Some results on k-free numbers

Transcription

1 Patrick Letendre 05 october 2013 Patrick Letendre Some results on k-free numbers 05 october / 17

2 A local point of view (1) Patrick Letendre Some results on k-free numbers 05 october / 17

3 A local point of view (1) Let where µ k (n) = Q k (x) := x µ k (n), n=1 { 1 if n is k-free 0 if n is not k-free. Patrick Letendre Some results on k-free numbers 05 october / 17

4 A local point of view (1) Let where µ k (n) = Q k (x) := x µ k (n), n=1 { 1 if n is k-free 0 if n is not k-free. For h N, we define g k (h) := max x 0 Q k(x + h) Q k (x). Patrick Letendre Some results on k-free numbers 05 october / 17

5 A local point of view (2) Patrick Letendre Some results on k-free numbers 05 october / 17

6 A local point of view (2) For some c 1,k, c 2,k > 0 we have c 1,k h 1/k (ln h) 1/k (ln ln h) 1 1/k g k(h) h ζ(k) c h 2/(k+1) 2,k (ln h). 2k/(k+1) Patrick Letendre Some results on k-free numbers 05 october / 17

7 A local point of view (2) For some c 1,k, c 2,k > 0 we have c 1,k h 1/k (ln h) 1/k (ln ln h) 1 1/k g k(h) h ζ(k) c h 2/(k+1) 2,k (ln h). 2k/(k+1) It is unclear what is the exact order for g k (h). We expect that for each ɛ > 0. g k (h) h ζ(k) h1/k+ɛ Patrick Letendre Some results on k-free numbers 05 october / 17

8 A local point of view (2) For some c 1,k, c 2,k > 0 we have c 1,k h 1/k (ln h) 1/k (ln ln h) 1 1/k g k(h) h ζ(k) c h 2/(k+1) 2,k (ln h). 2k/(k+1) It is unclear what is the exact order for g k (h). We expect that g k (h) h ζ(k) h1/k+ɛ for each ɛ > 0. Under a plausible conjecture we can show that for each ɛ > 0. g k (h) h ζ(k) h3/(2k+1)+ɛ Patrick Letendre Some results on k-free numbers 05 october / 17

9 A local point of view (3) Patrick Letendre Some results on k-free numbers 05 october / 17

10 A local point of view (3) On the opposite direction, we can find gaps of length ζ(k) k ln x (1 + o(1)) ln ln x as x. So at least everything in 0 Q k (x + h) Q k (x) h ζ(k) + c h 1/k 1,k (ln h) 1/k (ln ln h) 1 1/k happen. Patrick Letendre Some results on k-free numbers 05 october / 17

11 A first global point of view (1) Patrick Letendre Some results on k-free numbers 05 october / 17

12 A first global point of view (1) We want to know the normal behavior of Q k (x + h) Q k (x). Patrick Letendre Some results on k-free numbers 05 october / 17

13 A first global point of view (1) We want to know the normal behavior of Q k (x + h) Q k (x). We define N 1 h M k (N, h) := µ k (n + a) h 2 ζ(k). n=0 a=1 Patrick Letendre Some results on k-free numbers 05 october / 17

14 A first global point of view (2) Theorem. (Hall, L.) We have M k (N, h) = γ k h 1/k N + E k (h, N) where and E k (h, N) Nh 1/2k exp( c ln h) + Nh θ k +N 2/(k+1) h 2 2/(k+1) ln N + h 2 N 1/k ln N ζ(1/k 1) γ k := 2 1/k 1 ( p 1 1 p 2 2 p k + 2 p k+1 ). Patrick Letendre Some results on k-free numbers 05 october / 17

15 A first global point of view (3) Patrick Letendre Some results on k-free numbers 05 october / 17

16 A first global point of view (3) We can take θ 2 = 11 53, θ 3 = 1 6, θ 4 = , θ 5 = , θ 6 = θ k { 6 7k+6 for k {7, 8, 9, 10, 11} 1 for k 12 k+3. Patrick Letendre Some results on k-free numbers 05 october / 17

17 A first global point of view (3) We can take θ 2 = 11 53, θ 3 = 1 6, θ 4 = , θ 5 = , θ 6 = θ k { 6 7k+6 for k {7, 8, 9, 10, 11} 1 for k 12 k+3. With the exponent pairs conjecture, we have for each ɛ > 0. θ k = 1 2k ɛ Patrick Letendre Some results on k-free numbers 05 october / 17

18 A first global point of view (4) Patrick Letendre Some results on k-free numbers 05 october / 17

19 A first global point of view (4) For all but Nh 2ɛ value of x [0, N] N we have Q k (x + h) Q k (x) = for h N k 1 2k 1 /ln k 2k 1 N. h ζ(k) + O(h1/2k+ɛ ) Patrick Letendre Some results on k-free numbers 05 october / 17

20 A first global point of view (4) For all but Nh 2ɛ value of x [0, N] N we have Q k (x + h) Q k (x) = for h N k 1 2k 1 /ln k 2k 1 N. We also have the upper bound for h N 1/2. M k (h, N) Nh 2/k h ζ(k) + O(h1/2k+ɛ ) Patrick Letendre Some results on k-free numbers 05 october / 17

21 A first global point of view (4) For all but Nh 2ɛ value of x [0, N] N we have Q k (x + h) Q k (x) = for h N k 1 2k 1 /ln k 2k 1 N. We also have the upper bound for h N 1/2. M k (h, N) Nh 2/k h ζ(k) + O(h1/2k+ɛ ) There is a lot of overlapping of intervals in the definition of M k (N, h). Patrick Letendre Some results on k-free numbers 05 october / 17

22 A second global point of view (1) Patrick Letendre Some results on k-free numbers 05 october / 17

23 A second global point of view (1) For N = h(m + 1) and for each β [0, h 1] N we define M h T k (β, h, M) := µ k (nh + a + β) h 2 ζ(k). n=0 a=1 Patrick Letendre Some results on k-free numbers 05 october / 17

24 A second global point of view (1) For N = h(m + 1) and for each β [0, h 1] N we define T k (β, h, M) := We have the relation M h µ k (nh + a + β) h 2 ζ(k). n=0 a=1 h 1 T k (β, h, M) = M k (N, h). β=0 Patrick Letendre Some results on k-free numbers 05 october / 17

25 A second global point of view (2) Theorem. (L.) For N = h(m + 1) we have where T k (β, h, M) = γ k h 1/k (M + 1) + R k (h, N) R k (h, N) Mh θ k + Mh 1/2k p j (h,γ k (h)) +h 2 M 1/k ln N + hm 2/(k+1) ln N p h ( p k j p 1 j/2k ( ) p k/(k+1) ) Patrick Letendre Some results on k-free numbers 05 october / 17

26 A second global point of view (3) Patrick Letendre Some results on k-free numbers 05 october / 17

27 A second global point of view (3) We also have the upper bound T k (β, h, N) Mh 2/k for h N 1/3. Patrick Letendre Some results on k-free numbers 05 october / 17

28 A second global point of view (3) We also have the upper bound T k (β, h, N) Mh 2/k for h N 1/3. We have found a way to cut the interval [1, N] in small subsets and show that the average of µ k on those subsets is what we expect for most of them. Can we do the same for more general partition of [1, N] like µ k (n) I 2 n ζ(k) n 1 a I n where I n are intervals with h I n 2h, I n = [1, N], I n I m = n m. n Patrick Letendre Some results on k-free numbers 05 october / 17

29 Arithmetic progression (1) Patrick Letendre Some results on k-free numbers 05 october / 17

30 Arithmetic progression (1) We define where and Q k (a, q, x) := g k (a, q) := 1 qζ(k) W k (q, x) := x n=1 n a mod q q E k (a, q, x) 2 a=1 µ k (n) =: g k (a, q)x + E k (a, q, x) ( 1 1 ) 1 p k p q p q (q,p k ) a ( 1 (q, ) pk ). p k Patrick Letendre Some results on k-free numbers 05 october / 17

31 Arithmetic progression (2) Theorem. (L.) For x = qm, with M 1 and M R, we have W K (q, x) = γ k (q)qm 1/k + R k (q, x) where R k (q, x) 2 ω(q) q ( ) 2p p q +M θ k q ( p kθ k p q + l(kθ k )M 1/2k q p q ) + 2ω(q) x 1+1/k ln x. q ( ) p 1/2 Patrick Letendre Some results on k-free numbers 05 october / 17

32 Arithmetic progression (3) ζ(1/k 1) γ k (q) := 2 1/k 1 p α q (r 1)k<α rk r j=1 1 p jk 1 p α+r α/r (( 1 1 p ( p q 1 1 p 2 2 p k + 2 p k+1 )(( 1 1 p 2 2 p k + 2 p k+1 ( µ(p j ) 1 p 1 p p k+1 )) ([ ] 1 1 r p 1 p + 2 )) 2 p k+1 ) ) Patrick Letendre Some results on k-free numbers 05 october / 17

33 Arithmetic progression (4) Corollary. For 1 q x we have W k (q, x) x 2/k q 1 2/k. Patrick Letendre Some results on k-free numbers 05 october / 17

34 Applications Some results on k-free numbers Patrick Letendre Some results on k-free numbers 05 october / 17

35 Applications Some results on k-free numbers Evaluate for various functions. x µ k (n)f (n) n=1 Patrick Letendre Some results on k-free numbers 05 october / 17

36 Applications Evaluate for various functions. x µ k (n)f (n) n=1 Counting the solutions to the congruence f (x 1,..., x n ) 0 mod q for a polynomial f Z[x 1,..., x n ] when (x 1,..., x n ) [1, x] n with q x. Patrick Letendre Some results on k-free numbers 05 october / 17

37 References Some results on k-free numbers T. Cochrane and S. Shi, The congruence x 1 x 2 x 3 x 4 (mod m) and mean values of character sums, J. Number Theory 130 (2010), S. W. Graham, Moments of gaps between k-free numbers, J. Number Theory 44 (1993), R. R. Hall, Squarefree numbers on short intervals, Mathematika 29 (1982), P. Letendre, to appear. Patrick Letendre Some results on k-free numbers 05 october / 17