FOR WHICH p-adic INTEGERS x CAN k

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1 FOR WHICH p-adic INTEGERS x CAN x BE DEFINED? DONALD M. DAVIS Abstract. Let fn) = n. Frst, we show that f : N Q p s nowhere contnuous n the p-adc topology. If x s a p-adc nteger, we say that fx) s p-defnable f lm fx j ) exsts n Q p, where x j denotes the jth partal sum for x. We prove that f ) s p-defnable for all prmes p, and f p s odd, then s the only element of Z p N for whch fx) s p-defnable. For p = 2, we show that f s a postve nteger, then f ) s not 2-defnable, but that f the s n the bnary expanson of x are eventually very sparse, then fx) s 2-defnable. Some of our proofs requre that p satsfy one of two condtons. There are three small prmes whch do not satsfy the relevant condton, but our theorems can be proved drectly for these prmes. No other prme less than 00,000,000 fals to satsfy the condtons.. Statement of results Let N Z p Q p denote the natural numbers ncludng 0), p-adc ntegers, and p-adc numbers, respectvely, wth metrc d p x, y) = p νpx y). Here and throughout, ν p ) denotes the exponent of p n a ratonal or p-adc number. The functon f : N Q p defned by n fn) = n ) =0 has been studed n [2], [3], [6], and [7]. Throughout ths paper, f wll always refer to ths functon. Cursory calculatons suggested that perhaps ths f mght be contnuous n the 2-adc topology and extendable over Z 2. For example, Maple computes Proposton.. If ν 2 m + 2) = ν 2 n + 2) 4 and m, n < 8000, then ν 2 fm) fn)) = ν 2 m n) + 2ν 2 m + 2). Date: August 6, 202. Key words and phrases. bnomal coeffcents, p-adc ntegers Mathematcs Subject Classfcaton: 05A0, B65, D88.

2 2 DONALD M. DAVIS If n 4 mod 6 and n < 70, 000, then ν 2 fn)) = ν 2 n + 2). If ths perssted wthout the bounds, then f would extend to a contnuous functon on Z 2 {x : x 4 6), x 2}. However these bounds are not large enough to reveal the problems that can occur. For techncal reasons, some of our results nvolve the noton of Weferch prmes. Recall that a Weferch prme p s one for whch p 2 dvdes 2 p. The only nown Weferch prmes are 093 and 35. Let LW denote the set of Weferch prmes greater than 35. The name of ths set refers to large Weferch. As of May 202, t was nown that LW contans no ntegers less than [4]) Others of our results requre that p satsfes that for all n such that n p 2, ν p fn)). We wll call such prmes good. Paul Zmmermann has tested all p < 0 8 and found that n ths range there s only one case, p = 23 and n = 2, n whch ν p fn)) >, wth ν 23 f2)) = 2. Thus all prmes less than 00, 000, 000 are good except for p = 23. We wll prove the followng theorem n Secton 2. Its corollary s the frst of our man results. Theorem.2. a. If and e > max{, j + ν 2 j) : 0 < j < }, then ν 2 f2 e )) = + ν 2 ) e. b. If p s an odd prme whch s not a Weferch prme,, and e > max{, + ν p j) : 0 < j < }, then.3) ν p fp e )) = ν p ) + e. c. If p = 093 or 35,, and e > max{, 2 + ν p j) : 0 < j < }, then.4) ν p fp e )) = ν p ) + 2 e. Corollary.5. If p LW, then the functon f : N Q p s nowhere contnuous n the p-adc topology). Proof. We gve the proof when p = 2. The proof when p s odd s extremely smlar, usng.3) and.4).

3 FOR WHICH p-adic INTEGERS x CAN x BE DEFINED? 3 Let n N, and let ɛ > 0 be gven. We wll show there exsts m N such that d 2 m, n) < ɛ and d 2 fm), fn)) >. Choose e so that 2 e < ɛ and 2 e > n +. By Theorem.2, f L > max{j + ν 2 j) : 0 < j < 2 e n }, then ν 2 f2 L 2 e n))) = 2 e n + ν 2 n + ) L. Choose L large enough that ths s less than ν 2 fn)) and less than 0. Call ths value r wth r. Then ν 2 f2 L 2 e n)) fn)) = r and so d 2 f2 L 2 e n)), fn)) = 2 r 2, whle d 2 2 L 2 e n), n) = 2 e < ɛ. So m = 2 L 2 e n) has the desred propertes. Snce N s dense n Z p, ths of course mples that f cannot be extended to a functon f that s contnuous at even one pont of Z p. We state ths, but omt the elementary and standard proof. Corollary.6. If p LW and x Z p, t s mpossble to defne fx) so that for all sequences n n N such that n x we have fn ) fx). We remar that the summand functons f x) := ) x =! are xx ) x )) contnuous on Z p {0,..., }. Although fx) cannot be defned so that t wors ncely for all sequences of postve ntegers approachng x, t mght happen that t can be defned so that t s the lmt of the most natural sequence of postve ntegers approachng x, namely the fnte partal sums. Defnton.7. For a p-adc nteger x = ɛ p wth ɛ {0,..., p }, let x n = n ɛ p. Say fx) s p-defnable f lm fx n ) exsts n Q p. If so, defne fx) = =0 lm fx n ). =0 The followng result ponts to a smlarty and a dfference between the prme 2 and the odd prmes. Theorem.8. For all prmes p, f ) s p-defnable. If p = 2, then f ) = 0, whle f p s odd, then f ) mod p.

4 4 DONALD M. DAVIS Proof. We wll prove n Proposton 4.7 that ν 2 f2 e ) 2e, from whch the result for p = 2 s mmedate snce 2 e = ) e n the notaton of Defnton.7. Now let p be an odd prme. We wll prove n Proposton 2.6 that, for e, ν p fp e ) fp e )) e. Thus fp e ) s a Cauchy sequence, and snce p e = ) e, we deduce that f ) s p-defnable. Snce fp 0 ) =, we obtan fp e ) mod p for all e 0. We show now that, for all prmes p LW, fx) s not p-defnable f x s an nteger less than. Theorem.9. Let be a postve nteger and p LW. p-defnable. Then f ) s not Proof. We gve the argument when p s odd. The argument when p = 2 s extremely smlar. Let x =. Let e > max{, δ + ν p j) : 0 < j < }, where δ = f p s not Weferch, and δ = 2 f p {093, 35}. Then x e = p e and, by Theorem.2, ν p fx e )) = δ + ν p ) e. Hence, as e, d p fx e ), 0) = p e δ ν). Thus fx n ) s not a Cauchy sequence. The followng result apples only when p s odd; the analogous statement s not true when p = 2. It says that s the only element of Z p N for whch fx) s p-defnable when p s a good odd prme or 23. As noted earler, ths ncludes all odd prmes less than 00,000,000. Theorem.0. If p s a good odd prme or p = 23 and x Z p {, 0,, 2,...}, then fx) s not p-defnable. Proof. We wll prove n Secton 3 that f p s a good odd prme or 23 and c p and 0 < p e, then.) ν p fc p e + ) f)) 0. If x s as n the theorem, then there exst nfntely many e for whch.) apples drectly to say ν p fx e ) fx e )) 0. Thus d p fx e ), fx e )) and so fx n ) s not a Cauchy sequence.

5 FOR WHICH p-adic INTEGERS x CAN x BE DEFINED? 5 Our other man theorem says that fx) s 2-defnable for those x whose nfntely many s are eventually very sparse. Ths s a major dfference between the prme 2 and the odd prmes. Ths theorem wll be proved n Secton 4. Theorem.2. Suppose x = 2 e, e 0 < e <, and there exsts a postve =0 nteger N such that e > + 2 e for all N. Then fx) s 2-defnable. For example, s 2-defnable. =0 f ) Snce negatve ntegers are the 2-adc ntegers wth only a fnte number of 0 s n ther bnary expanson, Theorems.9 and.2 tell whether fx) s 2-defnable for x at two extremes. In Theorem 4.0 we descrbe 2-adc ntegers x wth nfntely many 0 s whch are eventually very sparse for whch fx) s not 2-defnable. It would be nterestng to now more completely whch 2-adc ntegers x have the property that fx) s 2-defnable. 2. Some p-exponents of fn) In ths secton, we prove Theorem.2 and a result, Proposton 2.6, whch was used above and wll be used agan later. We begn the proof of Theorem.2 wth a specal case. Lemma 2.. Let p be a prme not n LW, and e 2. Then { ν p fp e e 2) p {093, 35} 2)) = e ) otherwse.

6 6 DONALD M. DAVIS Proof. Let α p n) denote the sum of the coeffcents n the base-p expanson of n. For 0 p e 2, ν p ) e 2 p = p = p αp ) + α p p e 2 ) α p p e 2) ) αp ) + p )e α p + ) p )e ) ) = p αp + ) + p )ν p + )) +p )e α p + ) p )e + ) = ν p + ). Ths equals e when = c p e for c p, and s less than e for all other relevant. The result when p = 2 follows snce there s a sngle term of smallest exponent n the sum whch defnes f2 e 2). Now let p be odd. We wll show that for c p ) p e 2 2.2) /p e ) c+ c mod p. c p e Then we obtan fp e 2) = p ) c+ p e c + Ap) + c= where A, A j Z p. By a result of Esensten []) 2.3) p ) c+ c 2p 2 p c= mod p. j<e p j A j, If p s not a Weferch prme, then ν p 2p 2 p ) = 0 and so ν p fp e 2)) = e ). To prove 2.2), we note that the LHS equals mod p. p e 2) p e cp e ) 2 2 cp e )p e 2 cpe ) c) cp e There are cp e 2 of the fractons equal to, whch when multpled together and by c gve the desred ) c+ c. If p = 093 or 35, we consder p e fp e 2) mod p 2. Ths equals p2 )/2 ) p 2p e e 2. p e 2 =

7 FOR WHICH p-adic INTEGERS x CAN x BE DEFINED? 7 Smlarly to the above argument for 2.2), we can show that p e p e 2 p e 2 Maple computes p 2 )/2 = ) ) [ p ]+ p ) p mod p 2. [/p] ) [ p ]+ p ) p mod p 2 [/p] to equal when p = 093 and 5 35 f p = 35. Thus ν p p e fp e 2)) = when p {093, 35}. Proof of Theorem.2. The recursve formula 2.4) fn) = n + fn ) + 2n was proved n [2]. We nvert t to obtan 2.5) fn ) = fn) ) 2n n +. Our proof of Theorem.2 s by nducton on, wth Lemma 2. beng the case =. Assume the theorem has been proved for. Let n p = 2 n = p odd, not Weferch 2 p {093, 35}. Then, wth ν p u) = ν p u ) = 0, fp e ) = u pνp )+ p e = u p νp )+ e, snce e > ν p ) +. Now, usng 2.5), fp e ) = u p νp )+ e 2pe ) p e + = u p νp)+ e, wth ν p u ) = 0, snce ν p p e ) = ν p ) and ν p p e + ) = ν p ). The followng proposton was used n the proof of.8 and wll be used n the proof of 3.2. Proposton 2.6. For any odd prme p and e and c p, fcp e ) fcp e ) c 2 p )p e fcp e ) mod p e+.

8 8 DONALD M. DAVIS Proof. For any j p )/2, we have ) cp e ) cp e cp e + = )/ cp e ) )) cp e = p + 2j cp e p + 2j / cp e p + 2j p + 2j p + 2j p + 2j p + 2j / ) )) cp e p mod p e+. 2j ) cpe 2j The case c =, = 0 says ) p e ) p e / ) + pe p 2j 2j 2j 2j mod p e+. Combnng these, we obtan ) cp e ) cp e ) cp e ) p e ) p e ) + c + p + 2j p + 2j 2j 2j Summng ths over j gves p ) cp e cp e c p + = p c 2 p )p e cp e ) p e = mod p e+, usng Lemma 2.9 at the last step. Now sum over to obtan ) cp e 2.7) c 2 p )p e fcp e ) mod p e+. 0 p) Summng ) cp e pj cp e ) j over j gves ) cp e fcp e ) 0 mod p e+2 0 p) by Lemma 2.8. Add ths to 2.7) to obtan the clam of the proposton. mod p e+. The above proof requred the followng lemmas. Lemma 2.8. If < u < p and j < up e, and b = ap + u wth a 0, then ν bp ) e p pj bp ) ) e j e + 2.

9 x BE DEFINED? 9 FOR WHICH p-adic INTEGERS x CAN Proof. Note that ) bp e ) bp e pj j pj)pj ) = bp e pj)bp e pj )) bp e ) jj ) bp e j)bp e j )) bp e ). When the factors n the numerator and denomnator of the second quotent are multpled by p, they just gve the p-dvsble factors of the frst quotent. Thus the expresson equals ) bp e ) bp e j, where the product s taen over 0 < < pj wth 0 mod p. Snce ν bp ) e p j = 0 and the number of values of s even, ths has the same p-exponent as bp e ), and ths s dvsble by p e+2. To see ths, we wll show that p p p + ) p + bp e ) mod p e+2. = = The dfference, mod p e+2, s, up to unt multples, bp e σ p 2 p +,..., p + p ) + bp e ) 2 σ p 3 p +,..., p + p ) = bp e σ p 2,..., p ) + bp e p + bp e ) 2 )σ p 3,..., p ). Ths s 0 snce the coeffcent of x resp. x 2 ) n x + ) x + p ) s dvsble by p 2 resp. p). Indeed, the frst s p )! + + ), and ths s dvsble by p p2 for p > 3 by [5]. Mod p, the polynomal equals x 2 2 )x ) x 2 p 2 )2 ), and ts coeffcent of x 2 s congruent to Snce 2 p ) 2 mod p, 4 p )/2) 2 ths s congruent to 2 p = Lemma 2.9. For any odd prme p, = Proof. Notng that p e obtan p =, and ths s 0 mod p, also by [5]. 2 p p ) e p e 2 p ) mod p e+. p )/2 p ) e = j= ) ) mod p, and argung as n the proof of 2.6, we p e 2j p e 2j pe 2 p )/2 j= j mod p 2.

10 0 DONALD M. DAVIS Next note that, mod p p = ) = p )/2 j= p 2j ) 2j p )/2 ) Usng 2.3), we obtan the clam of the lemma. j= p )/2 p + j j. 2 j= 3. Results for good prmes and 23 In ths secton, we prove.) and hence Theorem.0), deferrng most detals of the proof for p = 23 to Secton 5. We wll prove the followng result n Secton 5. Proposton 3.. If p s any odd prme, c p, and ν p fc )) > 0, then fcp ) fc ) c 2 p )fc ) mod p 3. The followng corollary follows easly from ths and Proposton 2.6. Corollary 3.2. If p s an odd prme, ν p fc )) 2 and e 0, then ν p fcp e )) = ν p fc )). Proof. Snce 2 p s dvsble by p, the result when ν p fc )) follows by nducton on e usng Proposton 2.6. The result when ν p fc )) = 2 follows n the same way, usng Proposton 3. for the frst step. Now we can prove the followng result, whch mples.) for good prmes, snce ν p + ) e here. Proposton 3.3. If p s odd, ν p fc )), 0 < p e, and c p, then 3.4) ν p fcp e + ) f)) = e + ν p + ) + ν p fc )). Proof. Let e and c be fxed, and let ) := fcp e +) f). The proof s by nducton on, usng 2.4). From 2.4) we obtan fcp e ) = cpe + fcp e ), 2cp e whch, wth Corollary 3.2, yelds the clam for = 0. From 2.4), we also deduce 3.5) ) = cpe + + cp e ) f ). 2cp e + ) 2cp e + )

11 FOR WHICH p-adic INTEGERS x CAN x BE DEFINED? Assume the result for. Then the frst term n the RHS of 3.5) has exponent ν p + ) ν p ) + e + ν p ) + ν p fc )), whch s the desred value. The exponent of the second term n the RHS of 3.5) s e 2ν p ) + ν p f )) e 2ν p ) maxν ) p j j ). We wll now that the second term has larger exponent than the frst once we have shown that, for < p e and j, 2e > 2ν p ) + ν p + ) + ν ) p j. Ths follows easly from the fact that ν ) p j e νp ), snce t equals the number of carres n a base-p addton whose sum s. We wll prove the followng result n Secton 5. nformaton about the case e = yelds smlar nformaton for all e. The mplcaton of ths s that Lemma 3.6. If c, u p and ν p fc )) = 2, then for all e, the expresson fcp e + up e ) fup e ) s dvsble by p and ts resdue mod p 2 s ndependent of e. Now we can prove the followng analogue of Proposton 3.3. Proposton 3.7. Suppose ν p fc )) = 2 wth c p, and for all u whch satsfy u p and ν p fu )) = 0 we have ν p fcp + u ) fu )) = and 3.8) fcp + u ) fu )) c fu ) mod p. p u Then, for all e 3 and all satsfyng 0 < p e, 3.4) holds. Suppose p s a prme whch s not good but has ν p fc )) 2 for all c p and whenever ν p fc )) = 2 the hypotheses of Proposton 3.7 are satsfed. Then 3.4) holds for e 3 and hence so does.). Maple easly verfes that the condtons of Proposton 3.7 are satsfed when p = 23 and c = 3. Hence.), and thus also Theorem.0, holds when p = 23. Note that the proof of Theorem.0 only cares about large values of e n.), and so our restrcton to e 3 s not a problem. Proof of Proposton 3.7. As n the proof of Proposton 3.3, fx c and e, and let ) = fcp e + ) f). If we assume 3.4) holds for, the exponent of the frst

12 2 DONALD M. DAVIS term of 3.5) equals the desred value of ν p )). Now that νfc )) = 2, t can happen that the second term has the same exponent f = up e or up e, n whch case t could concevably happen that the exponent of the combnaton of the two terms s larger than that of the ndvdual terms. In the next paragraph, we wll use the hypothess to prove drectly that ν p up e )) and ν p up e )) have the desred values when u p. The valdty for 0) s verfed usng 2.4) as n the proof of 3.3. For 0 u p, the nducton from up e through u + )p e 2 wors just as t dd n the proof of 3.3. Thus the result s vald for all < p e, as clamed. That ν p up e )) has the desred value s mmedate from Lemma 3.6 and the hypothess that ν p u )) =. Mod p, 3.5) gves 2up e 2 up e ) p upe ) c u fupe ) fcp + u ) fu )) c fu ), p u usng Lemma 3.6 and Proposton 2.6 at the last step. By Proposton 3.7, the last expresson has ν p ) = 0, and so up e ) has the clamed value of e 2) prmary results In ths secton, we prove Theorem.2, whch s a major dfference between the stuaton when p = 2 and the odd prmes. We also state and prove a theorem, 4.0, about 2-adc ntegers x wth nfntely many, but sparse, 0 s for whch fx) s not 2-defnable. Our proof of Theorem.2 wll use the followng proposton. Proposton 4.. For n, ν 2 n j= 2j ) = 2ν2 n). Proof. The result when n s odd follows from the result for n, snce we are addng a number wth ν 2 = 0 to one wth ν 2 > 0. We wll use that n 4.2) ν 2 2j ) = ν2 σ n m 2m +, 2m + 3,..., 2n )), j=m+ where σ ) denotes an elementary symmetrc polynomal, and ts arguments are consecutve odd ntegers. The followng lemma about these wll be useful. We wll prove t after completng the proof of the proposton.

13 x BE DEFINED? 3 Lemma 4.3. For e, FOR WHICH p-adic INTEGERS x CAN 4.4) σ 2 e 2 e ), 2 e 3),...,,,..., 2 e 3, 2 e ) = 0, whle for e 2, 4.5) ν 2 σ 2 e 2 2 e ), 2 e 3),...,,,..., 2 e 3, 2 e )) = e. We frst prove the proposton when n = 2 e. Note that, mod 2 2e+, σ 2 e 2 e 2 e ),..., 2 e, 2 e +,..., 2 e + 2 e )) σ 2 e 2 e ),...,,,..., 2 e ) +2 2 e σ 2 e 2 2 e ),...,,,..., 2 e ) e σ 2 e 3 2 e ),...,,,..., 2 e ). The factors of 2 and 3 occur snce when factors are omtted, there are ways that the frst omsson could have been chosen. The thrd term s 0 mod 2 2e+ snce σ 2 e 3 2 e ),...,,,..., 2 e ) s the sum of ) 2 e odd numbers, and 2 e ) s even. By Lemma 4.3, the frst of the three terms s 0 and the second has ν 2 = 2 2e, mplyng the proposton when n = 2 e. We complete the proof by showng that valdty for n = 2 e 2a ) mples valdty 2 e 3 for n = 2 e 2a + ). Ths wll be done by showng 4.6) ν 2 2 e+2 a 2 e+ ) + + ) > 2e. 2 e+2 a + 2 e+ ) Ths s a sum of recprocals of consecutve odd ntegers. The LHS s ν 2 σ), where, mod 2 2e+4, σ = σ 2 e+ 2 e+2 a 2 e+ ),..., 2 e+2 a + 2 e+ )) 2 2 e+2 aσ 2 e+ 2 2 e+ ),...,,,..., 2 e+ ), argung smlarly to the prevous paragraph. By 4.5), ν 2 σ) e e. Proof of Lemma 4.3. These are the coeffcents of x and x 2 n 2 e 3 2 e x 2 2j ) 2 ). Thus 4.4) s clear, and the LHS of 4.5) equals ν 2 σ 2 e 2, 3 2,..., 2 e ) 2 )). We prove by nducton on e that ths equals e. It s easly checed when e = 2. Snce j=

14 4 DONALD M. DAVIS all arguments are odd, we need ν e ) 2 ) = e. Mod 2 e, /j 2 /2 e j) 2. Thus we need ν e ) 2 ) = e 2, and ths s the nducton hypothess. Our next step toward the proof of Theorem.2 s Proposton 4.7. If e 3, then ν 2 f2 e )) 2e. In fact, we conjecture that ν 2 f2 e )) = 3e 2 for e 4, but ths seems a good bt harder to prove, and not much more useful n provng somethng le Theorem.2. Proof of Proposton 4.7. For j 2 e 2, let p e,j := ) 2 e 2j + 2 ) e 2j + ) j+ 2 e ) j + 2 e j 2j )! = 2 e ) 2 e 2j )) 2 e 2 e 2j + ) j+ j )! 2 e ) 2 e j )) 2 e 2 e j = 2 e 2 e j 2 e j ) ) ) ) j 2 e ) 3 2e ) 2j 2e ) + ) )j+ = u 2 e ν 2j) + 2 e j ) + 22e A ) = u 2 2e ν 2j) j + 2e A). Here ν 2 u) = 0, and A Z 2 snce t s a combnaton of elementary symmetrc polynomals whose arguments are fractons wth odd denomnators. The dots n the second double) lne range over all ntegers n the range, whle n subsequent lnes the dots range over odd ntegers n the range. In gong from the second lne to the thrd, we have noted that the even factors n the frst fracton, after dvdng by 2, gve the factors of the second fracton. In gong from the thrd lne to the fourth, we have used that ) 2 e j s odd. Usng Proposton 4. and that e > ν2 j), we obtan ν 2 p e,j ) 2e.

15 x BE DEFINED? 5 FOR WHICH p-adic INTEGERS x CAN Smlarly we have p e,md := ) 2 e 2 e 2 e 2 e 2 = 2 e 2 e 2 ) ) 2 e ) 3 2e ) 2 e 2e ) ) = u 2 e e ) + 22e A ), satsfyng ν 2 p e,md ) 2e. Thus On the other hand, 2 e 2 j= p e,j = 2e 2 ν 2 j= 2 e 2 2 = p e,j + p e,md ) 2e. 2 e + 2 e 2 e 2 ) + 2 e 0 ), snce most of the terms wth alternatng sgns cancel. Hence 2 e 2 j= p e,j + p e,md = 2 e =0 2 ) e, whch s 2 f2e ). Thus ν 2 f2 e )) 2e, as clamed. Proof of Theorem.2. We wll use Proposton 4.7 and 2.4) to prove for 0 and 2 e >, 4.8) ν 2 f2 e + ) f)) e. Then, wth x as n Theorem.2, let E = 2 e. The hypothess of the theorem and 4.8) mply that for N, ν 2 fe ) fe )), and so dfe ), fe )) 2. Thus fe ) s a Cauchy sequence and so has a lmt n Q 2. Thus fx) s defnable. =0 Now we prove 4.8). Let e be fxed, and ) = f2 e + ) f). Usng Proposton 4.7, let f2 e ) = A 2 2e wth A Z 2. Usng 2.4) we obtan 0) = A2 e + )2 e and ) = 2e + + 2e ) f ). 22 e + ) 2 e + )

16 6 DONALD M. DAVIS Applyng ths teratvely, we obtan 4.9) ) = 2 e + + )A2 e j=0 2 e+j fj) j + )2 e + j + 2)2 e + j + ) ). Thus t suffces to prove ν 2 fj)) j + ν 2 j + ) 2 j + 2)). Ths can be easly checed for j 3. It s sharp for j = 0 and j = 2.) Snce ν 2 j ) [log2 j)], we deduce ν 2 fj)) [log 2 j)]. Snce j + [log 2 j)] + 2ν 2 j + ) + ν 2 j + 2) for j 4, as s easly proved, the desred result follows. Note how a comparson of 4.8) and Proposton 3.3 ponts to a huge dfference between the stuatons when p s an even or odd prme. The followng theorem descrbes 2-adc ntegers x wth nfntely many 0 s n ther bnary expanson, whch are eventually very sparse, for whch fx) s not 2-defnable. Theorem 4.0. Suppose x = 0 Then fx) s not 2-defnable. Proof. Let E = 2 e Theorem.2 e 2 e + e + e 0. 2 e wth e 0 > 0 and for all > 0 2 e. These are some of the partal sums for x. By =0 ν 2 fe )) = 2 e + e 0 e. =0 From ths, we deduce that ν 2 fe )) < ν 2 fe )) and hence ν 2 fe ) fe )) = ν 2 fe )). Thus dfe ), fe )) > dfe ), fe 2 )). These dstances form an ncreasng sequence of 2-powers, and so fe ) cannot be a subsequence of a Cauchy sequence. 5. Proofs relevant to the case ν p fc )) = 2 In ths secton, we prove Proposton 3. and Lemma 3.6. The followng lemmas wll be useful n the proof of Proposton 3..

17 x BE DEFINED? 7 FOR WHICH p-adic INTEGERS x CAN Lemma 5.. If fc ) 0 p), then c ) c 0 p). = Proof. We wll prove the stronger result that ) c = c )fc ). Ths s 2 deduced from the followng, where we use 2.4) n the last step. fc ) + ) c = c c + = cfc) ) = c c+fc ). 2c Lemma 5.2. If 0 C A < p and 0 D B < p, then, mod p 2, 5.3) Ap+B A B Cp+D) C) D) p A B C D C) D = B D +A C) = A B Proof. Snce, for any j, pj + p ) pj + ) p )! mod p 2, n evaluatng Cp + D)!A C)p + B D))! 5.4) mod p 2, Ap + B)! all the products whch appear n between two multples of p cancel out. We cancel out a factor of p from all multples of p n the numerator and denomnator of 5.4) and obtan that 5.4) s congruent to C!A C)! A! Cp + D) Cp + )A C)p + B D)) A C)p + ). Ap + B) Ap + ) = ). The second factor here equals D!B D)! B! E, where E = C p + ) C A C A C p + ) p + ) p + ) D B D Ap + ) Ap + ). B The LHS of 5.3) s congruent, mod p 2, to A B E C) D) ), and ths s congruent to the clamed expresson. Proof of Proposton 3.. We are assumng that c that for j p 5.5) c =0 2, p + 2j cp p+2j ) 2j c ) p 2j ) ) 0 mod p 2. 0 mod p. We wll prove

18 8 DONALD M. DAVIS Snce cp cp cp + = p + 2j p + 2j p + 2j) cp ), p+2j when 5.5) s summed over j and multpled by cp, we obtan p cp c c 5.6) c ) mod p 3. 0 p) = p By Lemma 2.8 cp c 5.7) 0 mod p 3. pj j =0 Addng 5.6) and the sum over j of 5.7) and usng Lemma 2.9 yelds the clam of the proposton. The proof of 5.5) begns by notng that the LHS equals c cp 2 5.8) cp p + 2j p ) c c ) p 2. =0 2j =0 Snce we are worng mod p 2 and assumng ) c 0 mod p, we may replace the p ) n ts coeffcent by. We wll show 5.9) c cp 2 p 2 c c + 0 mod p 2. p + 2j 2j =0 =0 =0 c Ths mples that cp 2 p+2j 0 p), and so whether ts coeffcent s or cp s rrelevant. Thus 5.9) mples 5.5). To prove 5.9), we note that tmes ts LHS s a sum of terms of the form of m 5.3) wth A = c, B = p 2, C =, and D = 2j. Let S m :=. By Lemma 5.2, the LHS of 5.9) equals tmes p ) p 2 c c ) S2j + c )S 2j p 2j c )S p 2 =0 = p ) p 2 c c K 2j + L), =0 =

19 FOR WHICH p-adic INTEGERS x CAN x BE DEFINED? 9 where K and L do not depend on. Ths s 0 mod p 2, usng Lemma 5. and the assumpton that ν p fc )) > 0. The remander of the paper s devoted to the proof of Lemma 3.6. Let D e := fcp e + up e ) fup e ). We wll show that ν p D e+ D e ) 2 for e. Usng 3.5), one easly shows that ν p D ). We deduce that ν p D e ) for all e. Let L e := <up e ) cp e + up e ) up e ), and H e := up e Then D e = L e + H e. We wll prove ν p L e+ L e ) 2 and ν p H e+ H e ) 2. We have 5.0) 5.) 5.2) = L e+ L e up e ) cp e+ + up e cp e + up e p =0 ) up e ) up e )) p <up e ) cp e+ + up e ) up e ) +. 0 p) ) cp e + up e Summands of 5.0) and 5.) are dvsble by p 2 by Lemma 2.8. Smlarly to the begnnng of the proof of 2.6, the sum 5.2) s a sum of terms of the form ) cp e+ + up e cp e+ + up e ) 5.3) upe up e 2j 2j 2j 2j wth j 0 mod p. Ths s clearly dvsble by p 2 when e >. If e =, the cp e+ -part s 0 mod p 2, and ) cp e+ +up e 2j up ) e 2j mod p, whch s good enough, snce there s an addtonal factor of p. Thus 5.3) s 0 mod p 2 n ths case, too.

20 20 DONALD M. DAVIS Fnally, 5.4) = 5.5) H e+ H e cp e +up e ) cp e+ + up e cp e + up e p + =up e up e 0 p) ) cp e+ + up e. The summands of 5.4) are dvsble by p e+. To see ths, we use the proof of Lemma 2.8 but note that the factor cp e +up e ) wll have νp ) =. Smlarly to the proof of 3., 5.5) s a sum over j of cp e+ + up e cp e +up e ) cp e+ + up e 2. cp e+ + up e p + 2j =up e Snce the terms n the sum certanly have ν p ), we are reduced to provng cp+u ) cp 2 + up 2 5.6) 0 mod p, and p + 2j =u cp2 +up 5.7) ν p =up We begn by provng 5.6). ) cp 3 + up 2 ) 2 0. p + 2j We splt the sum as c u ) cp 2 c p + u )p + p 2) + ap 2 + bp + 2j a= b=0 The frst sum, mod p, splts as c a= a=0 b=u b=0 ) cp 2 + u )p + p 2. ap 2 + bp + 2j c u a u b p 2. 2j ) The frst factor s fc) = fc ) c+ 0 mod p 2, and the other factors are p-ntegral. Wth 2c m S m = and l = 2j, we can prove, usng methods smlar to those n the proof = of 3., that the term n the second sum above s congruent, mod p to cp+u p 2 ap+b l + pbsl S p 2 l ) + u )S p 2 l S p 2 ))).

21 FOR WHICH p-adic INTEGERS x CAN x BE DEFINED? 2 Thus the second sum becomes, mod p, c p 5.8) α a=0 b=u wth α and β p-ntegral. cp+u c ap+b + pβ p a=0 b=u b ) cp+u ap+b We can prove that when b u, mod p 2, p ) cp+u ap+b u c b p +pasb c a u) b u) +c a )S u b+p cs u )), wth S m as above. Thus p tmes 5.8) s congruent, mod p 2, to c A a=0 c c a + Bp a=0 a ) c, a wth A and B p-ntegral. Snce ν p fc )) = 2 and usng Lemma 5., we obtan that ths expresson s dvsble by p 2 and hence 5.8) s dvsble by p, yeldng the clam. The proof of 5.7) s very smlar. In fact, the sum here s 0 mod p, stronger than requred. If we wrte the sum as p cp+u ) cp 3 + u )p 2 + p 2 2) tp =0 t=u 0 p + l) wth l = 2j as before, then ) p p+l behaves here as p 2 ) l dd before, and the t-sum s splt n the same way as the -sum before. References [] G. Esensten, Ene neue Gattung zahlentheoretscher Funtonen, welche von zwe Elementen abhängen und durch gewsse lneare Funtonal-Glechungen defnrt werden, Berchte der Könglch Preuschen Aademe der Wssenschaften zu Berln 5 850) [2] A. M. Rocett, Sums of the nverses of bnomal coeffcents, Fbo. Quar. 9 98) [3] B. Sury, Sum of the recprocals of the bnomal coeffcents, Eur. Jour. Comb ) [4] Wpeda artcle on Weferch prmes. [5] J. Wolstenholme, On certan propertes of prme numbers, Quar. Jour. Math ) [6] J.-H. Yang and F.-Z. Zhao, Certan sums nvolvng nverses of bnomal coeffcents and some ntegrals, Jour. Int. Sequ ) -. [7], The asymptotc expansons of certan sums nvolvng nverse of bnomal coeffcent, Int. Math. Forum 5 200)

22 22 DONALD M. DAVIS Department of Mathematcs, Lehgh Unversty, Bethlehem, PA 805, USA E-mal address: