Arithmetic Functions Evaluated at Factorials!

Transcription

1 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. 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.

4 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.

5 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.

6 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 1. 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

7 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).

8 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.

9 Lemma 1 (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.

10 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.

11 b σ(n!) = a m! 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.

12 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 1, 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.

13 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.

14 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.

15 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,...

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

17 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

18 Ideas for Lemma 6 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.

19 Ideas for Lemma 6 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.

20 Smaller primes p, i.e. p n/l Lemma 2. a, N Z, N = q r M, r 0 q Φ N (a) if and only if M = ord q (a) Lemma 3. ν q ( a N 1 ) log N + ord q(a) log a log q

21 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.

22 ( )) 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. 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).

23 ρ j,l = ρ j,l (q) = {t Z : 0 t q j 1, f l (t) 0 (mod q j )}. With some love and tender care: ν q ( n/l<p n l<l ( p N(p) ) 1 p 1 1 j J which gives the result. = l<l p I l ν q 2 I l ρ j,l φ(q j ) log ( I l /q j) + ( fl (p) ) J<j<K L ρ j,l ( )) Il q j + 1

24 The second part of the lemma? It s similar but less involved. Take L = n δ. Partition the interval I l into congruence classes of length q j. ( fl (p) ) ( ) Il q j + 1 ν q l<l p I l 1 j<k L l<l ρ j,l l<l 1 j<k l ρ j,l I l q j + ρ j,l. l<l 1 j<k L Lemma 4. ρ j,l 2 gcd ( φ(q j ), l+1 ) distinct roots modulo q j.

25 Lemma 5. l=1 gcd(φ(q j ), l) l 2 log log(q + 1) Applying Lemma 4 and Lemma 5 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 Main Lemma follows.

26 Thank you. Daniel Baczkowski USC