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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


  1. 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

  2. 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 .

  3. 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 .

  4. 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 .

  5. 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 .

  6. 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 .

  7. 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 .

  8. 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

  9. 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

  10. 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

  11. 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

  12. 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.

  13. 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.

  14. 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

  15. 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

  16. 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

  17. 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

  18. 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

  19. 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

  20. 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.

  21. 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.

  22. 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.

  23. 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

  24. 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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