On the Divisibility of Binomial Coefficients

Transcription

1 On the Dvsblty of Bnomal Coeffcents Sílva Casacuberta Pug Abstract We analyze an open problem n number theory regardng the dvsblty of bnomal coeffcents. It s conjectured that for every nteger n there exst prmes p and r such that f 1 k n 1 then the bnomal coeffcent ( n s dvsble by at least one of p or r. We prove the valdty of the conjecture n several cases and obtan nequaltes under whch the conjecture s satsfed. We relate the problem to Cramér s, Oppermann s and Remann s conjectures on prme gaps and study cases n whch the conjecture s true usng three prmes nstead of two. We also establsh four upper bounds on the mnmum number of prmes needed for the conjecture to be true. 1 Introducton Apart from ther many uses n varous felds of mathematcs, bnomal coeffcents dsplay nterestng dvsblty propertes. Kummer s [8] and Lucas [10] Theorems are two remarkable results relatng bnomal coeffcents and prme numbers. Kummer s Theorem provdes an easy way to determne the hghest power of a prme that dvdes a bnomal coeffcent, and Lucas Theorem yelds the remander of the dvson of a bnomal coeffcent by a prme number. Davs and Webb [4] found a generalzaton of Lucas Theorem for prme powers. Legendre [9] found two expressons for the largest power of a prme p that dvdes the factoral n! of a gven nteger n. However, some conjectures about bnomal coeffcents stll reman unproven. We focus on the followng condton consdered by Shareshan and Woodroofe n a recent paper [13]: Condton 1. For a postve nteger n, there exst prmes p and r such that, for all ntegers k wth 1 k n 1, the bnomal coeffcent ( n s dvsble by at least one of p or r. Ths condton leads to the followng queston: Queston 1.1. Does Condton 1 hold for every postve nteger n? In [13] t s conjectured that Condton 1 s true for all postve ntegers, yet there s no known proof. Shareshan and Woodroofe construct a chan of mplcatons n group theory that lead to Condton 1. We tackle ths problem usng manly number theory, although some lnks wth group theory are made. We also ntroduce the followng varaton of Condton 1, whch we study later n ths paper: 1

2 Defnton 1.2. A postve nteger n satsfes the N-varaton of Condton 1 f there exsts a set consstng of N dfferent prmes such that f 1 k n 1 then the bnomal coeffcent ( n s dvsble by at least one of the N prmes. Ths paper s organzed as follows. After provdng background nformaton n Secton 2, we prove that n satsfes Condton 1 f t s a product of two prme powers and also f t satsfes a certan nequalty regardng the largest prme smaller than n. Next we provde bounds related to the prme power dvsors of n and dscuss several cases n whch n satsfes Condton 1 dependng on the largest prme smaller than n/2. In Secton 6 and Secton 7 we use prme gap conjectures n order to settle some cases n whch a suffcently large nteger n satsfes Condton 1, and dscuss cases n whch n satsfes the 3-varaton of Condton 1. Fnally, n Secton 8 we provde upper bounds for a number N so that all ntegers n satsfy the N-varaton of Condton 1, followed by computatonal results and a generalzaton of Condton 1 to multnomals. 2 Background Three theorems about dvsblty of bnomal coeffcents and factorals are relevant for the proofs gven n ths paper. Theorem 2.1. (Kummer [8]) Let k and n be ntegers wth 0 k n. If α s a postve nteger and p a prme, then p α dvdes ( n f and only f α carres are needed when addng k and n k n base p. Theorem 2.2. (Lucas [10]) Let m and n be postve ntegers, let p be a prme, and let m = m k p k + m k 1 p k m 1 p + m 0 and n = n k p k + n k 1 p k n 1 p + n 0 be the base p expansons of m and n respectvely. Then ( m n) k ( m ) =0 n (mod p). It s mportant to notce that by conventon ( m n) = 0 f m < n. Hence, f any of the dgts of the base p representaton of m s 0 whereas the correspondng dgt of the base p representaton of k s not 0, then ( ) m k s dvsble by p because everythng s multpled by zero and by Lucas Theorem we have that ( ) m k 0 (mod p). Theorem 2.3. (Legendre [9]) If v p (n) denotes the maxmum power α of p such that p α dvdes n, then v p (n!) = n. k=1 p k Here x denotes the nteger part of x. Moreover, Legendre also showed that v p (n!) = n S p(n), p 1 where S p (n) denotes the sum of all the dgts n the base p expanson of n. 2

3 3 Some cases of n satsfyng Condton When n s a prme power Proposton 3.1. A postve nteger n satsfes the 1-varaton of Condton 1 wth p f and only f n = p α for some α > 0, for α N. α zeroes {}}{ Any k Proof. If n = p α, then the base p representaton of n s equal to 1 such that 1 k n 1 has at most α 1 zeroes n base p. Therefore, at least one of the dgts of the base p representaton of k s bgger than the correspondng dgt of ( n n base p (at least the leadng one). It then follows from Lucas Theorem that n ) k s dvsble by p. Otherwse, f n s not a prme power, then the th dgt of n n base p s not 0 for some value of. Thus, we can fnd at least one k such that the th dgt of k n base p s larger than 0. Hence, by Lucas Theorem ( n s not dvsble by p. Corollary 3.2. If n = p α + 1, then n satsfes Condton 1 wth p and any prme factor of n. Proof. The proof reles on the fact that ( ) ( m k + m ) ( k+1 = m+1 k+1) for all postve ntegers ( m and k. If m s a power of a prme p, then t follows from Proposton 3.1 that m ) ( k and m ) ( are dvsble by p f 1 k m 1. In these cases, because m+1 k+1 k+1) s the result of the sum of two multples of p, t also s a multple of p. When k = 1 or k = m, we have that ( ) m+1 k = m + 1, so any prme factor of m + 1 dvdes t. Corollary 3.3. If H s a proper subgroup of the alternatng group A n and n s a power of a prme p then the ndex [A n : H] s dvsble by p. Proof. As observed n [13], t s enough to prove ths clam when the subgroup H s maxmal, and n ths case, the ndex s ether ( n for some k or a multple of t, as shown n [13]. 3.2 When n s a product of two prme powers Proposton 3.4. If a postve nteger n s equal to the product of two prme powers p α 1 and p β 2, then n satsfes Condton 1 wth p 1 and p 2. Proof. Observe that lcm(p α 1, p β 2) = n. The base p 1 representaton of n ends n α zeroes and the base p 2 representaton of n ends n β zeroes. Because any postve k smaller than n cannot be dvsble by both p α 1 and p β 2, t s not possble that k fnshes wth α zeroes n base p 1 and β zeroes n base p 2. Thus, we can apply Lucas Theorem modulo the prme p 1 f p α 1 k or modulo the prme p 2 f p β 2 k. Corollary 3.5. Let n = p α 1 p β 2 and H be a proper subgroup of A n. Then the ndex [An : H] s dvsble by at least one of p 1 or p 2. 3

4 3.3 Consderng the closest prme to n Theorem 3.6. Let q be the largest prme smaller than n and let p a be any prme factor dvsor of n. If n q < p a, then n satsfes Condton 1 wth p and q. For the proofs of Theorem 3.4 and Corollary 3.6 we use the Bertrand-Chebyshev Theorem: Theorem 3.7. (Bertrand-Chebyshev [1]) For every nteger n > 3 there exsts a prme p such that n/2 < p < n. Proof of Theorem 3.4. We dstngush between two ntervals: the nterval (1, n q] and the nterval (n q, n]. Due to the symmetry of bnomal coeffcents, we only consder k n/2. By the Bertrand-Chebyshev Theorem, we know that there s at least one prme between n/2 and n, hence n/2 < q < n. Then, for all k, k < n/2 < q. The base q representaton of n s 1 q + (n q). Therefore, we do not need to consder the nterval (n q, n) because the last dgt of the base q representaton of any k > n q s larger than the last dgt of the base q representaton of n. Thus, by Lucas Theorem, the bnomal coeffcent ( n s dvsble by q. If there s no multple of p a n the nterval (1, n q), then by Lucas Theorem all the bnomal coeffcents ( n wth 1 k n/2 are dvsble by at least p or q. Moreover, equalty n Theorem 3.4 cannot hold because p a dvdes both p a and n, and hence q would not be a prme. Corollary 3.8. Let p a j j denote the largest prme power dvsor of an nteger n and q the closest prme to n. If n q < p a j j, then n satsfes Condton 1 wth p j and q. Note that f n satsfes Condton 1 then at least one of these two prmes has to be a prme factor of n, because otherwse ( n 1) = n s not dvsble by ether one of the two prmes. The only remanng cases are those n whch n q > p a and n s nether a prme nor a prme power. Let q 2 denote the largest prme smaller than n/2. By analyzng the ntegers that are part of these remanng cases, we notce that n usually satsfes Condton 1 wth the par formed by a prme factor of n and q 2. If we analyze the sx numbers smaller than 2,000 such that n q > p a, we see that the nequalty p a > n 2q 2 holds and q 2 and p satsfy Condton 1. Table 1 provdes evdence wth the only four numbers untl 1,000 that do not satsfy Condton 1 wth q and p. However, the sequence of all such ntegers s nfnte. The On-Lne Encyclopeda of Integer Sequences (OEIS) has accepted our submsson of ths sequence [2] wth the reference A

5 Number Prme factorzaton q q n q n 2q (1, n q] 2, 3, 4 2, 3, 4 2, 3, 4 2, 3, 4 (n q, n] 62, , , , 315 Pars that satsfy , 5-313, Table 1: Informaton about the four numbers below 1,000 that do not satsfy Condton 1 wth q and p. 4 Bounds for p a Before analyzng q and q 2 further, we establsh some bounds for p a n q > p a. Lemma 4.1. If n s not a prme and n q > p a, then pa < n/2. assumng that Proof. Usng the Bertrand-Chebyshev Theorem we see that n/2 > n q > 0. Also, n q > p a. Therefore, n/2 > pa. We can fnd an even lower bound for p a. In 1952, Nagura [11] showed that f n 25 then there s always a prme between n and (1 + 1/5)n. Therefore, we fnd that 5n/6 < q < n when n 30. Lemma 4.2. If n 30 s not a prme and n q > p a, then pa < n/6. The proof s the same as the one for Lemma 4.1. In 1976 Schoenfeld [14] showed that for n 2,010,760 there s always a prme between n and (1 + 1/16,597)n. Therefore, we know that f n > 2,010,882 then 16,597n 16,598 < q 2 < n. Shareshan and Woodroofe [13] checked computatonally that all ntegers smaller than 10 mllon satsfy Condton 1, whch means that we can apply Schoenfeld s bound. Lemma 4.3. If n s not a prme, n > 2,010,882 and n q > p a, then pa < n/16,598. The proof follows the same steps as the prevous two lemmas. Proposton 4.4. Let n = p a m. If n 2,010,882 and m < 16,598, then n satsfes Condton 1 wth p and q. Proof. By Schoenfeld s bound we know that n q < n/16,598. If m < 16,598, t means that p a > n/16,598. Thus, p a > n q and, by Theorem 3.4, q and p satsfy Condton 1. 5

6 5 When n q > p a > n 2q 2 In ths secton we analyze the ntegers n that satsfy the nequaltes n q > p a > n 2q 2, and we prove some cases n whch n satsfes Condton 1 wth p and q 2. The fact that we are consderng n 2q 2 comes from the base q 2 representaton of n. We dstngush between two cases: when k < q 2 and when k > q 2. The base q 2 representaton of n s 2 q 2 + (n 2q 2 ). The base q 2 representaton of k s 0 q 2 + k f k < q 2 and 1 q 2 + (k q 2 ) f k > q 2. Hence, there s no need to analyze the nterval (n 2q 2, q 2 ] because for all k such that n 2q 2 < k q 2, we can use Lucas Theorem to see that the bnomal coeffcent ( n s congruent to 0 modulo q2. Therefore, we only need to consder the nterval (q 2, n/2]. 5.1 If n s odd It s mportant to remark that f k s not a multple of p a ) then by Lucas Theorem s dvsble by p. Therefore, we only have to analyze the ntegers n (q 2, n/2] ( n k that are multples of p a. We then clam the followng: Theorem 5.1. If n s odd and n q > p a > n 2q 2, then n satsfes Condton 1 wth p and q 2. Proof. Snce n s odd, n/2 s not an nteger. Hence t s enough to prove that there s no multple of p aq q n the nterval (q r, n/2). We wll prove ths by contradcton. Thus assume that q r < λp aq q < n/2 for some nteger λ. Then λ (m 1)/2 f n = mp aq q, snce ((m 1)/2)p aq q s the largest multple of p aq q that s smaller than n/2 (note that m s odd because n s odd). Now from the nequalty ((m 1)/2)p aq q > q r t follows that n p aq q > 2q r and ths contradcts the assumpton that n 2q r < p aq q. 5.2 If n s even and p s not 2 Lemma 5.2. If n s even and p q 2, then the only multple of p aq q (q r, n/2] s n/2. n the nterval Proof. Snce p q 2, the nteger n/2 s stll a multple of p aq q. Hence we may wrte n/2 = λp aq q for some nteger λ. If there s another multple of p aq q between q r and n/2, then we have q r < (λ 1)p aq q < n/2, and ths mples that n/2 p aq q > q r. Hence n 2q r > 2p aq q > p aq q, whch s ncompatble wth our assumpton whch states that n 2q r < p aq q. Theorem 5.3. If 2 α s a prme power dvsor of n and 2 α satsfes n q > 2 α > n 2q 2, then n satsfes Condton 1 wth 2 and q 2. Proof. The nteger n has the factor 2 α n ts prme factorzaton, whch means that n/2 has the factor 2 α 1. The base 2 representaton of n has one more zero than the base 2 representaton of n/2, whch means that by Lucas Theorem ( n n/2) s congruent to 0 modulo 2 α. By Lemma 5.2, n/2 s the only multple of 2 α n the nterval (q 2, n/2]; hence the proof s complete. 6

7 5.3 If n s even and p s not 2 By Lemma 5.2 we only need to consder the central bnomal coeffcent ( n n/2), because the only multple of p a n the nterval (q 2, n/2] s n/2. We clam the followng proposton, usng Legendre s Theorem for ts proof. Proposton 5.4. The prme factor p dvdes ( n n/2) f and only f at least one of the fractons n/p α wth α 1 s odd. Proof. When we compare v p (n!) and v p ((n/2)!) we see that, for each α, n n/2 = 2 p α f n/p α s even. If n/p α s even for all α, we conclude that v p (n!) = 2v p ((n/2)!), and hence p does not dvde ( n n/2). However, f n/p α s odd, then p α n n/2 = Therefore, v p (n!) s greater than 2v p ((n/2)!). p α p α Corollary 5.5. Let S p (n) be the base p representaton of n. If n Sp(n) p 1 p dvdes ( n n/2). Corollary 5.5 s shown usng Legendre s formula. s odd then Corollary 5.6. If any of the dgts n the base p representaton of n/2 s larger than p /2, then the bnomal coeffcent ( n n/2) s dvsble by p. Corollary 5.7. If one of the dgts n the base p representaton of n s odd, then the prme p dvdes ( n n/2). Proof. The proofs of Corollares 5.6 and 5.7 are smlar. If a dgt of n/2 s larger than p /2, when we add n/2 to tself n base p to obtan n there at least one carry. Smlarly, f n has an odd dgt n base p, t means that there has been a carry when addng n/2 and n/2 n base p. By Kummer s Theorem wth k = n/2, f there s at least one carry when addng n/2 to tself n base p, then p dvdes the bnomal coeffcent ( n n/2). log(n) Proof. The largest α such that p α < n < p α+1 log(p Corollary 5.8. If p ) > n/2 and n q > p a > n 2q 2, then p dvdes ( ) n n/2 and therefore n satsfes Condton 1 wth p and q 2. s. Therefore, n Proposton 5.4, α s bounded by 1 α log(n) log(p ) exponent ( of p. If p n n/2) by Proposton 5.4. log(n) log(p ) log(n) log(p ). Also note that α a, where a s the > n/2 then n/p α = 1. Because ths s odd, p dvdes 7

8 5.4 Some cases n whch 2n mples n In ths secton we denote by p a k and q k any prme power factor of k and the largest prme smaller than k respectvely. For ntegers that satsfy the nequalty n q > p a > n 2q 2, we observe three cases n whch f 2n satsfes Condton 1 and p 2n 2, then n also satsfes Condton 1. Note that snce p s not 2, then p 2n = p n. Also, q 22n = q n. Therefore we clam: Clam 5.9. If 2n satsfes the nequalty 2n 2q 22n < 2n q 2n < p a 2n 2n, then n satsfes Condton 1 wth p and q. Proof. We rewrte the nequalty above as n q n < 2(n q n ) < 2n q 2n < p a n n. Therefore, n q n < p a n n, and, by Theorem 4.2, n satsfes Condton 1 wth the prmes p and q. Clam If 2n satsfes the nequalty 2n q 2n < 2n q 22n < p a 2n 2n, then n satsfes Condton 1 wth p and q. Clam If 2n satsfes the nequalty 2n 2q 22n satsfes Condton 1 wth p and q. < p a 2n 2n < 2n q 2n, then n The proofs of Clams 5.10 and 5.11 follow the same steps as the one of Clam Large multples of n satsfyng Condton 1 wth prme gap conjectures In ths secton we always denote the t th prme as ˆp t. 6.1 Cramér s Conjecture Conjecture 6.1. (Cramér [5]) There exst constants M and N such that f ˆp t N then ˆp t+1 ˆp t M(log ˆp t ) 2. We clam the followng: Proposton 6.2. If Cramér s conjecture s true, then for every postve nteger n and every prme p dvdng n, the number np k satsfes Condton 1 for all suffcently large values of k. Proof. Let M and N be the constants gven by Cramér s conjecture. Gven a postve nteger n whch s not a prme power and a prme dvsor p of n, we wrte n = mp a where p does not dvde m, and compare M(log nx) 2 wth p a x as x goes to nfnty. Usng L Hôptal s rule, we fnd that lm x p a x M(log nx) 2 = lm x p a nx 2Mn log nx = lm x = lm x p a nx 2Mn = lm x p a x 2M =. p a x 2M log nx 8

9 Therefore, p a x s bgger than M(log nx) 2 when x s suffcently large. Hence we can choose any k large enough so that p a+k > M(log np k ) 2 and furthermore, f q denotes the largest prme smaller than np k, then q N. Now, f r denotes the smallest prme larger than np k, we nfer that, f Cramér s conjecture holds, then, snce q N, Moreover np k q r q M(log q) 2. M(log q) 2 < M(log np k ) 2 < p a+k. Hence np k q < p a+k and, snce p a+k s the hghest power of p dvdng np k, Theorem 3.4 mples that np k satsfes Condton 1. Cramér s conjecture also proves the followng proposton: Proposton 6.3. Let m denote the number of dstnct prme factors of n. If Cramér s conjecture s true and n grows suffcently large keepng m fxed, then n satsfes Condton 1. Proof. If n has m dstnct prme factors, we defne the average prme factor of n as m n because f n were formed by m equal prme factors each one would equal m n. It s true that m n p a j j, where pa j j denotes the largest prme power dvsor of n. Hence we must see f M(log n) 2 < m n for large values of n. We apply agan L Hôptal s rule to compute the lmt lm x m nx M(log nx) 2 and we obtan that M(log n) 2 < m n holds when n s suffcently large. 6.2 Oppermann s Conjecture A weaker conjecture on prme gaps by Oppermann states the followng: Conjecture 6.4. (Oppermann [12]) For some constant M, ˆp t+1 ˆp t M ˆp t. Proposton 6.5. If Oppermann s conjecture s true, then for every postve nteger n and every prme p dvdng n, the number np k satsfes Condton 1 for all suffcently large values of k. Proof. The proof s smlar to the proof of Proposton 6.2. We apply L Hôptal s rule once to solve the ndetermnaton n lm x p a x M nx, where p a s the hghest power of p dvdng n. Snce the rato goes to nfnty our nequalty s satsfed, and by choosng x = p k wth k large enough the proof s complete. 9

10 6.3 Remann s Hypothess The followng conjecture s a consequence of Remann s Hypothess. Conjecture 6.6. (Remann [6]) For some constant M, ˆp t+1 ˆp t M(log ˆp t ) ˆp t. Ths bound can be used to prove the followng: Proposton 6.7. If Remann s conjecture s true, then for every postve nteger n and every prme p dvdng n, the number np k satsfes Condton 1 for all suffcently large values of k. Proof. We apply agan L Hôptal s rule to solve the ndetermnaton n lm x p a x M(log nx) nx The lmt goes to nfnty and hence, by choosng x = p k wth k large enough, the proof s complete. Due to the smlartes of the nequaltes, we skp the calculatons of Propostons 6.5 and Usng other prmes to satsfy Condton 1 In Sectons 3 and 5 we analyzed nequaltes nvolvng n q and n 2q 2. In general, for any postve nteger d, we can study the functon n dq d, where q d refers to the largest prme smaller than n/d (when wrtng q 1 we omtted the subndex 1). We consder the ntegers n that do not satsfy the nequalty p a > n 2q 2. Up to 1,000,000 there are only 88 ntegers that do not satsfy p a > n 2q 2. The On-Lne Encyclopeda of Integer Sequences (OEIS) has accepted our submsson of these numbers [3] wth the reference A Up to 1,000,000, there are 25 ntegers that do not satsfy the nequalty p a > n 3q 3 ; 7 ntegers that do not satsfy the nequalty p a > n 4q 4 ; 5 ntegers that do not satsfy the nequalty p a > n 5q 5, and only 1 nteger that does not satsfy the nequalty p a > n 6q 6. Fgure 1 shows the number of ntegers up to 1,000,000 that do not satsfy the nequalty p a > n dq d dependng on d. We also observe that the functon n dq d tends to 0 as d ncreases, whch means that t s lkely that at some pont the nequalty s acheved. Ths s explaned wth the propertes of the functon n/d, whch behaves n the same way as the functon 1/x except for the constant n. As d grows large, the dfference between n/d and n/(d + 1) grows smaller. Hence, the closest prme to n/d s the same one for all the n/d that are close. Then, when d ncreases, p d decreases much more slowly, and because t s multpled by d, whch grows lnearly, dp d tends to n. Fgure 2 shows how n dp d tends to 0 as d ncreases takng 330 as an example. All the ponts correspond to values of d such that p d satsfes Condton 1 wth another prme. Note that f n dp d s exactly zero then d s a dvsor of n such that n/d s a prme. Then there are two condtons that we use for p and q d to satsfy Condton 1. Condton 2. For any nteger n to satsfy Condton 1 wth p and q d we requre that p a > n dq d and n dq d < q d. 10

11 Fgure 1: Number of ntegers up to 1,000,000 that do not satsfy the nequalty p a > n dq d as a functon of d. Fgure 2: Decrease of n dp d for n = 330. When k s larger than p a, we rely on the fact that k s larger than n dq d to justfy that the bnomal coeffcent ( n s dvsble by qd usng Lucas Theorem unless f k s a multple of q d. However, f n dq d were larger than q d, when wrtng n n base q d the nequalty p a > n dq d would not hold. Lemma 7.1. If n 30 and d < 5, then n dq d < q d. Proof. By Lemma 4.2, f n 25, 5n/6d < q d < n/d. Therefore, n/6 > n dq d. 11

12 Now we need to show that q d > n dq d. It follows that n < q d + dq d and thus n < q d (1 + d). Usng Lemma 4.2, n < 5n(d + 1) 6d Therefore, 6d < 5d + 5 and we get that d < 5. < q d (1 + d). Lemma 7.2. If n 2,010,882 and d < 16,597, then n dq d < q d. The proof s the same one as the one for Lemma 7.1, except that by Lemma 4.3, the ntal nequalty s 16,597n/16,598d < q d < n/d. Corollary 7.3. The nteger d/2 q d s the largest multple of q d smaller than or equal to n/2. Proof. We apply the defnton of q d to obtan that n dq d. Assume, towards a contradcton, that n > q d (d+1). By Lemmas 7.1 and 7.2, n dq d < q d and therefore n < q d (d + 1). Ths contradcts the nequalty n > q d (d + 1). 7.1 The 3-varaton of Condton 1 In Secton 5 we proved that many ntegers that satsfy the nequaltes n q > p a > n 2q 2 also satsfy Condton 1 wth p and q 2. In ths secton we prove some cases n whch an nteger n satsfes the 3-varaton of Condton 1 (as stated n Defnton 1.2 n the Introducton). Theorem 7.4. If an even nteger n satsfes the nequalty n q > p a > n 2q 2 and p 2, then n satsfes the 3-varaton of Condton 1 wth p, q 2 and any prme that dvdes ( n n/2). Proof. In Secton 5.2 we show that f n satsfes the nequalty n q > p a > n 2q 2 and p s not 2, the only bnomal coeffcent we could not prove that was dvsble by ether p or q 2 s the central bnomal coeffcent. Thus, for such n to satsfy the 3-varaton of Condton 1 t suffces to add an extra prme that dvdes the central bnomal coeffcent Regardng the two hghest prme powers of n For any n, let q be the largest prme smaller than n, let p j be the prme factor of n such that p a j j s the largest prme power of n, and let p r be the prme factor of n such that p ar r s the second largest prme power dvsor of n. We then clam the followng: Proposton 7.5. If p a j j par r > n/6, then n satsfes the 3-varaton of Condton 1 wth p j, p r and q. 12

13 Proof. By Lucas Theorem, for any k such that 1 k p a j j, the bnomal coeffcent ( n ) k s dvsble by pj. For the same reason, by Lucas Theorem, for any k such that n q < k n/2 the bnomal coeffcent ( n s dvsble by pj. Then we need a prme that dvdes at least the bnomal coeffcents ( ) n a k wth p j j k n q such that k s a multple of p a j j. Now take p r as the thrd prme such that n mght satsfy the 3-varaton of Condton 1 wth p j, q and p r. For the same reasonng, n ths nterval we only consder the k that are multples of p ar r. The only k such that the bnomal coeffcent ( n s not dvsble by ether pj of p r are those k that are multples of both p a j j and p ar r. The least k that s multple of both prme powers s p a j j par r. By Lemma 4.2 we know that n q < n/6. Therefore, f p a j j par r > n/6, ths nteger s larger than n q and hence t s not part of the nterval that we are consderng. Thus, all the k lyng n the nterval p a j j k n q are such that the bnomal coeffcent ( n s dvsble by ether p j or p r. Moreover, usng the bounds descrbed n Lemma 4.2, we use the prmes p j, q and q d for n to satsfy the 3-varaton of Condton 1. Proposton 7.6. Let q d be the largest prme smaller than n/d. If q d > n/6, then n satsfes the 3-varaton of Condton 1 wth p j, q and q d. Proof. The prme q fals to dvde ( n only f 1 k n q. Smlarly, by Lucas Theorem, the prme q d fals to dvde ( n only f cqd k cq d + (n dq d ), where cq d refers to any postve multple of q d. Ths s because n dq d s the last dgt of the base q d representaton of n. But because by assumpton q d > n p j, the ntervals [1, n q] and [cq d, cq d + (n dq d )] are dsjont. 8 Bounds on the number of prmes needed to satsfy the N-varaton of Condton 1 For each postve nteger n, we are nterested n the mnmum number N of prmes such that n satsfes the N-varaton of Condton 1. In ths secton we provde four upper bounds for N. Because n all four bounds N s a functon of n, the sutablty of each bound depends on n; some bounds may be better for certan values of n. 8.1 Frst upper bound wth prme factors of n Clam 8.1. If n has m dfferent prme factors, then these prme factors satsfy the m-varaton of Condton 1. Proof. The proof s smlar to the one descrbed when n s a product of two prme powers. The smallest nteger dvsble by all the m prme powers of n s n. The base p representaton of all k < n has less zeroes than the base p representaton of n for at least one prme factor p of n. Usng Lucas Theorem, Clam 8.1 s proven. 13

14 8.2 Second upper bound wth d Proposton 8.2. Let q d be the largest prme smaller than n/d and let p a be any prme power dvsor of n such that p a > n dq d. If p a > q d + n dq d, then n satsfes the N-varaton of Condton 1 wth N = 2 + d/2. For the subsequent proofs we use the followng defnton: Defnton 8.3. Let cq d be any multple of q d and let β be n dq d. We call the nterval [cq d, cq d + β] a dangerous nterval. Note that for every tme that p a an extra prme. falls nto a dangerous nterval we need to add Proof. By Lucas Theorem all the bnomal coeffcents ( n are dvsble by qd except f k les n a dangerous nterval. In these dangerous ntervals we only consder the ntegers that are multples of p a because f k s not a multple of p a, then by Lucas Theorem the bnomal coeffcent ( ) n k s dvsble by p. Because p a > β we know that n any dangerous nterval there s at most one multple of p a. Ths means that the worst case s the one n whch there s a multple of p a n every dangerous nterval untl c d/2. Thus we need at most one extra prme each tme that there s a multple of p a n a dangerous nterval. Clam 8.4. If d < 5 and p a > q d + β, then n satsfes Condton 1 wth q d and p. Proof. If d < 5, then d/2 equals ether 1 or 2. If t equals one, then by assumpton p a > q d + β, whch means that no multple of p a falls n any dangerous nterval untl n/2. If d equals 2, then we need to check that 2p a > 2q d + β. Ths means that we want to see that the next multple of p a does not fall nto the second dangerous nterval. The mnmum value of p a such that our assumpton p a > q d + β holds s q d + β + 1. The next multple of q d + β + 1 s 2q d + 2β + 2. Ths last expresson s greater than 2q d + β, whch means that 2p a does not fall nto the second dangerous nterval. 8.3 Thrd upper bound In ths subsecton we consder the generalzaton of the cases that have been dscussed so far. Let d be a natural number and let q d be the largest prme number smaller or equal to n/d. Let β denote n dq d, let p a be any prme power dvsor of n, and let γ = p a cq d. In Sectons 8.3 and 8.4 we do not consder the cases n whch q d = p because the proofs hold by takng any other prme factor of n that s not p. Theorem 8.5. For all c 0, n satsfes the N-varaton of Condton 1 wth kγqd (c 1) N = 2 + β, γq d d where k =. 2q d γ 14

15 Proof. We frst consder the case n whch p a that p a form rp a = q d + γ and γ β. Ths means falls n the frst dangerous nterval. Any subsequent multple of p a s of the = rq d + rγ. Note that we only need to analyze rγ because ths s what falls n a dangerous nterval. determnes f p a Lemma 8.6. The prme power dvsor p a f rγ (mod q d ) β. falls nto a dangerous nterval f and only The proof of Lemma 8.6 comes from the defnton of a dangerous nterval (see Defnton 8.3). Now consder all the possble values of rγ modulo q d from γ untl γq d. Note the followng: Remark 8.7. The numbers γ and q d are always coprme. For the proof of the remark t suffces to see that q d s a prme number. Ths means that all the numbers from 1 to q d 1 appear exactly once n the nterval [γ, γq d ). Therefore, by Lemma 8.6 the number of ntegers that fall nto a dangerous nterval are those such that rγ (mod q d ) β. By Remark 8.7 we know there are only β such ntegers n the nterval [γ, γq d ). Thus, f γq d > d/2, we only need 2 + β prmes. We add 2 to β because we also need to count q d and p. Note that ths s an upper bound and therefore n some cases several of the prmes that we use for the dangerous ntervals are repeated. Now we consder the general case n whch p a = cq d + γ. We need to count the multples of γq d from cq d untl kγq d (k has the same defnton as n Theorem 7.1). Ths gves us the bound stated n Theorem 8.5. Note that, n Theorem 8.5, γ cannot be 0 because otherwse by defnton p would be equal to q d. Ths s a case that we are not consderng (see the begnnng of Secton 8.3). 8.4 Fourth upper bound wth Dophantne equatons We consder the Dophantne equaton p a k 1 q d α = δ, where 0 δ β. Let x = k 1 and let y = α. The general solutons of these Dophantne equaton dependng on the partcular solutons x 1 and y 1 are well-known: x = x 1 rq d y = y 1 + rp a Let ŷ(δ) denote the largest y d/2 dependng on δ. Note that for all y 1 (δ) we can add or subtract a certan number of p a untl we reach ŷ(δ). Theorem 8.8. All ntegers n satsfy the N-varaton of Condton 1 wth N = 2 + β ŷ(δ) δ=0 p a d 2 + (β + 1) 2p a. Proof. Note that the solutons of the Dophantne equaton correspond to all the cases n whch a multple of p a falls n some dangerous nterval. It s known that a Dophantne equaton ax+by = c has nfntely many solutons f gcd(a, b) dvdes c. 15

16 Therefore, for all δ such that 0 δ β there exsts a partcular soluton y 1 (δ) for y n Equaton 6 because gcd(p a, q d) = 1 (recall that we do not consder the case n whch p = q d ). Thus, for each ŷ(δ) we count the number of multples of p a n the nterval [1, ŷ(δ)]. Ths s the number of tmes that p a falls nto a dangerous nterval and hence we need to add an extra prme. We also add 2 to count p d and p. Moreover, note that by defnton ŷ(δ) d/2. Ths gves us the expresson stated n Theorem Computatonal results In order to obtan more nformaton about whch prmes make n satsfy Condton 1 we wrote some C++ programs. The results are presented n ths secton. 9.1 When we fx a prme In the orgnal artcle of Shareshan and Woodroofe [13], the authors computed the percentage of ntegers below 1,000,000 that satsfy Condton 1 f p 1 s fxed to be 2, and they found a percentage of 86.7%. We compute the percentage of ntegers untl 10,000 that satsfy Condton 1 fxng one prme to be not only 2 but also 3, 5, 7 and 11. Table 2 shows the number of ntegers below 10,000 that do not satsfy Condton 1 fxng one prme to be 2, 3, 5, 7 and 11 respectvely. It also shows the percentage of ntegers satsfyng Condton 1 fxng each prme. Fgure 3 shows the percentage of ntegers untl 10,000 that satsfy Condton 1 dependng on the fxed prme. Fxed prme Number of ntegers not satsfyng Percentage of ntegers satsfyng % 83.67% 73.74% 67.41% 58.20% Table 2: Number of ntegers that do not satsfy Condton 1 and percentage of ntegers that do satsfy Condton 1 fxng one prme untl 10, How many pars of prmes satsfy Condton 1 Gven a postve nteger n, multple pars of prmes p 1 and p 2 can satsfy Condton 1. We have found computatonally all the possble pars of prmes that satsfy Condton 1 wth a gven n 3,000. Ths fndngs helped us conjecture and then prove Theorem 3.4. Fgure 4 shows the data for n up to 3,000. We note four man tendences. The one wth the greatest slope corresponds to the one formed wth prme numbers and prme powers. Ths s explaned by Proposton 3.1. Because only one prme s needed to satsfy the 1-varaton of Condton 1 f n s a prme power, the other prme can be any prme smaller than n. Thus, ths frst tendency follows the functon f(n) = n/ log n dsregardng the prme powers [7]. The second greatest slope s formed wth even numbers that satsfy Condton 1 wth one prme beng 2. The thrd one s formed by numbers that satsfy Condton 1 wth one 16

17 Fgure 3: Percentage of ntegers untl 10,000 that satsfy Condton 1 fxng one prme to be 2, 3, 5, 7 and 11 respectvely. prme beng 3 and the followng one wth numbers that satsfy Condton 1 wth one prme beng 5. Fgure 4: Number of pars of prmes that satsfy Condton 1 dependng on the nteger n untl 3,000. In order to ft a functon for each curve, we approxmated the functon n/ log n for each branch usng Matlab, and we obtaned the followng functons: 17

18 Frst branch: 0.97n 0.96 (log n) 0.75 Second branch: 0.80n 0.96 (log n) 0.84 Thrd branch: 3.30n 1.14 (log n) 2.27 Fourth branch: 35.48n1.47 (log n) 4.81 Fgure 5 shows a plot of each separate branch wth ts correspondng curve. Fgure 5: The four branches of Fgure 4 separated and ftted wth a curve. 10 Multnomals We also consder a generalzaton of Condton 1 to multnomals. We nvestgate the followng condton that some nteger n mght satsfy: Condton 3. For a gven fxed nteger m there exst prmes p 1 and p 2 such that whenever k k m = n for 1 k n 1, ( ) n k 1,k 2,...,k m s dvsble by ether p1 or p 2. A very natural queston follows: Queston Does Condton 3 hold for all postve ntegers n? Here we show that Condton 1 mples Condton 3. We clam the followng: Proposton If n satsfes Condton 1 wth p 1 and p 2, then n also satsfes Condton 3 wth these two prmes and any m n. Proof. We assume that p 1 and p 2 satsfy Condton 1 for a gven n. We then take the multnomal ( ) n n! = k 1, k 2,..., k m k 1!k 2! k m! 18

19 wth the same n and any m n. We see that we can decompose the multnomal nto a product of m bnomals: n! k 1!k 2! k m! = n(n 1) (n k 1 + 1). k 1! (n k 1 )(n k 1 1) (n k 1 k 2 + 1) (k m 1 + k m )(k m 1 + k m 1) 1 k 2! (k m 1!k m! ( )( ) ( ) n n k1 1 km 1 + k m =. k 1 k 2 Because by assumpton ( ) m k 1 s dvsble by ether p1 or p 2, the prevous multnomal coeffcent s also dvsble by at least one of them. Ths decomposton can be used for any m and the frst bnomal coeffcent can be ( ) n k, k beng any of the k n the denomnator. Therefore, f Condton 1 s proven for bnomal coeffcents, then t automatcally holds for multnomal coeffcents. k m 11 Conclusons In ths paper we have obtaned results that sgnfcantly contrbute to the unsolved conjecture that motvated our research (see Condton 1 n the Introducton), whch was proposed n a recent artcle by Shareshan and Woodroofe [13]. After havng obtaned all these research results, we have analyzed how much we have contrbuted to the open problem addressed n ths paper. Up to 1,000,000, there are less than 50 numbers that do not ft nto any of the cases that we have solved. We consder ths to be a very substantal outcome. Moreover, our proofs concernng prme gap conjectures potentally have stronger mplcatons, as we beleve that we are very close to provng that all ntegers larger than a fxed constant satsfy Condton 1. Acknowledgements I would lke to express my grattude towards my mentor Oscar Mckeln, whose advce, help, and gudance have been absolutely crucal for hs work. I would also lke to thank my tutor Dr. John Rckert and Dr. Tanya Khovanova for ther valuable and nsghtful comments on ths project. I would lke to recognze Professors Dr. Davd Jerson and Dr. Slava Gerovtch for organzng and coordnatng the math mentorshps. I would also lke to thank Anq L, Mchelle Shen, Jordan Lee and Óscar Rvero for ther remarks. I would lke to acknowledge the Center for Excellence n Educaton, the Research Scence Program, and the Massachusetts Insttute of Technology for gvng me the once n a lfetme opportunty to conduct research durng the summer. Fnally, I would lke to recognze and thank Ms. Perceres, the Drector of Youth and Scence Programs, and Fundacó Catalunya la Pedrera for sponsorng me. 19

20 12 Appendx 12.1 Sequences of ntegers that do not satsfy the nequalty for n dp d In Secton 7 we mentoned that the set of ntegers that do not satsfy the nequalty for n dp d becomes smaller when d ncreases. In ths appendx we dsplay the frst terms of the sequence of ntegers that do not satsfy the nequalty n dp d < p a when d equals 1, 2, 3, 4 and 5. The On-Lne Encyclopeda of Integer Sequences has publshed our sequence n the cases when d equals 1 (see A290203) and when d equals 2 (see A290290). When d = 1: 126, 210, 330, 630, 1144, 1360, 2520, 2574, 2992, 3432, 3960, 4199, 4620, 5544, 5610, 5775, 5980, 6006, 6930, 7280, 8008, 8415, 9576, 10005, 10032, 12870, 12880, 13090, 14280, 14586, 15708, 15725, 16182, 17290, 18480, 18837, 19635, 19656, 20475, 20592, 22610, 24310, 25296, 25300, 25520, 25840, 27170, 27720, 27846, 28272, 28275, 29716, 30628, 31416, 31450, 31464, 31465, 32292, 34086, 34100, 34580, 35568, 35650, 35670, 35728, 36036, 36432, 37944, When d = 2: 3432, 5980, 12870, 12880, 13090, 14280, 14586, 20475, 28272, 28275, 31416, 31450, 34580, 35650, 39270, 45045, 45220, 72072, 76076, 96135, 97812, , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , , When d = 3: 3432, 31416, 34580, 35650, 39270, 96135, , , , , , , , , , , , , , , , , , , , , , , , , , , , , C++ code for fndng all the possble pars Here we provde the C++ code that we used to fnd all the possble pars of prmes that satsfy Condton 1 for each nteger. Ths code has been used to plot Fgure 4. The code for the data on how many ntegers satsfy Condton 1 f we fx one prme s qute smlar and s therefore not ncluded. 20

21 21

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