Partition Congruences in the Spirit of Ramanujan Yezhou Wang School of Mathematical Sciences University of Electronic Science and Technology of China yzwang@uestc.edu.cn Monash Discrete Mathematics Research Group Meeting Aug 22, 2016
Introduction Definition A partition of a positive integer n is a representation of n as a sum of positive integers, called parts , the order of which is irrelevant. Example The partitions of 4 are 4 = 4 = 3 + 1 = 2 + 2 = 2 + 1 + 1 = 1 + 1 + 1 + 1 .
Introduction Definition A partition of a positive integer n is a representation of n as a sum of positive integers, called parts , the order of which is irrelevant. Example The partitions of 4 are 4 = 4 = 3 + 1 = 2 + 2 = 2 + 1 + 1 = 1 + 1 + 1 + 1 .
Definition Let p ( n ) denote the number of partitions of n . The value of p ( n ) for 0 ≤ n ≤ 10 is shown below: n 0 1 2 3 4 5 6 7 8 9 10 p ( n ) 1 1 2 3 5 7 11 15 22 30 42 The number p ( n ) increases quite rapidly with n . For example, p (50) = 204,226 , p (100) = 190,569,292 , p (200) = 3,972,999,029,388 , p (1000) = 24,061,467,864,032,622,473,692,149,727,991 .
Definition Let p ( n ) denote the number of partitions of n . The value of p ( n ) for 0 ≤ n ≤ 10 is shown below: n 0 1 2 3 4 5 6 7 8 9 10 p ( n ) 1 1 2 3 5 7 11 15 22 30 42 The number p ( n ) increases quite rapidly with n . For example, p (50) = 204,226 , p (100) = 190,569,292 , p (200) = 3,972,999,029,388 , p (1000) = 24,061,467,864,032,622,473,692,149,727,991 .
Definition Let p ( n ) denote the number of partitions of n . The value of p ( n ) for 0 ≤ n ≤ 10 is shown below: n 0 1 2 3 4 5 6 7 8 9 10 p ( n ) 1 1 2 3 5 7 11 15 22 30 42 The number p ( n ) increases quite rapidly with n . For example, p (50) = 204,226 , p (100) = 190,569,292 , p (200) = 3,972,999,029,388 , p (1000) = 24,061,467,864,032,622,473,692,149,727,991 .
Definition Let p ( n ) denote the number of partitions of n . The value of p ( n ) for 0 ≤ n ≤ 10 is shown below: n 0 1 2 3 4 5 6 7 8 9 10 p ( n ) 1 1 2 3 5 7 11 15 22 30 42 The number p ( n ) increases quite rapidly with n . For example, p (50) = 204,226 , p (100) = 190,569,292 , p (200) = 3,972,999,029,388 , p (1000) = 24,061,467,864,032,622,473,692,149,727,991 .
How can we calculate p ( n )? Theorem (Euler) The generating function of p ( n ) satisfies ∞ ∞ 1 p ( n ) q n = � � 1 − q n . n = 0 n = 1 This is because 1 1 − q k = 1 + q k + q 2 k + q 3 k + · · · . For simplicity, we adopt the following notation ∞ � (1 − q n ) . ( q ; q ) ∞ = n = 1
How can we calculate p ( n )? Theorem (Euler) The generating function of p ( n ) satisfies ∞ ∞ 1 p ( n ) q n = � � 1 − q n . n = 0 n = 1 This is because 1 1 − q k = 1 + q k + q 2 k + q 3 k + · · · . For simplicity, we adopt the following notation ∞ � (1 − q n ) . ( q ; q ) ∞ = n = 1
How can we calculate p ( n )? Theorem (Euler) The generating function of p ( n ) satisfies ∞ ∞ 1 p ( n ) q n = � � 1 − q n . n = 0 n = 1 This is because 1 1 − q k = 1 + q k + q 2 k + q 3 k + · · · . For simplicity, we adopt the following notation ∞ � (1 − q n ) . ( q ; q ) ∞ = n = 1
How can we calculate p ( n )? Theorem (Euler) The generating function of p ( n ) satisfies ∞ ∞ 1 p ( n ) q n = � � 1 − q n . n = 0 n = 1 This is because 1 1 − q k = 1 + q k + q 2 k + q 3 k + · · · . For simplicity, we adopt the following notation ∞ � (1 − q n ) . ( q ; q ) ∞ = n = 1
Theorem (Euler’s Pentagonal Number Theorem) ∞ � ( − 1) n q n (3 n − 1) / 2 ( q ; q ) ∞ = n = −∞ ∞ � q n (3 n − 1) / 2 + q n (3 n + 1) / 2 � ( − 1) n � = 1 + n = 1 = 1 − q − q 2 + q 5 + q 7 − q 12 − q 15 + q 22 + q 26 − · · · . The numbers n (3 n − 1) / 2 are called pentagonal numbers. Figure: The pentagonal numbers 1 , 5 , 12 , 22.
Theorem (Euler’s Pentagonal Number Theorem) ∞ � ( − 1) n q n (3 n − 1) / 2 ( q ; q ) ∞ = n = −∞ ∞ � q n (3 n − 1) / 2 + q n (3 n + 1) / 2 � ( − 1) n � = 1 + n = 1 = 1 − q − q 2 + q 5 + q 7 − q 12 − q 15 + q 22 + q 26 − · · · . The numbers n (3 n − 1) / 2 are called pentagonal numbers. Figure: The pentagonal numbers 1 , 5 , 12 , 22.
Now we have ∞ ∞ � � p ( n ) q n ( − 1) n q n (3 n − 1) / 2 = 1 . n = −∞ n = 0 A recurrence formula for p ( n ) is obtained immediately � � � � �� n − k (3 k − 1) n − k (3 k + 1) � ( − 1) k − 1 p ( n ) = p + p . 2 2 n ≥ 1 An asymptotic expression for p ( n ) is Theorem (Hardy and Ramanujan, 1918) � 1 2 n p ( n ) ∼ √ exp π . 3 4 n 3
Now we have ∞ ∞ � � p ( n ) q n ( − 1) n q n (3 n − 1) / 2 = 1 . n = −∞ n = 0 A recurrence formula for p ( n ) is obtained immediately � � � � �� n − k (3 k − 1) n − k (3 k + 1) � ( − 1) k − 1 p ( n ) = p + p . 2 2 n ≥ 1 An asymptotic expression for p ( n ) is Theorem (Hardy and Ramanujan, 1918) � 1 2 n p ( n ) ∼ √ exp π . 3 4 n 3
Now we have ∞ ∞ � � p ( n ) q n ( − 1) n q n (3 n − 1) / 2 = 1 . n = −∞ n = 0 A recurrence formula for p ( n ) is obtained immediately � � � � �� n − k (3 k − 1) n − k (3 k + 1) � ( − 1) k − 1 p ( n ) = p + p . 2 2 n ≥ 1 An asymptotic expression for p ( n ) is Theorem (Hardy and Ramanujan, 1918) � 1 2 n p ( n ) ∼ √ exp π . 3 4 n 3
Ramanujan’s Famous Congruences Srinivasa Ramanujan (1887–1920) is acknowl- edged as an India’s greatest mathematical genius. He made substantial contributions to analytic number theory elliptic functions q -series
Ramanujan’s Famous Congruences Srinivasa Ramanujan (1887–1920) is acknowl- edged as an India’s greatest mathematical genius. He made substantial contributions to analytic number theory elliptic functions q -series
Ramanujan’s Famous Congruences Srinivasa Ramanujan (1887–1920) is acknowl- edged as an India’s greatest mathematical genius. He made substantial contributions to analytic number theory elliptic functions q -series
Theorem (Ramanujan, 1919) For all n ≥ 0 , p (5 n + 4) ≡ 0 (mod 5) , p (7 n + 5) ≡ 0 (mod 7) , p (11 n + 6) ≡ 0 (mod 11) . Ramanujan’s beautiful identities ∞ p (5 n + 4) q n = 5( q 5 ; q 5 ) 5 � ∞ , ( q ; q ) 6 ∞ n = 0 ∞ p (7 n + 5) q n = 7( q 7 ; q 7 ) 3 + 49 q ( q 7 ; q 7 ) 7 � ∞ ∞ . ( q ; q ) 4 ( q ; q ) 8 ∞ ∞ n = 0 No such simple identity exists for modulo 11.
Theorem (Ramanujan, 1919) For all n ≥ 0 , p (5 n + 4) ≡ 0 (mod 5) , p (7 n + 5) ≡ 0 (mod 7) , p (11 n + 6) ≡ 0 (mod 11) . Ramanujan’s beautiful identities ∞ p (5 n + 4) q n = 5( q 5 ; q 5 ) 5 � ∞ , ( q ; q ) 6 ∞ n = 0 ∞ p (7 n + 5) q n = 7( q 7 ; q 7 ) 3 + 49 q ( q 7 ; q 7 ) 7 � ∞ ∞ . ( q ; q ) 4 ( q ; q ) 8 ∞ ∞ n = 0 No such simple identity exists for modulo 11.
Theorem (Ramanujan, 1919) For all n ≥ 0 , p (5 n + 4) ≡ 0 (mod 5) , p (7 n + 5) ≡ 0 (mod 7) , p (11 n + 6) ≡ 0 (mod 11) . Ramanujan’s beautiful identities ∞ p (5 n + 4) q n = 5( q 5 ; q 5 ) 5 � ∞ , ( q ; q ) 6 ∞ n = 0 ∞ p (7 n + 5) q n = 7( q 7 ; q 7 ) 3 + 49 q ( q 7 ; q 7 ) 7 � ∞ ∞ . ( q ; q ) 4 ( q ; q ) 8 ∞ ∞ n = 0 No such simple identity exists for modulo 11.
Ramanujan’s Original Proof of p (5 n + 4) ≡ 0 (mod 5) By Fermat’s Little Theorem, we have (1 − q ) 5 ≡ 1 − q 5 (mod 5). Thus, ( q ; q ) 5 ∞ ≡ ( q 5 ; q 5 ) ∞ (mod 5) . Now we have ∞ p ( n ) q n + 1 = q ( q 5 ; q 5 ) ∞ � ( q 5 ; q 5 ) ∞ ≡ q ( q ; q ) 4 ∞ (mod 5) . ( q ; q ) ∞ n = 0 To prove that 5 | p (5 n + 4), we must show that the coe ffi cients of q 5 n + 5 in q ( q ; q ) 4 ∞ are multiples of 5. We now employ Jacobi’s identity ∞ � ( q ; q ) 3 ( − 1) k (2 k + 1) q k ( k + 1) / 2 . ∞ = k = 0
Ramanujan’s Original Proof of p (5 n + 4) ≡ 0 (mod 5) By Fermat’s Little Theorem, we have (1 − q ) 5 ≡ 1 − q 5 (mod 5). Thus, ( q ; q ) 5 ∞ ≡ ( q 5 ; q 5 ) ∞ (mod 5) . Now we have ∞ p ( n ) q n + 1 = q ( q 5 ; q 5 ) ∞ � ( q 5 ; q 5 ) ∞ ≡ q ( q ; q ) 4 ∞ (mod 5) . ( q ; q ) ∞ n = 0 To prove that 5 | p (5 n + 4), we must show that the coe ffi cients of q 5 n + 5 in q ( q ; q ) 4 ∞ are multiples of 5. We now employ Jacobi’s identity ∞ � ( q ; q ) 3 ( − 1) k (2 k + 1) q k ( k + 1) / 2 . ∞ = k = 0
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