Determining a Number Through its Sum of Divisors

Transcription

1 Determining a Number Through its Sum of Divisors Carter Smith with Alessandro Rezende De Macedo University of Texas at Austin arter Smith with Alessandro Rezende De Macedo Determining (University a Number of TexasThrough at Austin) its Sum of Divisors

2 Introduction of σ k For n = p e 1 1 pe pe n n we define σ k (n) = d k = d n d>0 n i=1 p k(e i+1) i 1 p k i 1 to be the sum of the k th power of divisors function.

3 Introduction of σ k For n = p e 1 1 pe pe n n we define σ k (n) = d k = d n d>0 n i=1 p k(e i+1) i 1 p k i 1 to be the sum of the k th power of divisors function. Note that σ 0 : N {0} N is the number of divisors function and σ 1 is the sum of divisors function.

4 Introduction of σ k For n = p e 1 1 pe pe n n we define σ k (n) = d k = d n d>0 n i=1 p k(e i+1) i 1 p k i 1 to be the sum of the k th power of divisors function. Note that σ 0 : N {0} N is the number of divisors function and σ 1 is the sum of divisors function. Question: When can we determine a number based off of its σ k value? arter Smith with Alessandro Rezende De Macedo Determining (University a Number of TexasThrough at Austin) its Sum of Divisors

5 When is σ k injective? 1 Input an upper bound n and a k value; 2 compute σ k (a) for all n in 2 a n; 3 remove duplicates from the list of σ k (a) outputs; 4 return the number of duplicates found.

6 When is σ k injective? 1 Input an upper bound n and a k value; 2 compute σ k (a) for all n in 2 a n; 3 remove duplicates from the list of σ k (a) outputs; 4 return the number of duplicates found. σ 0 (p) = 2 for all prime p arter Smith with Alessandro Rezende De Macedo Determining (University a Number of TexasThrough at Austin) its Sum of Divisors

7 When is σ k injective? 1 Input an upper bound n and a k value; 2 compute σ k (a) for all n in 2 a n; 3 remove duplicates from the list of σ k (a) outputs; 4 return the number of duplicates found. σ 0 (p) = 2 for all prime p σ 1 (14) = σ 1 (15) = 24 arter Smith with Alessandro Rezende De Macedo Determining (University a Number of TexasThrough at Austin) its Sum of Divisors

8 When is σ k injective? 1 Input an upper bound n and a k value; 2 compute σ k (a) for all n in 2 a n; 3 remove duplicates from the list of σ k (a) outputs; 4 return the number of duplicates found. σ 0 (p) = 2 for all prime p σ 1 (14) = σ 1 (15) = 24 σ 2 (6) = σ 2 (7) = 50 arter Smith with Alessandro Rezende De Macedo Determining (University a Number of TexasThrough at Austin) its Sum of Divisors

9 When is σ k injective? 1 Input an upper bound n and a k value; 2 compute σ k (a) for all n in 2 a n; 3 remove duplicates from the list of σ k (a) outputs; 4 return the number of duplicates found. σ 0 (p) = 2 for all prime p σ 1 (14) = σ 1 (15) = 24 σ 2 (6) = σ 2 (7) = 50 σ 3 (184, 926) = σ 3 (194, 315) = 7, 401, 260, 364, 550, 416. arter Smith with Alessandro Rezende De Macedo Determining (University a Number of TexasThrough at Austin) its Sum of Divisors

10 When is σ k injective? 1 Input an upper bound n and a k value; 2 compute σ k (a) for all n in 2 a n; 3 remove duplicates from the list of σ k (a) outputs; 4 return the number of duplicates found. σ 0 (p) = 2 for all prime p σ 1 (14) = σ 1 (15) = 24 σ 2 (6) = σ 2 (7) = 50 σ 3 (184, 926) = σ 3 (194, 315) = 7, 401, 260, 364, 550, 416. Computations show that σ 4 s interval of injectivity is 2 n 60, 000, 000.

11 All virtual memory has been exhausted so Magma cannot perform this statement.

12 All virtual memory has been exhausted so Magma cannot perform this statement.

13 Introduction of ψ k Consider ψ k : N N k+1 defined by ψ k (n) = (σ 0 (n), σ 1 (n),..., σ k (n))

14 Introduction of ψ k Consider ψ k : N N k+1 defined by ψ k (n) = (σ 0 (n), σ 1 (n),..., σ k (n)) Natural Question: For which k, if any, is ψ k injective?

15 Introduction of ψ k Consider ψ k : N N k+1 defined by ψ k (n) = (σ 0 (n), σ 1 (n),..., σ k (n)) Natural Question: For which k, if any, is ψ k injective? ψ 0 (n) is not since σ 0 (n) is not. ψ 1 (12) = ψ 1 (18) = (6, 42).

16 Introduction of ψ k Consider ψ k : N N k+1 defined by ψ k (n) = (σ 0 (n), σ 1 (n),..., σ k (n)) Natural Question: For which k, if any, is ψ k injective? ψ 0 (n) is not since σ 0 (n) is not. ψ 1 (12) = ψ 1 (18) = (6, 42). Is ψ 2 (n) injective?

17 Some Observations Note that when n = p e 1 1 pe pe n n, σ 0 (n) = (e 1 + 1)(e 2 + 1)... (e n + 1). arter Smith with Alessandro Rezende De Macedo Determining (University a Number of TexasThrough at Austin) its Sum of Divisors

18 Some Observations Note that when n = p e 1 1 pe pe n n, σ 0 (n) = (e 1 + 1)(e 2 + 1)... (e n + 1). ψ 2 (n) determines n when σ 0 (n) = 2. Why?

19 Some Observations Note that when n = p e 1 1 pe pe n n, σ 0 (n) = (e 1 + 1)(e 2 + 1)... (e n + 1). ψ 2 (n) determines n when σ 0 (n) = 2. Why? n = σ 1 1

20 Some Observations Note that when n = p e 1 1 pe pe n n, σ 0 (n) = (e 1 + 1)(e 2 + 1)... (e n + 1). ψ 2 (n) determines n when σ 0 (n) = 2. Why? n = σ 1 1 ψ 2 (n) determines n when σ 0 (n) = 3. Why?

21 Some Observations Note that when n = p e 1 1 pe pe n n, σ 0 (n) = (e 1 + 1)(e 2 + 1)... (e n + 1). ψ 2 (n) determines n when σ 0 (n) = 2. Why? n = σ 1 1 ψ 2 (n) determines n when σ 0 (n) = 3. Why? n = p 2 and σ 1 (n) = 1 + p + p 2. When σ 0 (n) = q for a prime q, σ 1 (n) determines n since n = p q 1 for some prime p.

22 The case when σ 0 (n) = 4 Note that if σ 0 (n) = 4 then n = p 3 or qr for primes p, q, r. If ψ 2 (n) were to fail to be injective, two cases would arise. For some m = p 3 and n = qr for distinct primes p, q, r we would have

23 The case when σ 0 (n) = 4 Note that if σ 0 (n) = 4 then n = p 3 or qr for primes p, q, r. If ψ 2 (n) were to fail to be injective, two cases would arise. For some m = p 3 and n = qr for distinct primes p, q, r we would have σ 1 (m) = 1 + p + p 2 + p 3 = 1 + q + r + qr = σ 1 (n) (1) σ 2 (m) = 1 + p 2 + p 4 + p 6 = 1 + q 2 + r 2 + q 2 r 2 = σ 2 (n) (2) arter Smith with Alessandro Rezende De Macedo Determining (University a Number of TexasThrough at Austin) its Sum of Divisors

24 The case when σ 0 (n) = 4 Note that if σ 0 (n) = 4 then n = p 3 or qr for primes p, q, r. If ψ 2 (n) were to fail to be injective, two cases would arise. For some m = p 3 and n = qr for distinct primes p, q, r we would have Mod q: σ 1 (m) = 1 + p + p 2 + p 3 = 1 + q + r + qr = σ 1 (n) (1) σ 2 (m) = 1 + p 2 + p 4 + p 6 = 1 + q 2 + r 2 + q 2 r 2 = σ 2 (n) (2) 2p 3 + 2p 4 + 2p 5 + p 2 + p 4 + p 6 = r 2 r 2 + 2p 3 + 2p 4 + 2p 5 = r 2 2p 3 + 2p 4 + 2p 5 = 0 2p 2 (p + p 2 + p 3 ) = 0 2p 2 r = 0. arter Smith with Alessandro Rezende De Macedo Determining (University a Number of TexasThrough at Austin) its Sum of Divisors

25 What about when σ 0 (n) = 8? Such a number is of the form p 7, pqr, or p 3 q for primes p, q, and r. That would give us five cases to check. n = p 7 1, m = p 2p 3 p 4 ; n = p 7 1, m = p3 2 p 3; n = p 1 p 2 p 3, m = p 3 4 p 5; n = p 1 p 2 p 3, m = p 4 p 5 p 6 ; n = p 3 1 p 2, m = p 3 3 p 4.

26 What about when σ 0 (n) = 8? Such a number is of the form p 7, pqr, or p 3 q for primes p, q, and r. That would give us five cases to check. n = p 7 1, m = p 2p 3 p 4 ; n = p 7 1, m = p3 2 p 3; n = p 1 p 2 p 3, m = p 3 4 p 5; n = p 1 p 2 p 3, m = p 4 p 5 p 6 ; n = p 3 1 p 2, m = p 3 3 p 4. If we were to use the same method on just one case then we would be forced to compute as well as (p 1 + p p3 1 + p4 1 + p5 1 + p6 1 + p7 1 )2 (p 2 + p 3 + p 4 + p 2 p 3 + p 2 p 4 + p 3 p 4 + p 2 p 3 p 4 ) 2.

27 Computations We have three separate programs to compute ψ 2 (n) when σ 0 = 8, 9 or 10. arter Smith with Alessandro Rezende De Macedo Determining (University a Number of TexasThrough at Austin) its Sum of Divisors

28 Computations We have three separate programs to compute ψ 2 (n) when σ 0 = 8, 9 or 10. When σ 0 (n) = 8, ψ 2 s interval of injectivity is at least 2 n arter Smith with Alessandro Rezende De Macedo Determining (University a Number of TexasThrough at Austin) its Sum of Divisors

29 Computations We have three separate programs to compute ψ 2 (n) when σ 0 = 8, 9 or 10. When σ 0 (n) = 8, ψ 2 s interval of injectivity is at least 2 n When σ 0 (n) = 9, ψ 2 s interval of injectivity is at least 2 n arter Smith with Alessandro Rezende De Macedo Determining (University a Number of TexasThrough at Austin) its Sum of Divisors

30 Computations We have three separate programs to compute ψ 2 (n) when σ 0 = 8, 9 or 10. When σ 0 (n) = 8, ψ 2 s interval of injectivity is at least 2 n When σ 0 (n) = 9, ψ 2 s interval of injectivity is at least 2 n When σ 0 (n) = 10, ψ 2 s interval of injectivity is at least 2 n arter Smith with Alessandro Rezende De Macedo Determining (University a Number of TexasThrough at Austin) its Sum of Divisors

31 Geometric Intuition Counterexamples can be thought of as integral solutions to the intersection of affine varieties.

32 Geometric Intuition Counterexamples can be thought of as integral solutions to the intersection of affine varieties. The intersection of higher dimensional varieties should have a better chance at having integral solutions, right?

33 Geometric Intuition Counterexamples can be thought of as integral solutions to the intersection of affine varieties. The intersection of higher dimensional varieties should have a better chance at having integral solutions, right? ψ 2 ( ) = ψ 2 ( ) = (32, , )

34 Further Questions Can we generalize σ k for larger integer rings? (Spira)

35 Further Questions Can we generalize σ k for larger integer rings? (Spira) Geometric approach using intersection of affine varieties.

36 Further Questions Can we generalize σ k for larger integer rings? (Spira) Geometric approach using intersection of affine varieties. What happens when we let k vary over all real numbers?

37 Further Questions Can we generalize σ k for larger integer rings? (Spira) Geometric approach using intersection of affine varieties. What happens when we let k vary over all real numbers? Is ψ k ever injective?

38 THANK YOU!

39 Sources Cooking with python, part 2 - OReilly media. (2005, June 23). Retrieved November 23, 2016, from Cooking With Python Dummit, D. S., and Foote, R. M. (2004). Abstract algebra. Hoboken, NJ: Wiley. Frohlich A., and Taylor, M. (1991). Algebraic number theory. Cambridge: Cambridge University Press. R. Spira, The Complex Sum of Divisors, The American Mathematical Monthly Vol. 68, No. 2 (1961),