H OW TO STUDY ARITHMETICAL FUNCTIONS ? O VERVIEW M AIN RESULTS F - PowerPoint PPT Presentation

O VERVIEW M AIN RESULTS F UTURE DIRECTION T HANK YOU ! V ARIANT OF A THEOREM OF E RD ˝ OS ON THE SUM - OF - PROPER - DIVISORS FUNCTION Hee-Sung Yang Joint work with Carl Pomerance Dartmouth College 11 January 2013

O VERVIEW M AIN RESULTS F UTURE DIRECTION T HANK YOU ! O UTLINE 1 O VERVIEW Introduction Paul Erd˝ os Herman te Riele 2 M AIN RESULTS Theoretical result Computational result 3 F UTURE DIRECTION 4 T HANK YOU !

O VERVIEW M AIN RESULTS F UTURE DIRECTION T HANK YOU ! I NTRODUCTION H OW TO STUDY ARITHMETICAL FUNCTIONS ?

O VERVIEW M AIN RESULTS F UTURE DIRECTION T HANK YOU ! I NTRODUCTION H OW TO STUDY ARITHMETICAL FUNCTIONS ? Study the distribution of the range of f

O VERVIEW M AIN RESULTS F UTURE DIRECTION T HANK YOU ! I NTRODUCTION H OW TO STUDY ARITHMETICAL FUNCTIONS ? Study the distribution of the range of f Or, study the “non-range” of f , i.e., which integers are not in f ’s range

O VERVIEW M AIN RESULTS F UTURE DIRECTION T HANK YOU ! I NTRODUCTION H OW TO STUDY ARITHMETICAL FUNCTIONS ? Study the distribution of the range of f Or, study the “non-range” of f , i.e., which integers are not in f ’s range

O VERVIEW M AIN RESULTS F UTURE DIRECTION T HANK YOU ! I NTRODUCTION H OW TO STUDY ARITHMETICAL FUNCTIONS ? Study the distribution of the range of f Or, study the “non-range” of f , i.e., which integers are not in f ’s range D EFINITION An integer m is called f-untouchable if there is no n such that f ( n ) = m . Equivalently, m is f -untouchable if m ∈ N \ f ( N ) .

O VERVIEW M AIN RESULTS F UTURE DIRECTION T HANK YOU ! P AUL E RD ˝ OS D ETOUR C ONJECTURE (G OLDBACH ) Every even number greater than or equal to 8 can be written as a sum of two distinct primes. According to this, we can deduce that s ( pq ) = s ∗ ( pq ) = p + q + 1, where p and q are distinct odd primes, will cover all the odd integers ≥ 9. A few things we can learn: Montgomery & Vaughan: The set of odd numbers not of the form p + q + 1 has density 0. It will be more exciting to study even numbers as far as s - (and s ∗ -) untouchables are concerned; Furthermore, examine even numbers if we want to find a positive proportion of s - (and s ∗ -) untouchables!

O VERVIEW M AIN RESULTS F UTURE DIRECTION T HANK YOU ! P AUL E RD ˝ OS OS AND s - UNTOUCHABLE NUMBERS E RD ˝ Erd˝ os Pál (1913 – 1996)

O VERVIEW M AIN RESULTS F UTURE DIRECTION T HANK YOU ! P AUL E RD ˝ OS OS AND s - UNTOUCHABLE NUMBERS E RD ˝

O VERVIEW M AIN RESULTS F UTURE DIRECTION T HANK YOU ! H ERMAN TE R IELE TE R IELE AND s ∗ - UNTOUCHABLE NUMBERS Herman te Riele (b. 1947)

O VERVIEW M AIN RESULTS F UTURE DIRECTION T HANK YOU ! H ERMAN TE R IELE TE R IELE AND s ∗ - UNTOUCHABLE NUMBERS In his doctoral thesis, he tried to tackle s ∗ -untouchables 1 Problem: integers of the form 2 w p ( w ≥ 1 , p an odd prime) 2 Problematic, as there are “too many” 2 w p ’s with s ∗ ( 2 w p ) ≤ x for 3 any x te Riele’s astute observation: de Polignac’s conjecture true ⇒ all 4 even numbers > 2 are s ∗ -touchable C ONJECTURE ( DE P OLIGNAC , 1849) Every odd number greater than 1 can be written in the form 2 k + p, where k ∈ Z + and p an odd prime. The conjecture proved to be false. In fact, Erd˝ os used the theory 5 of covering congruences to disprove this conjecture. This gave us the starting point.

O VERVIEW M AIN RESULTS F UTURE DIRECTION T HANK YOU ! H ERMAN TE R IELE TE R IELE AND s ∗ - UNTOUCHABLE NUMBERS In his doctoral thesis, he tried to tackle s ∗ -untouchables 1 Problem: integers of the form 2 w p ( w ≥ 1 , p an odd prime) 2 Problematic, as there are “too many” 2 w p ’s with s ∗ ( 2 w p ) ≤ x for 3 any x te Riele’s astute observation: de Polignac’s conjecture true ⇒ all 4 even numbers > 2 are s ∗ -touchable C ONJECTURE ( DE P OLIGNAC , 1849) Every odd number greater than 1 can be written in the form 2 k + p, where k ∈ Z + and p an odd prime. The conjecture proved to be false. In fact, Erd˝ os used the theory 5 of covering congruences to disprove this conjecture. This gave us the starting point.

O VERVIEW M AIN RESULTS F UTURE DIRECTION T HANK YOU ! T HEORETICAL RESULT M AIN RESULT

O VERVIEW M AIN RESULTS F UTURE DIRECTION T HANK YOU ! T HEORETICAL RESULT M AIN RESULT T HEOREM (P OMERANCE -Y., 2012) The lower density of the set U ∗ := N \ s ∗ ( N ) is positive.

O VERVIEW M AIN RESULTS F UTURE DIRECTION T HANK YOU ! T HEORETICAL RESULT O UTLINE OF OUR STRATEGY The set of positive lower density that we identify will be a subset 1 of the integers that are 2 mod 4.

O VERVIEW M AIN RESULTS F UTURE DIRECTION T HANK YOU ! T HEORETICAL RESULT O UTLINE OF OUR STRATEGY The set of positive lower density that we identify will be a subset 1 of the integers that are 2 mod 4. Derive an infinite arithmetic progression that are totally missed 2 by the numbers of the form s ∗ ( 2 w p ) (“Case 0”)

O VERVIEW M AIN RESULTS F UTURE DIRECTION T HANK YOU ! T HEORETICAL RESULT O UTLINE OF OUR STRATEGY The set of positive lower density that we identify will be a subset 1 of the integers that are 2 mod 4. Derive an infinite arithmetic progression that are totally missed 2 by the numbers of the form s ∗ ( 2 w p ) (“Case 0”) Now, tackle the remaining cases: 3 Case I: n ≡ 2 ( mod 4 ) , i..e, 2 � n Case II: n = 2 w p a ( a > 1 ) Case III: 4 | n , n has more than one odd prime factor Case IV: n is odd

O VERVIEW M AIN RESULTS F UTURE DIRECTION T HANK YOU ! T HEORETICAL RESULT O UTLINE OF OUR STRATEGY The set of positive lower density that we identify will be a subset 1 of the integers that are 2 mod 4. Derive an infinite arithmetic progression that are totally missed 2 by the numbers of the form s ∗ ( 2 w p ) (“Case 0”) Now, tackle the remaining cases: 3 Case I: n ≡ 2 ( mod 4 ) , i..e, 2 � n Case II: n = 2 w p a ( a > 1 ) Case III: 4 | n , n has more than one odd prime factor Case IV: n is odd We see that Cases II, III, and IV aren’t that exciting. 4

O VERVIEW M AIN RESULTS F UTURE DIRECTION T HANK YOU ! T HEORETICAL RESULT O UTLINE OF OUR STRATEGY The set of positive lower density that we identify will be a subset 1 of the integers that are 2 mod 4. Derive an infinite arithmetic progression that are totally missed 2 by the numbers of the form s ∗ ( 2 w p ) (“Case 0”) Now, tackle the remaining cases: 3 Case I: n ≡ 2 ( mod 4 ) , i..e, 2 � n Case II: n = 2 w p a ( a > 1 ) Case III: 4 | n , n has more than one odd prime factor Case IV: n is odd We see that Cases II, III, and IV aren’t that exciting. 4 However, Case I is more interesting. 5

O VERVIEW M AIN RESULTS F UTURE DIRECTION T HANK YOU ! T HEORETICAL RESULT C ASE 0: n = 2 w p Every w ∈ Z satisfies at least one of the following six congruences: w ≡ 1 ( mod 2 ) , w ≡ 1 ( mod 3 ) w ≡ 2 ( mod 4 ) , w ≡ 4 ( mod 8 ) w ≡ 8 ( mod 12 ) , w ≡ 0 ( mod 24 ) . Now, for each modulus m ∈ { 2 , 3 , 4 , 8 , 12 , 24 } , we find a prime q so that 2 m ≡ 1 ( mod q ) . For z := s ∗ ( 2 w p ) = 1 + 2 w + p we have: 2 w mod q m q z mod q Conclusion z ≡ p z �≡ 0 ( mod 3 ) or p = 3 2 3 2 z ≡ 3 + p z �≡ 3 ( mod 7 ) or p = 7 3 7 2 z ≡ p z �≡ 0 ( mod 5 ) or p = 5 4 5 − 1 − 1 z ≡ p z �≡ 0 ( mod 17 ) or p = 17 8 17 z ≡ − 3 + p z �≡ − 3 ( mod 13 ) or p = 13 − 4 12 13 z ≡ 2 + p z �≡ 2 ( mod 241 ) or p = 241 24 241 1

O VERVIEW M AIN RESULTS F UTURE DIRECTION T HANK YOU ! T HEORETICAL RESULT C ASE 0: n = 2 w p Every w ∈ Z satisfies at least one of the following six congruences: w ≡ 1 ( mod 2 ) , w ≡ 1 ( mod 3 ) w ≡ 2 ( mod 4 ) , w ≡ 4 ( mod 8 ) w ≡ 8 ( mod 12 ) , w ≡ 0 ( mod 24 ) . Now, for each modulus m ∈ { 2 , 3 , 4 , 8 , 12 , 24 } , we find a prime q so that 2 m ≡ 1 ( mod q ) . For z := s ∗ ( 2 w p ) = 1 + 2 w + p we have: 2 w mod q m q z mod q Conclusion z ≡ p z �≡ 0 ( mod 3 ) or p = 3 2 3 2 z ≡ 3 + p z �≡ 3 ( mod 7 ) or p = 7 3 7 2 z ≡ p z �≡ 0 ( mod 5 ) or p = 5 4 5 − 1 − 1 z ≡ p z �≡ 0 ( mod 17 ) or p = 17 8 17 z ≡ − 3 + p z �≡ − 3 ( mod 13 ) or p = 13 − 4 12 13 z ≡ 2 + p z �≡ 2 ( mod 241 ) or p = 241 24 241 1

O VERVIEW M AIN RESULTS F UTURE DIRECTION T HANK YOU ! T HEORETICAL RESULT C ASE 0: n = 2 w p Every w ∈ Z satisfies at least one of the following six congruences: w ≡ 1 ( mod 2 ) , w ≡ 1 ( mod 3 ) w ≡ 2 ( mod 4 ) , w ≡ 4 ( mod 8 ) w ≡ 8 ( mod 12 ) , w ≡ 0 ( mod 24 ) . Now, for each modulus m ∈ { 2 , 3 , 4 , 8 , 12 , 24 } , we find a prime q so that 2 m ≡ 1 ( mod q ) . For z := s ∗ ( 2 w p ) = 1 + 2 w + p we have: 2 w mod q m q z mod q Conclusion z ≡ p z �≡ 0 ( mod 3 ) or p = 3 2 3 2 z ≡ 3 + p z �≡ 3 ( mod 7 ) or p = 7 3 7 2 z ≡ p z �≡ 0 ( mod 5 ) or p = 5 4 5 − 1 − 1 z ≡ p z �≡ 0 ( mod 17 ) or p = 17 8 17 z ≡ − 3 + p z �≡ − 3 ( mod 13 ) or p = 13 − 4 12 13 z ≡ 2 + p z �≡ 2 ( mod 241 ) or p = 241 24 241 1