On rational numbers associated with arithmetic functions evaluated at factorials

Transcription

1 On rational numbers associated with arithmetic functions evaluated at factorials Dan Baczkowski (joint work with M. Filaseta, F. Luca, and O. Trifonov)

2 (F. Luca) Fix r Q, there are a finite number of positive integers n and m for which f(n!) = r m! where f is one of τ, φ, or σ.

3 Theorem 1. Let f denote one of the arithmetic functions τ, φ or σ, and let k be a fixed positive integer. Then there are finitely many positive integers n, m, a and b such that b f(n!) = a m!, gcd(a, b) = 1 and ω(ab) k.

4 Theorem 1. Let f denote one of the arithmetic functions τ, φ or σ, and let k be a fixed positive integer. Then there are finitely many positive integers n, m, a and b such that b f(n!) = a m!, gcd(a, b) = 1 and ω(ab) k. I.e. the total number of distinct primes dividing the numerator and denominator of the fraction obtained by reducing the quotient f(n!)/m! tends to infinity as the product nm tends to infinity.

5 Theorem 2. There are finitely many positive integers a, b, n and m such that b τ(n!) = a m!, gcd(a, b) = 1, ω(b) m 1/4 and P 0 (a) log n 22, where P 0 (a) denotes the least prime not dividing a.

6 Theorem 2. There are finitely many positive integers a, b, n and m such that b τ(n!) = a m!, gcd(a, b) = 1, ω(b) m 1/4 and P 0 (a) log n 22, where P 0 (a) denotes the least prime not dividing a. Theorem 3. and for n > 1 b φ(n!) = a m!, gcd(a, b) = 1 and max{ω(a), ω(b)} n 7 log n.

7 Theorem 4. Fix ε > 0. Then there are finitely many positive integers a, b, n and m such that b σ(n!) = a m!, gcd(a, b) = 1, ω(ab) n 0.2 ε.

8 Let q be a prime and ν q (N) denote the exponent of q in the prime factorization of N.

9 Let q be a prime and ν q (N) denote the exponent of q in the prime factorization of N. σ(n!) = p n σ ( p ν p(n!) )

10 Let q be a prime and ν q (N) denote the exponent of q in the prime factorization of N. σ(n!) = p n σ ( p ν p(n!) ) = p n p ν p(n!)+1 1 p 1

11 Let q be a prime and ν q (N) denote the exponent of q in the prime factorization of N. σ(n!) = p n σ ( p ν p(n!) ) = p n p ν p(n!)+1 1 p 1 Lemma 6. If 0 < ɛ < 1/5 and q is a prime n 1/5 ɛ, then ( ) n log log(q + 1) (i.) ν q σ(n!). q log n

12 Let q be a prime and ν q (N) denote the exponent of q in the prime factorization of N. σ(n!) = p n σ ( p ν p(n!) ) = p n p ν p(n!)+1 1 p 1 Lemma 6. If 0 < ɛ < 1/5 and q is a prime n 1/5 ɛ, then ( ) n log log(q + 1) (i.) ν q σ(n!). q log n If 0 < δ ( < 1/3 and q is a prime, then (ii.) ν q σ ( p ν p(n!) )) n log log(q + 1) q n 1 δ p n + n3δ log n. log q

13 Let q be a prime and ν q (N) denote the exponent of q in the prime factorization of N. σ(n!) = p n σ ( p ν p(n!) ) = p n p ν p(n!)+1 1 p 1 Lemma 6. If 0 < ɛ < 1/5 and q is a prime n 1/5 ɛ, then ( ) n log log(q + 1) (i.) ν q σ(n!). q log n If 0 < δ ( < 1/3 and q is a prime, then (ii.) ν q σ ( p ν p(n!) )) n log log(q + 1) q n 1 δ p n + n3δ log n. log q

14 Proof of Theorem for σ: b σ(n!) = a m!, gcd(a, b) = 1 and ω(ab) n 0.2 ε

15 Proof of Theorem for σ: b σ(n!) = a m!, gcd(a, b) = 1 and ω(ab) n 0.2 ε The proof: Show there is a N such that n N holds.

16 Proof of Theorem for σ: b σ(n!) = a m!, gcd(a, b) = 1 and ω(ab) n 0.2 ε The proof: Show there is a N such that n N holds. then we can deduce that b σ(n!) has a bounded number of distinct prime factors (depending only on N)

17 Proof of Theorem for σ: b σ(n!) = a m!, gcd(a, b) = 1 and ω(ab) n 0.2 ε The proof: Show there is a N such that n N holds. then we can deduce that b σ(n!) has a bounded number of distinct prime factors (depending only on N) implying that m is bounded and, hence, that there are only a finite number of possibilities for the value of a/b = f(n!)/m!.

18 Proof of Theorem for σ: b σ(n!) = a m!, gcd(a, b) = 1 and ω(ab) n 0.2 ε The proof: Show there is a N such that n N holds. then we can deduce that b σ(n!) has a bounded number of distinct prime factors (depending only on N) implying that m is bounded and, hence, that there are only a finite number of possibilities for the value of a/b = f(n!)/m!. Given that gcd(a, b) = 1, we can then deduce that there are a finite number of possibilities for the quadruple (n, m, a, b).

19 b σ(n!) = a m! Set c = 1/5 2ɛ where 0 < ɛ < 1/10. Assume n is sufficiently large and ω(ab) n c.

20 b σ(n!) = a m! Set c = 1/5 2ɛ where 0 < ɛ < 1/10. Assume n is sufficiently large and ω(ab) n c. First, we consider the case that ω ( σ(n!) ) 2n c.

21 b σ(n!) = a m! Set c = 1/5 2ɛ where 0 < ɛ < 1/10. Assume n is sufficiently large and ω(ab) n c. First, we consider the case that ω ( σ(n!) ) 2n c. Then there exists n c distinct primes p dividing σ(n!) and not dividing ab.

22 b σ(n!) = a m! Set c = 1/5 2ɛ where 0 < ɛ < 1/10. Assume n is sufficiently large and ω(ab) n c. First, we consider the case that ω ( σ(n!) ) 2n c. Then there exists n c distinct primes p dividing σ(n!) and not dividing ab. Among the first 2n c primes, prime q s.t. q σ(n!) but q ab.

23 b σ(n!) = a m! Set c = 1/5 2ɛ where 0 < ɛ < 1/10. Assume n is sufficiently large and ω(ab) n c. First, we consider the case that ω ( σ(n!) ) 2n c. Then there exists n c distinct primes p dividing σ(n!) and not dividing ab. Among the first 2n c primes, prime q s.t. q σ(n!) but q ab. Moreover, q n c+ɛ n 1/5 ɛ.

24 b σ(n!) = a m! Set c = 1/5 2ɛ where 0 < ɛ < 1/10. Assume n is sufficiently large and ω(ab) n c. First, we consider the case that ω ( σ(n!) ) 2n c. Then there exists n c distinct primes p dividing σ(n!) and not dividing ab. Among the first 2n c primes, prime q s.t. q σ(n!) but q ab. Moreover, q n c+ɛ n 1/5 ɛ. Since q does not divide ab, we have ν q ( σ(n!) ) = νq (m!) m q 1.

25 Lemma 6 (i) now implies b σ(n!) = a m! m n log log n log n.

26 Lemma 6 (i) now implies b σ(n!) = a m! m n log log n log n. The case when ω ( σ(n!) ) < 2n c also gives this. Indeed, m log m π(m) = ω(m!) ω( b σ(n!) ) ω(b)+ω ( σ(n!) ) n c implying that m n c log n.

27 b σ(n!) = a m! Observe that log σ(n!) log(n!) n log n

28 b σ(n!) = a m! Observe that log σ(n!) log(n!) n log n and now log(m!) m log m n log log n.

29 b σ(n!) = a m! Observe that log σ(n!) log(n!) n log n and now Hence, log(m!) m log m n log log n. log a = log(b/m!) + log σ(n!) n log n.

30 Fix 0 < δ < 1/3. Let a = ( p ν p (n!) ) and a = gcd ( a, σ(n!)/a ). p n 1 δ σ

31 Fix 0 < δ < 1/3. Let a = ( p ν p (n!) ) and a = gcd ( a, σ(n!)/a ). p n 1 δ σ Clearly, a a a.

32 Fix 0 < δ < 1/3. Let a = ( p ν p (n!) ) and a = gcd ( a, σ(n!)/a ). p n 1 δ σ Clearly, a a a. Notice that n ν p (n!) = p u u=1 < u=1 n p u = n p 1

33 Fix 0 < δ < 1/3. Let a = ( p ν p (n!) ) and a = gcd ( a, σ(n!)/a ). p n 1 δ σ Clearly, a a a. Notice that n ν p (n!) = p u n p 1 ν p(n!) < log a u=1 n p 1 so that < u=1 p n 1 δ ν p (n!) log p n p u = n p 1 ( 1 δ ) n log n.

34 b σ(n!) = a m! n log n log a log a +log a (1 δ)n log n+ q a ν q (a ) log q.

35 b σ(n!) = a m! n log n log a log a +log a (1 δ)n log n+ q a ν q (a ) log q. From Lemma 6, q a ν q (a ) log q

36 b σ(n!) = a m! n log n log a log a +log a (1 δ)n log n+ q a ν q (a ) log q. From Lemma 6, q a ν q (a ) log q is q a q n c+ɛ n log log n q log n log q + ( n log log(q + 1) q q a q>n c+ɛ ) log q+n 3δ log n.

37 So, δn log n is q a q n c+ɛ n log log n q log n log q + ( n log log(q + 1) q q a q>n c+ɛ ) log q+n 3δ log n.

38 So, δn log n is q a q n c+ɛ n log log n q log n log q + ( n log log(q + 1) q q a q>n c+ɛ For the first sum on the right, we have ) log q+n 3δ log n. q a q n c+ɛ n log log n q log n n log log n log q log n log q q q n c+ɛ n log log n.

39 For the second sum, we use that the number of terms is bounded by ω(a). q a q>n c+ɛ n log log(q + 1) q log q ω(a) n log log(nc+ɛ + 1) n c+ɛ log n c+ɛ

40 For the second sum, we use that the number of terms is bounded by ω(a). q a q>n c+ɛ n log log(q + 1) q log q ω(a) n log log(nc+ɛ + 1) n c+ɛ n c n log log n n c+ɛ log n n log n c+ɛ

41 For the second sum, we use that the number of terms is bounded by ω(a). q a q>n c+ɛ n log log(q + 1) q log q ω(a) n log log(nc+ɛ + 1) n c+ɛ log n c+ɛ n c n log log n n c+ɛ log n n and n q a q>n c+ɛ 3δ log n ω(a)n 3δ log n.

42 b σ(n!) = a m! δn log n n log log n + n + ω(a)n 3δ log n.

43 b σ(n!) = a m! δn log n n log log n + n + ω(a)n 3δ log n. Consequently, ω(a)n 3δ log n δn log n.

44 b σ(n!) = a m! δn log n n log log n + n + ω(a)n 3δ log n. Consequently, ω(a)n 3δ log n δn log n. Taking δ = 4/15 < 1/3, the left-hand side is n and we reach the desired contradiction. Hence,...

45 the proof is complete... uhhh... third dimension

46 Preliminaries for the function σ Recall: ν q (N) denotes the exponent of q in the prime factorization of N. σ(n!) = p νp(n!)+1 1 p 1 p n

47 Preliminaries for the function σ Recall: ν q (N) denotes the exponent of q in the prime factorization of N. σ(n!) = p νp(n!)+1 1 p 1 p n Let Φ N (x) denote the Nth cyclotomic polynomial.

48 Preliminaries for the function σ Recall: ν q (N) denotes the exponent of q in the prime factorization of N. σ(n!) = p νp(n!)+1 1 p 1 p n Let Φ N (x) denote the Nth cyclotomic polynomial. x N 1 = d N Φ d (x).

49 Preliminaries for the function σ Recall: ν q (N) denotes the exponent of q in the prime factorization of N. σ(n!) = p νp(n!)+1 1 p 1 p n Let Φ N (x) denote the Nth cyclotomic polynomial. GOAL: approximate ν q (σ(n!)) x N 1 = d N Φ d (x).

50 Preliminaries for the function σ Recall: ν q (N) denotes the exponent of q in the prime factorization of N. σ(n!) = p νp(n!)+1 1 p 1 p n Let Φ N (x) denote the Nth cyclotomic polynomial. x N 1 = d N Φ d (x). GOAL: approximate ν q (σ(n!)) HOW: analyze the highest power of a given prime q that can divide an expression of the form a N 1

51 Lemma 1. a, N Z, N = q r M, r 0 q Φ N (a) if and only if M = ord q (a)

52 Lemma 1. a, N Z, N = q r M, r 0 q Φ N (a) if and only if M = ord q (a) Lemma 2. f(x) = x l + x l x + 1 has 2 gcd ( φ(q j ), l + 1 ) distinct roots modulo q j.

53 Lemma 1. a, N Z, N = q r M, r 0 q Φ N (a) if and only if M = ord q (a) Lemma 2. f(x) = x l + x l x + 1 has 2 gcd ( φ(q j ), l + 1 ) distinct roots modulo q j. Lemma 3. ν q ( a N 1 ) log N + ord q(a) log a log q

54 Lemma 1. a, N Z, N = q r M, r 0 q Φ N (a) if and only if M = ord q (a) Lemma 2. f(x) = x l + x l x + 1 has 2 gcd ( φ(q j ), l + 1 ) distinct roots modulo q j. Lemma 3. Lemma 4. ( ν q a N 1 ) log N + ord q(a) log a log q gcd(φ(q j ), l) l 2 log log(q + 1) l=1

55 Proof of Lemma 4: Notice that for a positive integer r, l L gcd(r, l) l 2 = t r l L gcd(r,l)=t t l 2

56 Proof of Lemma 4: Notice that for a positive integer r, l L gcd(r, l) l 2 = t r l L gcd(r,l)=t t l 2 = t r s L/t gcd(r/t,s)=1 1 s 2 t

57 Proof of Lemma 4: Notice that for a positive integer r, l L gcd(r, l) l 2 = t r l L gcd(r,l)=t t l 2 = t r s L/t gcd(r/t,s)=1 1 s 2 t ζ(2) t r 1 t.

58 Proof of Lemma 4: Notice that for a positive integer r, l L gcd(r, l) l 2 = t r l L gcd(r,l)=t t l 2 = t r s L/t gcd(r/t,s)=1 1 s 2 t ζ(2) t r 1 t. When r = φ(q j ) = q j 1 (q 1), the right-hand side above is σ(q 1) < ζ(2) q 1 i=0 1 q i

59 Proof of Lemma 4: Notice that for a positive integer r, l L gcd(r, l) l 2 = t r l L gcd(r,l)=t t l 2 = t r s L/t gcd(r/t,s)=1 1 s 2 t ζ(2) t r 1 t. When r = φ(q j ) = q j 1 (q 1), the right-hand side above is σ(q 1) < ζ(2) q 1 i=0 1 log log(q + 1) qi since σ(n) N log log N.

60 Lemma 5. Let log q n = log n/ log q. Then l gcd(φ(q K ), l) 3q 4 (log q n) 2 τ(q 1) + (log q n) 9. l q 2 log q n

61 Lemma 5. Let log q n = log n/ log q. Then l gcd(φ(q K ), l) 3q 4 (log q n) 2 τ(q 1) + (log q n) 9. l q 2 log q n Lemma 6. If 0 < ɛ < 1/5 and q is a prime n 1/5 ɛ, then ( ) n log log(q + 1) (i) ν q σ(n!). q log n

62 Lemma 5. Let log q n = log n/ log q. Then l gcd(φ(q K ), l) 3q 4 (log q n) 2 τ(q 1) + (log q n) 9. l q 2 log q n Lemma 6. If 0 < ɛ < 1/5 and q is a prime n 1/5 ɛ, then ( ) n log log(q + 1) (i) ν q σ(n!). q log n If 0 < δ < 1/3 and q is a prime, then ( (ii) ν q σ ( p ν p(n!) )) n 1 δ p n n log log(q + 1) q + n3δ log n. log q

63 2 Proof. e(p) = νp(n!), N (p) = e(p) + 1, and L = q logq n

64 Proof. e(p) = ν p (n!), N(p) = e(p) + 1, and L = q 2 log q n σ(n!) = σ(p e(p) ) σ(p e(p) ) p n/l n/l<p n

65 Proof. e(p) = ν p (n!), N(p) = e(p) + 1, and L = q 2 log q n σ(n!) = = p n/l p n/l σ(p e(p) ) p N(p) 1 p 1 n/l<p n n/l<p n σ(p e(p) ) p N(p) 1 p 1,

66 Proof. e(p) = ν p (n!), N(p) = e(p) + 1, and L = q 2 log q n σ(n!) = = p n/l p n/l σ(p e(p) ) p N(p) 1 p 1 n/l<p n n/l<p n σ(p e(p) ) p N(p) 1 p 1, Estimate the contribution of factors of q arising from σ ( p e(p)) separately depending on whether p n/l or p > n/l.

67 Combine estimates for V = V(q) = ν q ( p n/l ) σ(p e(p) ) = ν q ( p n/l p N(p) ) 1 p 1 and V = V (q) = ν q ( n/l<p n ) σ(p e(p) ) = ν q ( n/l<p n p N(p) ) 1. p 1

68 From Lemma 3, for any prime q we have that V p n/l log N(p) + ord q (p) log p, log q

69 From Lemma 3, for any prime q we have that V p n/l log N(p) + ord q (p) log p, log q We use that N(p) n, which follows easily from e(p) = u=1 n p u < u=1 n p u = n p 1 n, and that ord q (p) divides φ(q).

70 V p n/l p n/l log N(p) + ord q (p) log p log q log n log q + p n/l q log p log q

71 V p n/l p n/l log N(p) + ord q (p) log p log q log n log q + π( n/l ) log n log q p n/l + q log q q log p log q p n/l log p

72 V p n/l p n/l log N(p) + ord q (p) log p log q log n log q + π( n/l ) log n log q q n L log q p n/l + q log q n q log n. q log p log q p n/l log p

73 Larger primes p, i.e. p > n/l For each positive integer l < L, we consider the contribution of q s from σ ( p e(p)) with p I l = (n/(l + 1), n/l].

74 Larger primes p, i.e. p > n/l For each positive integer l < L, we consider the contribution of q s from σ ( p e(p)) with p I l = (n/(l + 1), n/l]. Fix such an l and a prime p I l.

75 Larger primes p, i.e. p > n/l For each positive integer l < L, we consider the contribution of q s from σ ( p e(p)) with p I l = (n/(l + 1), n/l]. Fix such an l and a prime p I l. The definition of L implies p > n. Since p I l, we obtain N(p) = n/p + 1 = l + 1.

76 Larger primes p, i.e. p > n/l For each positive integer l < L, we consider the contribution of q s from σ ( p e(p)) with p I l = (n/(l + 1), n/l]. Fix such an l and a prime p I l. The definition of L implies p > n. Since p I l, we obtain N(p) = n/p + 1 = l + 1. Let f l (x) = x l +x l 1 + +x 2 +x+1. Then σ(p e(p) ) = f l (p).

77 Larger primes p, i.e. p > n/l For each positive integer l < L, we consider the contribution of q s from σ ( p e(p)) with p I l = (n/(l + 1), n/l]. Fix such an l and a prime p I l. The definition of L implies p > n. Since p I l, we obtain N(p) = n/p + 1 = l + 1. Let f l (x) = x l +x l 1 + +x 2 +x+1. Then σ(p e(p) ) = f l (p). Observe that this polynomial defining σ(p e(p) ) does not change as p varies over the primes in I l.

78 ( )) ν q (σ p e(p) p I l = p I l ν q ( fl (p) ) = j 1 p I l f l (p) 0 (mod q j ) 1.

79 ( )) ν q (σ p e(p) p I l = p I l ν q ( fl (p) ) = j 1 p I l f l (p) 0 (mod q j ) 1. Let J = log q (log n) + 1 so that q J = q log n. For each l < L, we obtain from q n 1/5 ɛ that I l q J n L 2 q J

80 ( )) ν q (σ p e(p) p I l = p I l ν q ( fl (p) ) = j 1 p I l f l (p) 0 (mod q j ) 1. Let J = log q (log n) + 1 so that q J = q log n. For each l < L, we obtain from q n 1/5 ɛ that I l q J n L 2 q J n q 5 (log q n) 2 log n

81 ( )) ν q (σ p e(p) p I l = p I l ν q ( fl (p) ) = j 1 p I l f l (p) 0 (mod q j ) 1. Let J = log q (log n) + 1 so that q J = q log n. For each l < L, we obtain from q n 1/5 ɛ that I l q J n L 2 q J n q 5 (log q n) 2 log n n5ɛ log 3 n.

82 ρ j,l = ρ j,l (q) = {t Z : 0 t q j 1, f l (t) 0 (mod q j )}. Via Lemma 2, we have that ρ j,l 2 gcd ( φ(q j ), l + 1 ).

83 ρ j,l = ρ j,l (q) = {t Z : 0 t q j 1, f l (t) 0 (mod q j )}. Via Lemma 2, we have that ρ j,l 2 gcd ( φ(q j ), l + 1 ). For each a {0, 1,..., q j 1} with j J and f l (a) 0 (mod q j ), we obtain from a version of the Brun-Titchmarsh inequality that as I l /q j, we have π ( n/l; q j, a ) π ( n/(l+1); q j, a ) ( 2+o(1) ) I l φ(q j ) log ( I l /q j).

84 Now, we consider the j > J. Recall that N(p) = l + 1 for each p I l. Define K = K l = (l + 1) log q n.

85 Now, we consider the j > J. Recall that N(p) = l + 1 for each p I l. Define K = K l = (l + 1) log q n. It s easy to see that K can be used as an upper bound on the j appearing in ( )) ν q (σ p e(p) = 1. p I l j 1 p I l f l (p) 0 (mod q j )

86 Now, we consider the j > J. Recall that N(p) = l + 1 for each p I l. Define K = K l = (l + 1) log q n. It s easy to see that K can be used as an upper bound on the j appearing in ( )) ν q (σ p e(p) = 1. p I l j 1 q j divides f l (p) for some p I l p I l f l (p) 0 (mod q j )

87 Now, we consider the j > J. Recall that N(p) = l + 1 for each p I l. Define K = K l = (l + 1) log q n. It s easy to see that K can be used as an upper bound on the j appearing in ( )) ν q (σ p e(p) = 1. p I l j 1 p I l f l (p) 0 (mod q j ) q j divides f l (p) for some p I l = q j (p N 1)/(p 1)

88 Now, we consider the j > J. Recall that N(p) = l + 1 for each p I l. Define K = K l = (l + 1) log q n. It s easy to see that K can be used as an upper bound on the j appearing in ( )) ν q (σ p e(p) = 1. p I l j 1 p I l f l (p) 0 (mod q j ) q j divides f l (p) for some p I l = q j (p N 1)/(p 1) = q j < p N

89 Now, we consider the j > J. Recall that N(p) = l + 1 for each p I l. Define K = K l = (l + 1) log q n. It s easy to see that K can be used as an upper bound on the j appearing in ( )) ν q (σ p e(p) = 1. p I l j 1 p I l f l (p) 0 (mod q j ) q j divides f l (p) for some p I l = q j (p N 1)/(p 1) = q j < p N = j < N log q p N log q n = (l + 1) log q n = K.

90 Fix l and consider the positive integer values of j such that J < j < K.

91 Fix l and consider the positive integer values of j such that J < j < K. For each such j, we proceed as before by counting the number of primes p I l such that q j divides f l (p).

92 Fix l and consider the positive integer values of j such that J < j < K. For each such j, we proceed as before by counting the number of primes p I l such that q j divides f l (p). We simply use that the number of primes p I l for which q j divides f l (p) is ( ) Il ρ j,l q j + 1. Recall ρ j,l = {t Z : 0 t q j 1, f l (t) 0 (mod q j )}.

93 Altogether, we deduce that V = ν q ( l<l n/l<p n ( 1 j J p N(p) ) 1 p 1 = l<l 2 I l ρ j,l φ(q j ) log ( I l /q j) + p I l ν q ( fl (p) ) J<j<K L ρ j,l ( )) Il q j + 1

94 Altogether, we deduce that V = ν q ( l<l n/l<p n ( 1 j J p N(p) ) 1 p 1 = l<l 2 I l ρ j,l φ(q j ) log ( I l /q j) + p I l ν q ( fl (p) ) J<j<K L ρ j,l ( )) Il q j + 1 In turn estimate the three double sums.

95 Altogether, we deduce that V = ν q ( l<l n/l<p n ( 1 j J p N(p) ) 1 p 1 = l<l 2 I l ρ j,l φ(q j ) log ( I l /q j) + p I l ν q ( fl (p) ) J<j<K L ρ j,l ( )) Il q j + 1 In turn estimate the three double sums. Three sums?

96 l<l 1 j J 2 I l ρ j,l φ(q j ) log ( I l /q j)

97 l<l 1 j J 2 I l ρ j,l φ(q j ) log ( I l /q j) 1 j J n gcd ( φ(q j ), l + 1 ) l<l φ(q j )(l + 1) 2 log ( I l /q j)

98 l<l 1 j J 2 I l ρ j,l φ(q j ) log ( I l /q j) 1 j J 1 j J l<l n gcd ( φ(q j ), l + 1 ) φ(q j )(l + 1) 2 log ( I l /q j) n log log(q + 1) φ(q j ) log n

99 l<l 1 j J 2 I l ρ j,l φ(q j ) log ( I l /q j) 1 j J 1 j J j=1 l<l n gcd ( φ(q j ), l + 1 ) φ(q j )(l + 1) 2 log ( I l /q j) n log log(q + 1) φ(q j ) log n n log log(q + 1) q j log n

100 l<l 1 j J 2 I l ρ j,l φ(q j ) log ( I l /q j) 1 j J 1 j J j=1 l<l n gcd ( φ(q j ), l + 1 ) φ(q j )(l + 1) 2 log ( I l /q j) n log log(q + 1) φ(q j ) log n n log log(q + 1) q j log n n log log(q + 1). q log n

101 I l ρ j,l q j J<j<K L l<l J<j<K L l<l n gcd ( φ(q j ), l + 1 ) q j (l + 1) 2

102 I l ρ j,l q j J<j<K L l<l J<j<K L j>j l<l n log log(q + 1) q j n gcd ( φ(q j ), l + 1 ) q j (l + 1) 2

103 I l ρ j,l q j J<j<K L l<l J<j<K L j>j l<l n log log(q + 1) q j n gcd ( φ(q j ), l + 1 ) q j (l + 1) 2 n log log(q + 1) q J n log log(q + 1). q log n Recall that q J = q log n.

104 Recall K l = (l + 1) log q n, and observe that ρ j,l 2 gcd ( φ(q j ), l + 1 ) 2 gcd ( φ ( q K l ), l + 1 ).

105 Recall K l = (l + 1) log q n, and observe that ρ j,l 2 gcd ( φ(q j ), l + 1 ) 2 gcd ( φ ( q K l ), l + 1 ). l<l J<j<K l ρ j,l l<l J<j<K l gcd ( φ ( q K l ), l + 1 )

106 Recall K l = (l + 1) log q n, and observe that ρ j,l 2 gcd ( φ(q j ), l + 1 ) 2 gcd ( φ ( q K l ), l + 1 ). l<l ρ j,l J<j<K l gcd l<l J<j<K l ( φ ( q K l ), l + 1 ) K l gcd ( φ ( q K l ), l + 1 ) l<l

107 Recall K l = (l + 1) log q n, and observe that ρ j,l 2 gcd ( φ(q j ), l + 1 ) 2 gcd ( φ ( q K l ), l + 1 ). l<l ρ j,l J<j<K l l<l l<l J<j<K l gcd ( φ ( q K l ), l + 1 ) K l gcd ( φ ( q K l ), l + 1 ) (log q n) l<l (l + 1) gcd ( φ ( q K l ), l + 1 )

108 Recall K l = (l + 1) log q n, and observe that ρ j,l 2 gcd ( φ(q j ), l + 1 ) 2 gcd ( φ ( q K l ), l + 1 ). l<l ρ j,l J<j<K l l<l l<l J<j<K l gcd ( φ ( q K l ), l + 1 ) K l gcd ( φ ( q K l ), l + 1 ) (log q n) l<l (l + 1) gcd ( φ ( q K l ), l + 1 ) q 4 τ(q 1)(log q n) 3 + (log q n) 10.

109 Thus for fixed ɛ > 0 and q n 1/5 ɛ, l<l J<j<K l ρ j,l q 4 τ(q 1)(log q n) 3 + (log q n) 10 n 1 ɛ /q n/(q log n).

110 Thus for fixed ɛ > 0 and q n 1/5 ɛ, l<l J<j<K l ρ j,l q 4 τ(q 1)(log q n) 3 + (log q n) 10 n 1 ɛ /q n/(q log n). Altogether, for q n 1/5 ɛ we have ν q ( σ(n!) ) = V + V n log log(q + 1). q log n

111 The second part of the lemma?

112 The second part of the lemma? Actually, its similar but much simpler.

113 The second part of the lemma? Actually, its similar but much simpler. Take L = n δ. Partition the interval I l into congruence classes of length q j.

114 The second part of the lemma? Actually, its similar but much simpler. Take L = n δ. Partition the interval I l into congruence classes of length q j. For each l < L, we consider all possible values of 1 j K l together.

115 The second part of the lemma? Actually, its similar but much simpler. Take L = n δ. Partition the interval I l into congruence classes of length q j. For each l < L, we consider all possible values of 1 j K l together. ν q l<l p I l ( fl (p) ) ρ j,l l<l 1 j<k l ( ) Il q j + 1

116 The second part of the lemma? Actually, its similar but much simpler. Take L = n δ. Partition the interval I l into congruence classes of length q j. For each l < L, we consider all possible values of 1 j K l together. ν q l<l p I l ( fl (p) ) 1 j<k L l<l ρ j,l l<l 1 j<k l ρ j,l I l q j ( ) Il q j ρ j,l. l<l 1 j<k L

117 Applying Lemma 2 and Lemma 4 to the first double sum on the right-hand side above and using that ρ j,l l to the latter, ν q ( n 1 δ <p n p N(p) ) 1 p 1 = l<n δ p I l ν q ( fl (p) )

118 Applying Lemma 2 and Lemma 4 to the first double sum on the right-hand side above and using that ρ j,l l to the latter, ν q ( n 1 δ <p n p N(p) ) 1 p 1 = l<n δ n log log(q + 1) q p I l ν q ( fl (p) ) + n3δ log n. log q The lemma follows.

119 Thank you. Daniel Baczkowski USC