So we do ot have ay restrictios o r, excet for a ossible uer boud deedig o. If the umerator of d + d + + d +2 = 3d2 +6d +2 d(d +)(d +2) r is less tha where d =, the we deduce by the secod art of Theorem 2 that α / N. Hece it is eough to check the rimes where > 3d 2 +6d +2. Observe that 3r 2 + 2r 2α +( ) < 0 imlies > 3d 2 +6d +2. This iequality is satisfied whe the α iequalities 2 α + 3 <r< 2 + α + 3 hold. Thus r < α + 3 2leads to h (r) the fact that h (r) is ot a iteger. Now we give a more geeral result similar to the revious examle where we ca comare the ower of combiatorial ad aalytic methods o o-itegeress roblem of hyerharmoic umbers. Let = s α for some α, s N ad a rime. For s = ad eve values of s, we kow by Theorem 8 that h (r) is ot a iteger for all values

23 H. Göral, D.C. Sertbaş / Joural of Number Theory 7 (207) ( ) of r. Also if s = 3the we ca take r = O 2α due to the Examle 22. For α =, the uer boud for r ca be give as O ( ). A slightly weaker result ca be obtaied ) from aalytic methods uder the Riema Hyothesis, i.e. r ca be take as O ( 3 2 ɛ for all ɛ > 0. Therefore it is better to use combiatorial methods istead of aalytic oes whe s 3ad α =. However, whe we take a odd s > 3, we wo t get ay better results comared to the aalytic case. To exlai this fact more recisely, first recall that z = s H = sp H d(d +) (d + s ), where P deotes the roduct ad H deotes the harmoic mea of d, d +,..., d + s. Note that sp H is a iteger ad d = r as i Examle 22. By the well-kow bouds α o H, we ca say that the umerator of z is equal to sp H sp d = s(d +) (d + s ). So if > ( s(d +) ) (d + s ) we obtai that h (r) / N. This imlies that d ca be take of size O s r, for a fixed s. Sice d =, we deduce that α ( ) ) r = O α+ s = O ( + α(s ). Ufortuately this boud does ot lead to better results comared to the aalytic case, eve if we take α =. Note that the order of the magitude ca be obtaied from the aalytic aroach is.475, due to Theorem 0; whereas the corresodig boud here is at most + α(s ).25. Therefore it is ecessary to use differet methods i order to obtai better bouds for r. 4. Algebraic methods ad comutatioal results I this sectio, we rove Theorem 3 ad Theorem 4. The followig is Theorem 3 from the itroductio. Theorem 23. Let = k α be a odd iteger where is a rime, α ad r is give. Defie a = k 2, c = r ad α a a F k (x) := (x j) x i. i= a j= a Assume that ν (F k (c + a)) ν (k!). The the corresodig hyerharmoic umber h (r) is ot a iteger. I articular if c + a is ot a root of F k (x) i modulo, the h (r) / N.

24 58 H. Göral, D.C. Sertbaş / Joural of Number Theory 7 (207) Proof. By Lemma 2, we kow that I α(, r) = = k. Cosider α ( + h (r) = ) (h +r h r ) = N D cα (c +) α (c + k ) α k α 2 α k α +r (c + i) α + l=r α l l, where ν (N) = ν (D). To see this fact, first ote that Nis the multilicatio of the elemets i I(, r) which are ot divisible by α whereas Dis the multilicatio of the elemets i I(, ) which are ot divisible by α. Recall by Lemma 2 that for all β<αwe have I β(, r) = = k α β. This imlies that the umber of elemets β i the set J β(, r) = { a I(, r) β a } = { a I(, r) β a } \ { a I(, r) β+ a } is equal to k α β k α β. Sice I β (, ) = = k α β, we see that J β β(, ) = k α β k α β. Therefore for ay β<α, the umber of terms which are exactly divisible by β i the multilicatio of Nad Dare the same. Cacellatio of these terms leads to the fact that ν (N) = ν (D). Now defie q = N D, q 2 = +r l=r α l ad C = c(c +) (c + k ). As we have show earlier, we have ν (q ) = 0. Note that ν (C) 0sice C Z. Moreover we have ν (q 2 ) > α, as the lowest ower of i the sum of q 2 is α. Due to the o-archimedia roerty of -adic valuatio we obtaied the metioed result. Observe that h (r) = q α k! l k k (c + j) c + i j=0 + C q q 2. k! By the defiitio of F k (x) ad a, we have k k F k (x + a) = (x + j). x + i This imlies that { ( ) ( )} ν (h (r) q F k (c + a) C q q 2 ) mi ν α,ν k! k! j=0

25 H. Göral, D.C. Sertbaş / Joural of Number Theory 7 (207) by the o-archimedia roerty ( of the -adic ) valuatio. ( ) Recall that ν (h (r) )is equal to either oe of them uless ν q F k (c+a) α k! = ν C q q 2 k!. Suose that ν (F k (c + a)) ν (k!). The the -adic valuatio of the first term is ( ) q F k (c + a) ν α = ν (q )+ν (F k (c + a)) ν (k!) ν ( α ) α, (4.) k! sice ν (q ) = 0. Due to the same reaso, we get ( ) C q q 2 ν = ν (C) + ν (q 2 ) ν (k!). k! Observe that C c (c +) (c + k ) = = k! 2 k ( ) c + k Z. k Therefore we have ν (C) ν (k!). This idicates that the -adic valuatio of the secod term satisfies ( ) C q q 2 ν = ν (C) + ν (q 2 ) ν (k!) ν (q 2 ) k! ( ) q F k (c + a) > α ν α, (4.2) k! due to (4.). Hece we obtai by the o-archimedia roerty that ν (h (r) )is equal to the miimum of these two terms. By (4.2) we also deduce that ( ) ν (h (r) q F k (c + a) )=ν α α ; k! ad thus h (r) / N. For the secod case, assume that c + a is ot a root of F k (x). This imlies that F k (c + a) 0 mod ; ad hece ν (F k (c + a)) =0. Sice k! is a iteger, we have ν (k!) 0 = ν (F k (c + a)). Thus we coclude by the first art of the theorem that the corresodig h (r) is ot a iteger. Note that the last art of Theorem 8 ca also be show by usig Theorem 23. To do so, take k =. I that case a = 0 ad the olyomial F (x) is the costat olyomial. Thus for ay iteger c ad P we have ν (F (c)) =0. We also have ν (k!) = ν () = 0, which imlies that ν (F k (c)) ν (k!). Sice the coditio give i Theorem 23 is satisfied, we coclude that h (r) α is ot a iteger for ay r N. I order to show that h (r) is ot a iteger, it is eough to fid oe such rime i the factorizatio of that satisfies the roerties give i Theorem 23. I other words, a iteger hyerharmoic umber h (r) has to satisfy the followig roerty: for all

26 520 H. Göral, D.C. Sertbaş / Joural of Number Theory 7 (207) that divides ad for all α ν (), we have F k (c + a) 0 mod where k =, ad α the itegers c ad a are as i Theorem 23. Heuristically if has l may distict rime α α divisors, say = l l, the the robability that F k (c + a) 0 mod i is i for i {,...,l}, k = β i ad β α i. Ad hece the robability that h (r) is ot a iteger is. So it is very ulikely that h(r) is a iteger, as it was roved i Theorem. Obviously oe ca observe that if is sufficietly large for a give r, i.e. > F k (c +a), the the corresodig hyerharmoic umber h (r) is ot a iteger. Recall that this situatio was ivestigated i Examle 22 ad its geeralizatio just after this examle. Also ote that, if F k (x) has o roots modulo the give rime the agai for ay r N we deduce by Theorem 23 that h (r) is ot a iteger. This fact will be useful to rove the followig corollary which is a art of Theorem 4: Corollary 24. Let r, α be ay atural umbers. If is a rime such that ±5 mod 2, the h (r) 3 / N. Aart from that if α 7,, 3, 4, 9, 2, 22, 23, 26, 28, 3, 33, 38, 39, 4, 42, 44, 46, 52, 53, 56, mod 45, 57, 6, 62, 63, 66, 67, 69, 76, 78, 79, 82, 83, 84, 88, 89, 92, 93, 99, 0, 03, 04, 06, 07, 2, 4, 7, 9, 22, 23, 24, 26, 3, 32, 34, 38 (4.3) the h (r) 5α / N. Proof. I order to show this corollary, we will rove that the corresodig olyomial F k (x) for k =3, 5 has o roots i modulo. For the first art of the corollary, ote that F 3 (x) = 3x 2. So F 3 (x) has o roots i modulo if ad oly if 3x 2 0 mod x 2 3 mod, where 3 deotes the multilicative iverse of 3 i F. This idicates that F 3 (x) has o roots i {0,..., } if ad oly if 3 is ot a square modulo, which meas ( ) that 3 3 is ot a square modulo. So i order to use Theorem 23, it is eough to have = ( ) where deotes the Legedre symbol. By the well-kow quadratic recirocity law (see [6, Proositio II.2.5]), we kow that ( ) 3 ( =( ) 2. (4.4) 3) If 5 mod 2, the ( ) 2 =ad ( ) ( 3 = 2 ( ) 3) =. So by (4.4), we deduce 3 that =, ad hece h (r) 3 / N. Similarly for a rime 7 mod 2, we have α ( ) 2 = ad ( ) ( 3 = ) (r) 3 =. Agai by (4.4), we deduce that h 3α / N.

27 H. Göral, D.C. Sertbaş / Joural of Number Theory 7 (207) For the secod case, observe that F 5 (x) = 5x 4 5x 2 +4 = g(y) where g(y) = 5y 2 5y +4 ad y = x 2. Note that F 5 (x) has o roots i F = {0,..., }, if the ( discrimiat ) ( ) ( Δof ) g(y) is ot a square. Sice Δ = 45 =5 29, we obtai that =. I order to get 45 is ot a square modulo, it is eough to have ( ) ( ) =. Recall that by the quadratic recirocity law, we have 5 29 ( ) q =( ) 2 q 2 ( ), (4.5) q where ) q is ) either 5 or 29. Sice q mod 4 for q =5, 29, we deduce by (4.5) that =. So we have two differet cases: either ( q ( q ( 5) ( = = or 29) ( 5) ( = = 29) holds. Note that each of these cases leads to a relatioshi betwee modulo 5 ad 29. By usig the list of squares i modulo 5, 29 ad the well-kow Chiese remaider theorem, oe ca obtai the modular equivaleces give i (4.3). Observe that the modular equivaleces give i Corollary 24 caot be obtaied by usig combiatorial methods. Remark 25. We ca also comute F k (x) iteratively. To obtai that oe ca defie the olyomial G k (x) := a (x i), i= a where a = k 2. Accordig to this defiitio G k(x) is othig but x a i= (x2 i 2 ). By usig this, we ca deduce that Therefore we have F k (x) = a i= a G k (x) x i. G k (x) =(x 2 a 2 ) G k 2 (x), F k (x) =(x 2 a 2 ) F k 2 (x)+2xg k 2 (x). These equalities lead to the comutatio of F k (x) for the cases k =7, 9: F 7 (x) =7x 6 70x x 2 36, F 9 (x) =9x 8 20x x x

28 522 H. Göral, D.C. Sertbaş / Joural of Number Theory 7 (207) Table The list of the values of x d, d ad r d = d d, for a give d 6. d x d d r d For those cases we have to solve cubic ad quartic recirocities i the corresodig field F, resectively. This requires a lot of comutatioal cases ad we will ot elaborate it i this aer. Now we ca combie what we have achieved so far ad the followig result is the rest of Theorem 4: Corollary 26. Let < 32 be ay iteger. The for all r N we have h (r) / N. Moreover for all > ad r N, if r 6 the the corresodig hyerharmoic umber h (r) iteger. is ot a iteger. I fact for all >, if r the h (r) is ot a Proof. To see the first art of the corollary recall that if is eve or a rime ower, the h (r) is ot a iteger due to the Theorem 8. The oly odd atural umbers which are less tha 33 ad ot rime owers are 5 ad 2. Note that these umbers are of the form 3 where ±5 mod 2. By Corollary 24 we deduce that h (r) 5 ad h(r) 2 are ot itegers also for ay r N. Hece the first art of the corollary follows immediately. We use Theorem 0 to rove the secod art, i.e. for a give d we fid the ossible smallest d such that for all d ad r d we have h (r) / N. To do so, we first calculate the corresodig (( lower ) boud ] x d such that for all x x d there is a rime lyig i the iterval 2d+7 x, x. I order to fid these x d s, we use the bouds give i [3, Theorem.9]. After that for each d 6, we comute the corresodig value of d where d = 9 7 x d. Accordig to our comutatios, we ca decrease the value of d for d =, 2. I fact, d ca be take as 53 ad 27 for d = ad d = 2, resectively. Table shows the list of values of d for give d s u to 6. Accordig to this table, it is eough to check the lattice oits i R 2 where 696, r.36. By usig Theorem 8 ad Corollary 24, we elimiate the values of where