ON THE DISTANCE BETWEEN SMOOTH NUMBERS

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1 #A25 INTEGERS (20) ON THE DISTANCE BETWEEN SMOOTH NUMBERS Jean-Marie De Koninc Département e mathématiques et e statistique, Université Laval, Québec, Québec, Canaa jm@mat.ulaval.ca Nicolas Doyon Département e mathématiques et e statistique, Université Laval, Québec, Québec, Canaa nicolas.oyon@mat.ulaval.ca Receive: 3/2/0, Revise: /29/0, Accepte: 2/7/, Publishe: 4/29/ Abstract Let P (n) stan for the largest prime factor of n 2 an set P (). For each integer n 2, let δ(n) be the istance to the nearest P (n)-smooth number, that is, to the nearest integer whose largest prime factor is no larger than that of n. We provie a heuristic argument showing that /δ(n) (4 log o())x as x. Moreover, given an arbitrary real-value arithmetic function f, we stuy the behavior of the more general function δ f (n) efine by δ f (n) min mn, f(m) f(n) n m for n 2, an δ f (). In particular, given any positive integers a < b, we show that a n<b /δ f (n) 2(b a)/3 an that if f(n) f(a) for all n [a, b[, then a<n<b δ f (n) (b a) log(b a)/(2 log 2).. Introuction A smooth number (or a friable number) is a positive integer n whose largest prime factor is small compare to n. Hence, given an integer B 2, we say that an integer n is B-smooth if all its prime factors are B. Let P (n) stan for the largest prime factor of n (with P () ). For each integer n 2, let δ(n) be the istance to the nearest P (n)-smooth number, that is, to the nearest integer whose largest prime factor is no larger than that of n. In other wors, δ(n) : min n m. mn P (m) P (n) Research wor supporte in part by a grant from NSERC.

2 INTEGERS (20) 2 Equivalently, if we let Ψ(x, y) : #{n x : P (n) y}, then δ(n) is the smallest positive integer δ such that either one of the following two equalities occur: Ψ(n + δ, P (n)) Ψ(n, P (n)), Ψ(n, P (n)) Ψ(n δ, P (n)). For convenience, we set δ(). In particular, The first 40 values of δ(n) are δ(2 a ) 2 a for each integer a. (),,,2,,2,,4,,,,3,,,,8,,2,,2,,,,3,,,3,,,2,,6,,,,4,,,,4. We call δ(n) the inex of isolation of n an we say that an integer n is isolate if δ(n) 2 an non-isolate if δ(n). Finally, an integer n is sai to be very isolate if δ(n) is large. It follows from () that the most isolate number x is the largest power of 2 not exceeing x, which implies in particular that δ(n) n/2 for all n 2. Remar. One might thin, as a rule of thumb, that the smaller P (n) is, the larger δ(n) will be, that is, that smooth numbers have a large inex of isolation. But this is not true for small values of n: for instance, n has a small P (n) an nevertheless δ(n), since n , n However, for large values of n, one can say that smooth numbers o inee have a large inex of isolation. Inee, one can prove (see Lemma 3) that, given B 3 fixe, there exist a constant c c(b) > 0 an a number n 0 n 0 (B) such that δ(n) > n (log n) c for all B-smooth integers n n 0.

3 INTEGERS (20) 3 2. Preliminary Observations an Results It is clear that δ(p) for each prime p an also that if p is o, then δ(p 2 ). Each of the following also hols: δ(2p) for p 5, δ(3p) for p 3, δ(4p) 2 for p 5, δ(5p) for p 2, δ(6p) 2 for p 3, δ(7p) 2 for p 2, δ(8p) 3 for p 7, δ(9p) 3 for p 2, δ(0p) 2 for p 2. The above are easily proven. For instance, to prove the secon statement, observe that if p (mo 4), then 3p + 0 (mo 4), in which case P (3p + ) < p, while if p 3 (mo 4), then 3p 0 (mo 4), in which case P (3p ) < p, so that in both cases δ(3p). Observe also that given any prime number p, if a is an integer such that P (a) p, then δ(ap) a, because P (ap a) P (a(p )) max(p (a), P (p )) p. If follows from this simple observation that δ(n) δ(ap (n)) a n P (n) Moreover, one can easily show that if P (n) 2 n, then δ(n) Definition. For each integer n, let (n) : n δ(). (n 2). (2) n P (n) 2. Trivially we have (n) τ(n). Lemma. If n is a power of 2, then (n) n. On the other han, for all n > such that P (n) 3, we have (n) < n. (3)

4 INTEGERS (20) 4 Proof. The first assertion is obvious since for each integer α, we have (2 α ) α i 2i 2 α. Now consier the case when n is not a power of 2. First, it is easy to show that if P (n) 3, then (3) hols. Inee, if n 2 α 3 β for some integers α 0 an β, then in light of (2), we have (n) n + 2 P () + n P ()2 α 2 i + 3 i n 3 P () + n P ()3 P () + 2 α + 3 (σ(n) σ(2α )) 2 α + (2 α+ ) 3β+ (2 α+ ) 3 2 n 3β n < n, which proves that (3) hols if P (n) 3. Hence, from here on, we shall assume that P (n) 5. We shall use inuction on the number of istinct prime factors of n in orer to prove that n < n. (4) P () First observe that the above inequality is true if ω(n). Inee, in this case, we have n p b. It is clear that n P () + + p + p p b + pb p < + pb p b n, which will clearly establish (3). Let us now assume that the result hols for all n such that ω(n) r an prove that it oes hol for n such that ω(n) r. Tae such an integer n with being the unique positive integer such that P (n) n. We then have

5 INTEGERS (20) 5 n P () n/p (n) n/p (n) n/p (n) P () + P () + i n/p (n) n/p (n) + P (n) P (n) P (n)i P (P (n) i ) n/p (n) n/p (n) P () + ( + P (n) P (n) )σ(n/p (n) ). (5) n σ(n) Using the ientity σ P (n) + P (n) + P (n) 2 an the inuction argument, it follows from (5) P (n) that n P () < n + P (n) P (n) + P (n) + P (n) P (n) σ(n) < n P (n) + σ(n) P (n). (6) On the other han, it is clear that for any integer n > with P (n) 5 an ω(n) 2, σ(n) n < p n p p < 3 4 P (n). Using this in (6), we obtain, since P (n) 5, that n P () < n n < n, which completes the proof of (4) an thus of (3). Lemma 2. The following are true: (i) #{n x : δ(n) } x 2 for all x 2; (ii) x 2 < x for all x 4; δ(n) (iii) δ(n) δ(m) n m ; (iv) m n min(δ(m), δ(n)); (v) c (x) : #{n x : δ(n) } x ; (vi) #{n x : δ(n) y} x y.

6 INTEGERS (20) 6 Proof. Since it is clear that if δ(n) 2, then δ(n ) δ(n + ), (i) follows immeiately, along with (ii). On the other han, (iii) follows from the efinition of δ(n). From (iii), we easily euce (iv) an (v). To prove (vi), we procee as follows. Fix < y x an assume that : #{n x : δ(n) y} > x/y. Let y n < n 2 <... < n be the integers n i x such that δ(n i ) y. By the Pigeonhole Principle, there exist n r an n s with r < s such that n s n r < y. Without any loss in generality, one can assume that P (n r ) P (n s ), in which case we have δ(n s ) n s n r < y, a contraiction. Lemma 3. Let B 3 be a fixe integer. Then there exist a constant c c(b) > 0 an a number n 0 n 0 (B) such that δ(n) > n (log n) c for all B-smooth integers n n 0. Proof. The result follows almost immeiately from an estimate of Tijeman [7] who showe, using the theory of logarithmic forms of Baer (see Theorem 3. in the boo of Baer []), that if n < n 2 <... represents the sequence of B-smooth numbers, then there exist positive constants c (B) an c 2 (B) such that n i (log n i ) n n i c(b) i+ n i (log n i ), c2(b) where c 2 (B) π(b) c (B). Observe that Langevin [6] later provie explicit values for the constants c (B) an c 2 (B). 3. Probabilistic Results In 978, Erős an Pomerance [4] showe that the lower ensity of those integers n for which P (n) < P (n + ) (or P (n) > P (n + )) is positive. Most liely, this ensity is 2, but this fact remains an open problem. In 200, Balog [2] showe that the number of integers n x with P (n ) > P (n) > P (n + ) (7) is x an observe that unoubtely the ensity of those integers n such that (7) hols is equal to 6. To establish our next result, we shall mae the following reasonable assumption.

7 INTEGERS (20) 7 Hypothesis A. Fix an arbitrary integer 2 an let n be a large number. Let a, a 2,..., a be any permutation of the numbers 0,, 2,...,. Then, Prob[P (n + a ) < P (n + a 2 ) <... < P (n + a )]!. Theorem 4. Assuming Hypothesis A an given any integer, the expecte 2 proportion of integers n for which δ(n) is equal to 4 2. In particular, the proportion of non-isolate numbers is 2 3. Proof. Fix. Let E be the expecte proportion of integers n for which δ(n). Given a large integer n, the probability that δ(n) > is equal to the probability that min(p (n ± ),..., P (n ± )) > P (n). Uner Hypothesis A, this probability is equal to. This implies that 2 + E P (δ(n) > ) P (δ(n) > ) , which completes the proof of the theorem. Remar. Let S (x) : #{n x : δ(n) } an choose x 0 6. Then, we obtain the following numerical evience a S (x) b [x 2/(4 2 )] a/b Theorem 5. Assuming Hypothesis A, lim 2(2 log 2 ) x x δ(n) Proof. Accoring to Theorem 4 (prove assuming Hypothesis A), x #{n x : δ(n) } ( + o()) (x ).

8 INTEGERS (20) 8 Therefore, given a fixe large integer N, we have, as x, δ(n) N N δ(n) + [x/2] N+ δ(n) 2x (4 2 ( + o()) + O ) ( + o())xt (N) + O(xT 2 (N)), >N x 2 (8) say, where we use Lemma 2(v). First, one can show that T (N) N 2 (4 2 2(2 log 2 ) + O ) N (N ). (9) To prove (9), we procee as follows. Assume for now that N 3 (mo 4). Then, N using the estimate log N + γ + O(/N) as N (where γ is Euler s constant), we have 2 T (N) N N N +2 N N N N log 2 + O(/N) + 2 N j(n+)/2 log 2 + O(/N) + log 2 + O(/N) 2 log 2 + O(/N), (mo 4), thus estab- which proves (9). A similar argument hols if N 0, or 2 lishing (9). On the other han, so that >N 2 < N t 2 t N, + 2N + 2j + + 2N + T 2 (N) < N. (0)

9 INTEGERS (20) 9 Now let ε > 0 be arbitrarily small an let N [/ε] +. We then have, using (9) an (0) in (8), 2(2 log 2 )x + O(εx) + O(εx) δ(n) which completes the proof of the theorem. (x ), Remar. Using a computer, one obtains that 0 9 δ(n) n The Isolation Inex With Respect to a Given Function The function δ can also be efine relatively to any real-value arithmetic function f as δ f (n) : min mn f(m) f(n) n m (n 2) with δ f (). Examples Let f(n) be any monotonic function. Then, δ f (n), n 2. Let f(n) ω(n) : p n. Then, the first 40 values of δ ω (n) are:,,,,,,,,,,2,,2,,,,,,2,,,,2,,2,,2,,2,,,,,,,,4,,,. Let f(n) Ω(n) : p α n α. Then, the first 40 values of δ Ω (n) are:,,,,2,,2,,,,2,,2,,,,2,,2,,,,4,,,,,,2,,2,,,,,,4,,,. Let f(n) τ(n) :. Then, the first 40 values of δ τ (n) are: n,,,,2,,2,,2,,2,,2,,,,2,,2,,,,4,,2,,,,2,,2,,,,,,4,,,. Remar 6. It turns out that Hypothesis A hols unconitionally when one replaces the function P (n) by the function ω(n) or Ω(n) or τ(n), an therefore that the

10 INTEGERS (20) 0 equivalent of Theorem 4 for either of these three functions is true without any conitions, that is, that for any fixe positive integer, x #{n x : δ 2 ω(n) } o() the same being true for Ω(n) or τ(n) in place of ω(n). (x ), To prove our claim, we first nee to prove the following two propositions. Proposition 7. Let a, a 2,..., a be any istinct integers an let z, z 2,..., z be arbitrary real numbers. Then, lim x x # n x : ω(n + a j) log log n < z j, j Φ(z j ), () log log n where Φ(z) 2π z e t2 /2 t. j Proof. Given a real-value vector (t, t 2,..., t ), consier the function H(n) t j ω(n + a j ). j We will now apply Proposition 2 an Theorem of Granville an Sounararajan [5], where instea of consiering the function f p (n) p p if p n, otherwise, we use the function tr p j f p (n) t j if p n + a r, p j t j otherwise, where it is clear that, except for a finite number of primes p, each prime p ivies n + a r for at most one a r. Using this newly efine function f p (n) an following exactly the same steps as in the proof of Proposition 2 of Granville an Sounararajan, we obtain that lim n x x # x : H(n) j t j log log n < z log log n Φ z j t2 j. (2) In other wors, H(n) has a Gaussian istribution with mean value j t j log log n an stanar eviation j t2 j log log n. Because of the moments of a Gaussian

11 INTEGERS (20) istribution, statement (2) is equivalent to lim x x H(n) j t m j log log n G(m) log log n where, for each positive integer m, j m/2 (2j ) if m is o, G(m) 0 if m is even. 0 u i m, i,..., u + +u m lim x x m! u! u! j t uj j j t 2 j m/2 (m, 2,...), By expaning the left-han sie of (3) using the multinomial theorem, we may rewrite it (for each positive integer m) as uj ω(n + aj ) log log n. (4) log log n By consiering (4) as a function of t, t 2,..., t an comparing the coefficients with those on the right-han sie of (3), we obtain, for each positive integer m, m! uj ω(n + aj ) log log n (m/2)! lim G(m) u! u! x x log log x (u /2)! (u /2)! or equivalently lim x x j j ω(n + aj ) log log n log log x uj j (3) G(u j ). (5) Since the right-han sie of (5) correspons to the centere moments of a multivariate inepenent Gaussian istribution, the valiity of () follows, thereby completing the proof. Proposition 8. Let g stan for any of the functions ω, Ω or τ. Let a, a 2,..., a be any permutation of the integers 0,,...,. Then, lim x x #{n x : g(n + a j) < g(n + a j+ ), j,..., }!. Proof. In the case g ω, the result follows from Proposition 7, namely by simple integration of (). As for g Ω, observe that it is easy to show that lim lim #{n x : Ω(n) ω(n) > K} 0. (6) K x x Hence, using (6) an integrating (), Proposition 7 hols for g Ω. Finally, since Proposition 7 hols for g ω an g Ω, the inequality 2 ω(n) τ(n) 2 Ω(n) implies that it also hols for g τ, thus completing the proof.

12 INTEGERS (20) 2 Theorem 9. Let f(n) ω(n) or Ω(n) or τ(n). Then, lim 2(2 log 2 ). x x δ f (n) Proof. In light of Remar 6 an of Proposition 8, the result is immeiate. Theorem 0. Let a < b be positive integers. For any real-value arithmetic function f an any interval I [a, b[ of length N b a, n I δ f (n) 2 3 N 2 3. Proof. We conuct the proof using inuction on N. First observe that Theorem 0 hols for small values of N. For instance, if N, n I δ f (n) δ f (a) > If N 2, then since at least one of δ f (n) an δ f (n + ) must be, it follows that n I δ f (n) δ f (a) + δ f (a + ) > > We will now assume that the result hols for every integer smaller than N an prove that it must therefore hol for N. We shall o this by istinguishing three possible cases: (i) either δ f (a) or δ f (b ) ; (ii) Case (i) is not satisfie an there exists a positive integer ]a, b 2[ such that both δ f () an δ f ( + ) ; (iii) neither of the two previous cases hols. In Case (i), we can assume without any loss of generality that δ f (a), in which case δ f (n) + δ f (n). n I Using our inuction hypothesis we get that + a+ n<b a+ n<b δ f (n) (N ) N 3 > 2 3 N 2 3,

13 INTEGERS (20) 3 proving the theorem in Case (i). Suppose now that Case (ii) is satisfie. Then, there exists an integer ]a, b 2[ such that δ f () δ f ( + ), in which case n I δ f (n) a n< δ f (n) n<b δ f (n). Again, using our inuction hypothesis we have that the right-han sie is larger or equal to 2 3 ( a) (b 2) N 2 3, proving the theorem in Case (ii). We now consier Case (iii). In this situation, N has to be o, since the sum starts with the term /δ f (a) < an ens with the term /δ f (b ) <, because every secon value of δ f (n) must be. Assume that a is o, in which case b is o. The case a an b even can be treate in a similar way. Hence, is at o integers n that δ f (n) >, in which case we must have both f(n ) > f(n) an f(n + ) > f(n). Now recall the efinition δ f (n) : min mn m n ; now since f(m) f(n) the integer m at which this minimum occurs must be o (since for the other m s, the even ones, we have δ f (m) ), it follows that for an o n 2j +, we have δ f (2j + ) 2 min j, j f(2+) f(2j+) so that we may write n [a,b[ n o δ f (n) j [ a 2, b 2 [ j [ a 2, b 2 [ j [ a 2, b 2 [ j [ a 2, b 2 [ δ f (2j + ) min j f(2+) f(2j+) min j g() g(j) j j δ g (j), (7) where we have set g(j) : f(2j + ). Now since the interval [ a 2, b 2 [ is of length N + < N, we can apply our inuction hypothesis an write that the last expression in (7) is no larger than N

14 INTEGERS (20) 4 It follows from this that n I δ f (n) N N (N ), 3 as neee to be prove. This completes the proof of the theorem. In the statement of Theorem 0, is there any hope that one coul replace the constant 2 3 by a larger one? The answer is no, as the following result shows. Theorem. Let α 0,q (n), α,q (n), α 2,q (n),..., α,q (n) be the igits of n when written in base q, that is, n α j,q (n)q j. Then the function f g q efine by g q (n) j0 α j,q (n)q j (8) j0 has the following property: lim x x δ f (n) q q +. (9) Remar. Observe that g q (n) is the number obtaine by writing the basis q igits of n in reverse orer. In a sense, the result claims that the function g 2 is the one that provies the minimal value for the sum of the reciprocals of the inex of isolation. Remar 2. Let n be written in basis q 2, that is, n : α j,q (n)q j. j0 Let m be the smallest integer such that α m,q (n) is greater than zero. Then, uner the assumption that n is not a perfect power of q, it is easy to verify that δ gq (n) q m. On the other han, if n is a perfect power of q, say n q, we have δ gq (n) q (q ).

15 INTEGERS (20) 5 Proof of Theorem. We shall only consier the case q 2, since the general case can be treate similarly. We observe that δ g2 (n) if an only if n is o. More generally, in light of Remar 2, we have We can therefore write δ g2 (n) 2 if an only if [log x/ log 2] δ g2 (n) 0 n is an o integer ( 0). 2 2 # n x : n 2 (mo 2). It is easy to see that so that # n x : n 2 (mo 2) x + O(), 2+ [log x/ log 2] δ f (n) x O x + O(), 0 0 which proves (9) in the case q 2, thus completing the proof of the theorem. Theorem 3. For real numbers y, w such that 2 3 y w, there exists an arithmetic function f such that lim inf x x y an lim sup δ f (n) x x w. (20) δ f (n) Proof. The function /δ g2 value of the reciprocal of the inex of isolation of any monotone function h is. We shall construct a function that behaves piecewise lie g 2 an piecewise lie h so that the mean value of the reciprocal of its isolation inex will be a ponere mean of 2 3 has a mean value of 2 3, while it is clear that the mean an. We first efine real numbers s, t [0, ] in such a way that s + ( s) 2 3 y an t + ( t) 2 3 w. Now consier the intervals I j : [2 2j, 2 2j+ [ (j, 2, 3,...). For j even an for each integer m [, 2 2j j (2 2j )], efine the families of subintervals K j,m an L j,m as follows: K j,m : [2 2j + (m )2 j, 2 2j + (m + s) 2 j [ an L j,m : [2 2j + (m + s)2 j, 2 2j + m2 j [,

16 INTEGERS (20) 6 so that I j 2 2j j (2 2j ) m (K j,m L j,m ). For j o, we replace the number s by t in the efinition of the subintervals K j,m an L j,m. For simplicity, we efine implicitly A j,m, B j,m, C j,m, D j,m as an K j,m [A j,m, B j,m [ L j,m [C j,m, D j,m [. We are now reay to efine a function f satisfying (20). We efine f piecewise in the following manner. If n K j,m, then set while if n L j,m, set Assume first that j is even. Then, δ f (n) n I j m f(n) n A j,m, f(n) g 2 (n C j,m ). n K j,m δ f (n) + m n L j,m δ f (n) (B j,m A j,m + O()) (D j,m C j,m + O()) m m s2 j + O() ( s)2j + O() m m y2 j + O() m 2 2j+ y 2 2j+ 2 2j + O. (2) If j is o, we obtain in a similar fashion n I j 2 j δ f (n) w 2 2j+ 2 2j + O 2 2j+ 2 j. (22) Let x be a large real number. Let j be the largest integer such that 2 2j < x an let m be the largest integer such that D j,m x. We then have δ f (n) n<2 2j δ f (n) + δ n I j f (n) + 2 2j δ f (n) Σ + Σ 2 + Σ 3, (23)

17 INTEGERS (20) 7 say. On the one han, we trivially have Σ O while assuming j even, we have, using (2), Σ 2 y 2 2j + O 2 2j, (24) 2 2j 2 j. (25) Finally, Σ 3 m m n L j,m δ f (n) + m m n K j,m δ f (n) + O 2 j, which yiels, in light of (22) (since j is o), Σ 3 w x 2 2j + O 2 j + m. (26) Using (24), (25) an (26) in (23), we get δ f (n) y22j + w x 2 2j + O which completes the proof of the theorem. x, log x 5. The Mean Value of the Inex of Isolation While the mean value of the reciprocal of the inex of isolation gives information on the local behavior of a function f, the mean value of the inex of isolation itself gives information on the very isolate numbers. Theorem 4. Let f be a real-value arithmetic function. Let a an b be two positive integers such that b a N. Suppose furthermore that for all m [a, b[, f(m) f(a). Then δ f (n) N log N 2 log 2. (27) n ]a,b[ Proof. We prove Theorem 4 by using inuction on N. The result hols for N, because the left-han sie of (27) is 0 (since the interval of summation contains no integers), yieling the inequality 0 0. It also hols for N 2, because in this case we have δ f (a + ), yieling the inequality. So, let us assume that (27) is

18 INTEGERS (20) 8 true for all intervals ]a, b[ with a b N for some integer N >. We shall prove that uner the hypothesis that b a N + an that for all m ]a, b[, f(m) f(a) an f(m) f(b), we have a+ n b δ f (n) N 2 log N log 2. (28) We prove (28) by using inuction on N. Clearly the result is true for N an N 2. Assume that it is true for N an let us prove that it is true for N. Define n as an integer such that n ]a, b[ an such that for all m ]a, b[, f(m) f(n ). We thus have δ f (n) min(n a, b n )+ δ f (n)+ δ f (n). a+ n b Using our inuction hypothesis, we have an a+ n n n + n b a+ n n δ f (n) n a log(n a) 2 log 2 δ f (n) b n 2 log(b n ). log 2 n + n b Without any loss of generality, we can assume that n ]a, a + (b a)/2], so that a+ n b δ f (n) n a + n a log(n a) + b n 2 log 2 2 log(b n ). (29) log 2 Assuming for now that n is a real variable, an taing the erivative of the righthan sie of (29) with respect to n, we obtain + log(n a) 2 log 2 log(b n ). 2 log 2 Since the secon erivative is positive, the right-han sie of (29) reaches its maximum value at the en points, that is, either when n a + or n (b + a)/2. In the first case, we get a+ n b δ f (n) (N ) log(n ) 2 log 2 N log N 2 log 2. In the secon case, that is, when n (b + a)/2, we obtain a+ n b δ f (n) N 2 + N log(n/2) 2 log 2 N log N 2 log 2 N log N 2 log 2,

19 INTEGERS (20) 9 thus proving (28) in all cases. We are now reay to complete the proof of Theorem 4, that is, to remove the conition f(m) f(b). For this, we use inuction. Let n 0 ]a, b[ be an integer such that for all m ]a, b[, f(m) f(n 0 ). We can write δ f (n) δ f (n) + n 0 a + δ f (n). a+ n b a+ n n 0 Using our inuction hypothesis, we obtain an a+ n n 0 n 0+ n b From the last three estimates, it follows that a+ n b δ f (n) (n 0 a) log(n 0 a) 2 log 2 δ f (n) (b n 0) log(b n 0 ). 2 log 2 δ f (n) n 0 a + (n 0 a) log(n 0 a) 2 log 2 n 0+ n b + (b n 0) log(b n 0 ). (30) 2 log 2 Proceeing as we i to estimate the right-han sie of (29), we obtain that the right-han sie of (30) is less than N log N, which completes the proof of the 2 log 2 theorem. Theorem 5. As x, δ ω (n) x log log x. Proof. For each x 2, δ ω (n) ω(n) 0 log log x δ ω (n) ω(n) δ ω (n) + >0 log log x ω(n) δ ω (n). (3) Clearly, for any fixe, ω(n) δ ω (n) 2x.

20 INTEGERS (20) 20 Therefore, 0 log log x ω(n) δ ω (n) 20x log log x. (32) Let x be large an fixe, an consier the set S : {n x : ω(n) > 0 log log x}. Write S as the union of isjoint intervals S I I 2... I, an let j stan for the length of the interval I j. We have from Theorem 4 that n I j δ ω (n) j log j 2 log 2. (33) On the other han, one can show that j #S j ω(n)>0 log log x o x log x (x ) (34) (see, for instance, relation (8) in De Koninc, Doyon an Luca [3]). It follows from (33) an (34) that δ ω (n) n S j j log j 2 log 2 log x x 2 log 2 o log x o(x) (x ). (35) Substituting (32) an (35) in (3) completes the proof of the theorem. Remar. Using a computer, one can observe that, for x 0 9, x log log x δ ω (n) Theorem 6. Let the function g q be efine as in (8). Then n N δ gq (n) (q )N log N q log q + O(N), so that the function g 2 is the function for which the sums of the inex of isolation is maximal. Proof. Let n be written in base q 2, that is, n : α j,q (n)q j. j0

21 INTEGERS (20) 2 In light of Remar 2, we get, letting 0 be the largest integer such that q 0 N, δ gq (n) δ gq(n) + δ gq(n) n N n N n N nq nq q m #{n N : δ gq (n) q m } + q (q ) m log N/ log q m log N/ log q q N q q m N + O() + (q ) q qm+ q N q N log N q log q + O(N) + (q )q0 q q N log N q log q + O(N), thus completing the proof of the theorem. 6. Computational Data an Open Problems If n n stans for the smallest positive integer n such that then we have the following table: δ(n) δ(n + )... δ(n + ), (36) n n Some open problems concerning the sequence n,, 2, 3,..., are the following:. Prove that n exists for each integer Estimate the size of n as a function of. Also, is it true that n! for each integer 5? 3. Prove that for any fixe 3, there are infinitely many integers n such that (36) is satisfie. The fact that the matter is settle for 2 follows immeiately from the Balog result state at the beginning of Section 3. Interesting questions also arise from the stuy of the function (n) first mentione in Section 2. For instance, let m stan for the smallest number m for

22 INTEGERS (20) 22 which (m) (m + )... (m + ). (37) Then m 2 4, with (4) (5) 4; m 3 33, with (33) (34) (35) 4; m , with (m 4 + i) 2 for i 0,, 2, 3; m , with (m 5 + i) 4 for i 0,, 2, 3, 4; if m 6 exists, then m 6 > Specific questions are the following:. Prove that m exists for each integer Estimate the size of m as a function of. 3. Prove that for any fixe 3, there are infinitely many integers m such that (37) is satisfie? Acnowlegement. The authors woul lie to than the referee for some very helpful suggestions which le to the improvement of this paper. References [] A. Baer, Transcenental Number Theory, Cambrige University Press, Lonon, 975. [2] A. Balog, On triplets with escening largest prime factors, Stuia Sci. Math. Hungar. 38 (200), [3] J.M. De Koninc, N. Doyon an F. Luca, Sur la quantité e nombres économiques, Acta Arithmetica 27 (2007), no.2, [4] P. Erős an C. Pomerance, On the largest prime factors of n an n+, Aeq. Math. 7 (978), [5] A. Granville an K. Sounararajan, Sieving an the Erős-Kac theorem in Equiistribution in number theory, An introuction, NATO Sci. Ser. II Math. Phys. Chem. vol. 237 (2007), [6] M. Langevin, Quelques applications e nouveaux résultats e Van er Poorten, Séminaire Delange-Pisot-Poitou, 7 e année (975/76), Théorie es nombres: Fasc. 2, Exp.No. G2, pp. Secrétariat Math., Paris, 977. [7] R. Tijeman, On the maximal istance between integers compose of small primes, Compositio Math. 28 (974),