Modular forms, combinatorially and otherwise p. 1/103 Modular forms, combinatorially and otherwise David Penniston
Sums of squares Modular forms, combinatorially and otherwise p. 2/103
Modular forms, combinatorially and otherwise p. 2/103 Sums of squares Which primes can be written as a sum of two integer squares?
Modular forms, combinatorially and otherwise p. 3/103 2 = 1 2 + 1 2 5 = 1 2 + 2 2 13 = 2 2 + 3 2 17 = 1 2 + 4 2 29 = 2 2 + 5 2
A prime p is equal to a sum of two squares if and only p = 2 or p 1 (mod 4) Modular forms, combinatorially and otherwise p. 4/103
Modular forms, combinatorially and otherwise p. 5/103 0 2 = 0 0 (mod 4) 1 2 = 1 1 (mod 4) 2 2 = 4 0 (mod 4) 3 2 = 9 1 (mod 4)
Modular forms, combinatorially and otherwise p. 5/103 0 2 = 0 0 (mod 4) 1 2 = 1 1 (mod 4) 2 2 = 4 0 (mod 4) 3 2 = 9 1 (mod 4) s 2 {0, 1} (mod 4)
Modular forms, combinatorially and otherwise p. 5/103 0 2 = 0 0 (mod 4) 1 2 = 1 1 (mod 4) 2 2 = 4 0 (mod 4) 3 2 = 9 1 (mod 4) s 2 {0, 1} (mod 4) s 2 + t 2 {0, 1, 2} (mod 4)
How many squares suffice? Modular forms, combinatorially and otherwise p. 6/103
Modular forms, combinatorially and otherwise p. 6/103 How many squares suffice? s 2 {0, 1, 4} (mod 8)
Modular forms, combinatorially and otherwise p. 6/103 How many squares suffice? s 2 {0, 1, 4} (mod 8) s 2 + t 2 + u 2 {0, 1, 2, 3, 4, 5, 6} (mod 8)
Modular forms, combinatorially and otherwise p. 7/103 Four squares theorem (Legendre) Every positive integer can be written as a sum of four integer squares
Modular forms, combinatorially and otherwise p. 8/103 Theta function θ(q) = s Z q s2
Modular forms, combinatorially and otherwise p. 8/103 Theta function θ(q) = s Z q s2 = 1 + 2q + 2q 4 + 2q 9 +
Modular forms, combinatorially and otherwise p. 9/103 θ(q) 2 = s,t Z q s2 +t 2 = 1 + 4q + 4q 2 + 4q 4 + 8q 5 +
Modular forms, combinatorially and otherwise p. 9/103 θ(q) 2 = s,t Z q s2 +t 2 = 1 + 4q + 4q 2 + 4q 4 + 8q 5 + = n=0 r 2 (n)q n r 2 (n) = # of reps of n as a sum of two integer squares
Modular forms, combinatorially and otherwise p. 10/103... + 4q 4 + 8q 5 +... 4 = (±2) 2 + 0 2 = 0 2 + (±2) 2 5 = (±1) 2 + (±2) 2 = (±2) 2 + (±1) 2
Modular forms, combinatorially and otherwise p. 11/103 Four squares theorem The coefficient of q n in θ(q) 4 is positive for every n 1
Modular forms, combinatorially and otherwise p. 11/103 Four squares theorem The coefficient of q n in θ(q) 4 is positive for every n 1 θ(q) 4 = 1 + 8q + 24q 2 + 32q 3 + 24q 4 + 48q 5 +
θ analytically Modular forms, combinatorially and otherwise p. 12/103
Modular forms, combinatorially and otherwise p. 12/103 θ analytically q = e 2πiz
Modular forms, combinatorially and otherwise p. 12/103 θ analytically q = e 2πiz θ(z + 1) = θ(z)
Modular forms, combinatorially and otherwise p. 12/103 θ analytically q = e 2πiz θ(z + 1) = θ(z) θ( 1/4z) = 2iz θ(z)
Modular forms, combinatorially and otherwise p. 13/103 Partitions A partition of n is a nonincreasing sequence of positive integers whose sum is n
Modular forms, combinatorially and otherwise p. 14/103 Partitions of 4 4 3 + 1 2 + 2 2 + 1 + 1 1 + 1 + 1 + 1
p(n) := number of partitions of n Modular forms, combinatorially and otherwise p. 15/103
Modular forms, combinatorially and otherwise p. 15/103 p(n) := number of partitions of n p(4) = 5
Modular forms, combinatorially and otherwise p. 16/103 n p(n) 10 42 20 627 30 5604 40 37338 50 204226
p(0) p(1) p(2) p(3) p(4) p(5) p(6) p(7) p(8) p(9) p(10) p(11) p(12) p(13) p(14) p(15) p(16) p(17) p(18) p(19) p(20) p(21) p(22) p(23) p(24) p(25) p(26) p(27) p(28) p(29) p(30) p(31) p(32) p(33) p(34) p(35) p(36) p(37) p(38) p(39) p(40) p(41) p(42) p(43) p(44) p(45) p(46) p(47) p(48) p(49) Modular forms, combinatorially and otherwise p. 17/103
1 1 2 3 5 7 11 15 22 30 42 56 77 101 135 176 231 297 385 490 627 792 1002 1255 1575 1958 2436 3010 3718 4565 5604 6842 8349 10143 12310 14883 17977 21637 26015 31185 37338 44583 53174 63261 75175 89134 105558 124754 147273 173525 Modular forms, combinatorially and otherwise p. 18/103
1 1 2 3 5 7 11 15 22 30 42 56 77 101 135 176 231 297 385 490 627 792 1002 1255 1575 1958 2436 3010 3718 4565 5604 6842 8349 10143 12310 14883 17977 21637 26015 31185 37338 44583 53174 63261 75175 89134 105558 124754 147273 173525 Modular forms, combinatorially and otherwise p. 19/103
Modular forms, combinatorially and otherwise p. 20/103 p(4),p(9),p(14),p(19),...,p(49) are all divisible by 5
Modular forms, combinatorially and otherwise p. 20/103 p(4),p(9),p(14),p(19),...,p(49) are all divisible by 5 Is p(5n + 4) divisible by 5 for every n 0?
Ramanujan congruences Modular forms, combinatorially and otherwise p. 21/103
Modular forms, combinatorially and otherwise p. 22/103 Ramanujan congruences For every n 0, p(5n + 4) 0 (mod 5)
Modular forms, combinatorially and otherwise p. 22/103 Ramanujan congruences For every n 0, p(5n + 4) 0 (mod 5) p(7n + 5) 0 (mod 7)
Modular forms, combinatorially and otherwise p. 22/103 Ramanujan congruences For every n 0, p(5n + 4) 0 (mod 5) p(7n + 5) 0 (mod 7) p(11n + 6) 0 (mod 11)
Other congruences for p(n)? Modular forms, combinatorially and otherwise p. 23/103
Modular forms, combinatorially and otherwise p. 24/103 Other congruences for p(n)? (Atkin-O Brien) For every n 0, p(157525693n + 111247) 0 (mod 13)
Modular forms, combinatorially and otherwise p. 24/103 Other congruences for p(n)? (Atkin-O Brien) For every n 0, p(157525693n + 111247) 0 (mod 13) (p(111247) is a number with well over 300 digits)
Modular forms, combinatorially and otherwise p. 25/103 (K. Ono) For every prime m 5, there exist positive integers A and B such that for every n 0, p(an + B) 0 (mod m).
Generating functions Modular forms, combinatorially and otherwise p. 26/103
Modular forms, combinatorially and otherwise p. 27/103 Generating functions p(0) p(1) p(2) p(3) p(4)
Modular forms, combinatorially and otherwise p. 27/103 Generating functions p(0) p(1) p(2) p(3) p(4) p(0) + p(1)q + p(2)q 2 + p(3)q 3 + p(4)q 4 +
Modular forms, combinatorially and otherwise p. 27/103 Generating functions p(0) p(1) p(2) p(3) p(4) p(0) + p(1)q + p(2)q 2 + p(3)q 3 + p(4)q 4 + = 1 + q + 2q 2 + 3q 3 + 5q 4 + 7q 5 +...
Modular forms, combinatorially and otherwise p. 27/103 Generating functions p(0) p(1) p(2) p(3) p(4) p(0) + p(1)q + p(2)q 2 + p(3)q 3 + p(4)q 4 + = 1 + q + 2q 2 + 3q 3 + 5q 4 + 7q 5 +... = n=0 p(n)q n
5q 4 Modular forms, combinatorially and otherwise p. 28/103
Modular forms, combinatorially and otherwise p. 28/103 5q 4 = q 4 + q 4 + q 4 + q 4 + q 4
Modular forms, combinatorially and otherwise p. 28/103 5q 4 = q 4 + q 4 + q 4 + q 4 + q 4 = q 4 + q 3+1 + q 2+2 + q 2+1+1 + q 1+1+1+1
Modular forms, combinatorially and otherwise p. 28/103 5q 4 = q 4 + q 4 + q 4 + q 4 + q 4 = q 4 + q 3+1 + q 2+2 + q 2+1+1 + q 1+1+1+1 = q 4 + q 3 q 1 + q 2+2 + q 2 q 1+1 + q 1+1+1+1
Modular forms, combinatorially and otherwise p. 29/103 (1 + q 1 + q 1+1 + q 1+1+1 + ) (1 + q 2 + q 2+2 + ) (1 + q 3 + q 3+3 + )
Modular forms, combinatorially and otherwise p. 29/103 (1 + q 1 + q 1+1 + q 1+1+1 + ) (1 + q 2 + q 2+2 + ) (1 + q 3 + q 3+3 + ) = n=0 p(n)q n
Modular forms, combinatorially and otherwise p. 30/103 p(n)q n = n=0 (1 + q 1 + q 1+1 + q 1+1+1 + ) (1 + q 2 + q 2+2 + ) (1 + q 3 + q 3+3 + )
Modular forms, combinatorially and otherwise p. 30/103 p(n)q n = n=0 (1 + q 1 + q 1+1 + q 1+1+1 + ) (1 + q 2 + q 2+2 + ) (1 + q 3 + q 3+3 + ) = ( 1 ) ( 1 ) ( 1 ) ( 1 ) 1 q 1 q 2 1 q 3 1 q 4
Modular forms, combinatorially and otherwise p. 31/103 Generating function for p(n) n=0 p(n)q n = n=1 ( 1 ) 1 q n
Modular forms, combinatorially and otherwise p. 32/103 Dedekind s eta function η(z) = q 1/24 (1 q n ) n=1 (q := e 2πiz )
η(z + 1) = e πi/12 η(z) Modular forms, combinatorially and otherwise p. 33/103
Modular forms, combinatorially and otherwise p. 33/103 η(z + 1) = e πi/12 η(z) η( 1/z) = z i η(z)
Modular forms, combinatorially and otherwise p. 33/103 η(z + 1) = e πi/12 η(z) η( 1/z) = z i η(z) η 24 (z + 1) = η 24 (z)
Modular forms, combinatorially and otherwise p. 33/103 η(z + 1) = e πi/12 η(z) η( 1/z) = z i η(z) η 24 (z + 1) = η 24 (z) η 24 ( 1/z) = z 12 η 24 (z)
Modular forms, combinatorially and otherwise p. 34/103 z + 1 = 1z + 1 0z + 1 ( 1 1 0 1 )
Modular forms, combinatorially and otherwise p. 34/103 z + 1 = 1z + 1 0z + 1 1 z = 0z 1 1z + 0 ( 1 1 0 1 ( 0 1 1 0 ) )
Modular forms, combinatorially and otherwise p. 34/103 z + 1 = 1z + 1 0z + 1 1 z = 0z 1 1z + 0 ( 1 1 0 1 ( 0 1 1 0 ) ) ( 1 1 0 1 ) and ( 0 1 1 0 ) generate SL 2 (Z)
Modular forms, combinatorially and otherwise p. 35/103 Integer weight modular forms A modular form of weight k on Γ = SL 2 (Z) is a holomorphic function f : H C such that for every ( ) a b Γ, c d ( ) az + b f = (cz + d) k f(z) cz + d and f is holomorphic at the cusp of Γ.
Modular forms, combinatorially and otherwise p. 35/103 Integer weight modular forms A modular form of weight k on Γ = SL 2 (Z) is a holomorphic function f : H C such that for every ( ) a b Γ, c d ( ) az + b f = (cz + d) k f(z) cz + d and f is holomorphic at the cusp of Γ. We denote the C-vector space of such functions by M k (Γ), and the subset of forms that vanish at the cusp by S k (Γ).
Modular forms, combinatorially and otherwise p. 36/103 Examples η 24 (z) M 12 (Γ)
Modular forms, combinatorially and otherwise p. 36/103 Examples η 24 (z) M 12 (Γ) E 4 (z) = 1 + 240 n=0 σ 3 (n)q n M 4 (Γ)
Modular forms, combinatorially and otherwise p. 36/103 Examples η 24 (z) M 12 (Γ) E 4 (z) = 1 + 240 E 6 (z) = 1 + 504 n=0 n=0 σ 3 (n)q n M 4 (Γ) σ 5 (n)q n M 6 (Γ)
Modular forms, combinatorially and otherwise p. 36/103 Examples η 24 (z) M 12 (Γ) E 4 (z) = 1 + 240 E 6 (z) = 1 + 504 n=0 n=0 σ 3 (n)q n M 4 (Γ) σ 5 (n)q n M 6 (Γ) σ k (n) = d n d k
- M k (Γ) is a finite dimensional vector space Modular forms, combinatorially and otherwise p. 37/103
Modular forms, combinatorially and otherwise p. 37/103 - M k (Γ) is a finite dimensional vector space - M k (Γ) is spanned by {E i 4E j 6 4i + 6j = k}
Modular forms, combinatorially and otherwise p. 38/103 Modular forms with character If f ( ) az + b cz + d = χ(d)(cz + d) k f(z) for all ( a b c d ) Γ with N c, we write f M k (Γ 0 (N),χ).
Modular forms, combinatorially and otherwise p. 39/103 Example θ(z) 4 M 2 (Γ 0 (4))
Modular forms, combinatorially and otherwise p. 39/103 Example θ(z) 4 M 2 (Γ 0 (4)) dim(m 2 (Γ 0 (4))) = 2
Modular forms, combinatorially and otherwise p. 39/103 Example θ(z) 4 M 2 (Γ 0 (4)) dim(m 2 (Γ 0 (4))) = 2 G 2 (z) = 1 24 + n=1 σ 1 (n)q n
Modular forms, combinatorially and otherwise p. 39/103 Example θ(z) 4 M 2 (Γ 0 (4)) dim(m 2 (Γ 0 (4))) = 2 G 2 (z) = 1 24 + n=1 σ 1 (n)q n basis: {G 2 (z) 2G 2 (2z),G 2 (2z) 2G 2 (4z)}
θ(z) 4 = 8[G 2 (z) 2G 2 (2z)] + 16[G 2 (2z) 2G 2 (4z)] Modular forms, combinatorially and otherwise p. 40/103
Modular forms, combinatorially and otherwise p. 40/103 θ(z) 4 = 8[G 2 (z) 2G 2 (2z)] + 16[G 2 (2z) 2G 2 (4z)] = 8G 2 (z) 32G 2 (4z)
Modular forms, combinatorially and otherwise p. 40/103 θ(z) 4 = 8[G 2 (z) 2G 2 (2z)] + 16[G 2 (2z) 2G 2 (4z)] = 8G 2 (z) 32G 2 (4z) r 4 (n) = 8 d n,4 d d
Modular forms, combinatorially and otherwise p. 40/103 θ(z) 4 = 8[G 2 (z) 2G 2 (2z)] + 16[G 2 (2z) 2G 2 (4z)] = 8G 2 (z) 32G 2 (4z) r 4 (n) = 8 d n,4 d d > 0
Modular forms, combinatorially and otherwise p. 41/103 ( 1 1 0 1 ) Γ 0 (N), χ(1) = 1
Modular forms, combinatorially and otherwise p. 41/103 ( 1 1 0 1 ) Γ 0 (N), χ(1) = 1 f(z + 1) = f(z)
Modular forms, combinatorially and otherwise p. 41/103 ( 1 1 0 1 ) Γ 0 (N), χ(1) = 1 f(z + 1) = f(z) f(z) = n=0 a(n)q n
Restricted partition functions Modular forms, combinatorially and otherwise p. 42/103
Modular forms, combinatorially and otherwise p. 43/103 Restricted partition functions Suppose I am only interested in partitions into distinct parts:
Modular forms, combinatorially and otherwise p. 43/103 Restricted partition functions Suppose I am only interested in partitions into distinct parts: (1 + q 1 +q 1+1 + q 1+1+1 + ) (1 + q 2 +q 2+2 + ) (1 + q 3 +q 3+3 + )
Modular forms, combinatorially and otherwise p. 44/103 The generating function in this case is the infinite product (1 + q 1 )(1 + q 2 )(1 + q 3 )(1 + q 4 )
Modular forms, combinatorially and otherwise p. 44/103 The generating function in this case is the infinite product (1 + q 1 )(1 + q 2 )(1 + q 3 )(1 + q 4 ) = (1 + q n ) n=1
Suppose I am only interested in partitions where no summand exceeds 3: Modular forms, combinatorially and otherwise p. 45/103
Modular forms, combinatorially and otherwise p. 45/103 Suppose I am only interested in partitions where no summand exceeds 3: ( 1 ) ( 1 )( 1 ) ( 1 ) 1 q 1 q 2 1 q 3 1 q 4
Modular forms, combinatorially and otherwise p. 46/103 The generating function in this case is the finite product ( 1 1 q ) ( 1 1 q 2 )( 1 1 q 3 )
Modular forms, combinatorially and otherwise p. 47/103 l-regular partitions A partition is called l-regular provided that none of its summands is divisible by l
Modular forms, combinatorially and otherwise p. 47/103 l-regular partitions A partition is called l-regular provided that none of its summands is divisible by l b l (n) := number of l-regular partitions of n
Modular forms, combinatorially and otherwise p. 48/103 5 4 + 1 3 + 2 3 + 1 + 1 2 + 2 + 1 2 + 1 + 1 + 1 1 + 1 + 1 + 1 + 1
Modular forms, combinatorially and otherwise p. 49/103 5 4 + 1 3 + 2 3 + 1 + 1 2 + 2 + 1 2 + 1 + 1 + 1 1 + 1 + 1 + 1 + 1
Modular forms, combinatorially and otherwise p. 49/103 5 4 + 1 3 + 2 3 + 1 + 1 2 + 2 + 1 2 + 1 + 1 + 1 1 + 1 + 1 + 1 + 1 b 2 (5) = 3
Generating function for b 2 (n)? Modular forms, combinatorially and otherwise p. 50/103
Modular forms, combinatorially and otherwise p. 50/103 Generating function for b 2 (n)? ( 1 ) ( 1 )( 1 ) ( 1 ) 1 q 1 q 2 1 q 3 1 q 4
Modular forms, combinatorially and otherwise p. 51/103 b 2 (n)q n = n=0 ( 1 ) ( 1 1 q 1 q 3 ) ( 1 ) 1 q 5
Modular forms, combinatorially and otherwise p. 51/103 b 2 (n)q n = n=0 ( ) ( ) ( ) 1 1 1 1 q 1 q 3 1 q 5 ( ) 1 q 2n = 1 q n n=1
(Euler) The number of 2-regular partitions of n is equal to the number of partitions of n into distinct parts Modular forms, combinatorially and otherwise p. 52/103
Modular forms, combinatorially and otherwise p. 52/103 (Euler) The number of 2-regular partitions of n is equal to the number of partitions of n into distinct parts n=0 b 2 (n)q n = n=1 ( ) 1 q 2n 1 q n
Modular forms, combinatorially and otherwise p. 52/103 (Euler) The number of 2-regular partitions of n is equal to the number of partitions of n into distinct parts n=0 b 2 (n)q n = n=1 ( ) 1 q 2n 1 q n = (1 + q n ) n=1
Modular forms, combinatorially and otherwise p. 53/103 b l (n)q n = n=0 n=1 ( ) 1 q ln 1 q n
Given positive integers l and m, exactly when is b l (n) divisible by m? How often is b l (n) divisible by m? Modular forms, combinatorially and otherwise p. 54/103
Modular forms, combinatorially and otherwise p. 54/103 Given positive integers l and m, exactly when is b l (n) divisible by m? How often is b l (n) divisible by m? δ l (m,x) := #{1 n X b l(n) 0 X (mod m)}
Modular forms, combinatorially and otherwise p. 55/103 δ l (m, 10 6 ) m = 5 7 11 l = 5.645.161.091 7.408.741.091 11.642.145.635
Modular forms, combinatorially and otherwise p. 55/103 δ l (m, 10 6 ) m = 5 7 11 l = 5.645.161.091 7.408.741.091 11.642.145.635 m = 5 7 11 l = 5.200.143.091 7.200.143.091 11.200.143.091
(P.) Suppose 3 l 23 and p 5 are distinct primes. Then for every j 1 we have: Modular forms, combinatorially and otherwise p. 56/103
Modular forms, combinatorially and otherwise p. 56/103 (P.) Suppose 3 l 23 and p 5 are distinct primes. Then for every j 1 we have: If p l 1, then lim inf X δ l(p j,x) p + 1 2p.
Modular forms, combinatorially and otherwise p. 56/103 (P.) Suppose 3 l 23 and p 5 are distinct primes. Then for every j 1 we have: If p l 1, then If p l 1, then lim inf X δ l(p j,x) p + 1 2p. lim inf X δ l(p j,x) p 1 p.
Modular forms, combinatorially and otherwise p. 57/103 (Serre) Suppose m and k are positive integers and f(z) = a(n)q n S k (Γ 0 (N),χ) Z[[q]]. Then for almost all n, a(n) 0 (mod m).
Modular forms, combinatorially and otherwise p. 58/103 η(lz) η(z) = ql/24 n=1 (1 qln ) q 1/24 n=1 (1 qn )
Modular forms, combinatorially and otherwise p. 58/103 η(lz) η(z) = ql/24 n=1 (1 qln ) q 1/24 n=1 (1 qn ) l 1 n+ = b l (n)q 24 n=0
Modular forms, combinatorially and otherwise p. 58/103 η(lz) η(z) = ql/24 n=1 (1 qln ) q 1/24 n=1 (1 qn ) l 1 n+ = b l (n)q 24 n=0 is NOT a modular form
Modular forms, combinatorially and otherwise p. 59/103 Given l and p distinct odd primes, there exists a positive integer α such that f l,p (z) = η(lz) η(z) η(24α 1)p (lpz)η (24α+1)p (pz) is a modular form.
Modular forms, combinatorially and otherwise p. 60/103 Twisting forms If f(z) = a(n)q n M k (Γ 0 (N),χ) and ψ is a character modulo M, then
Modular forms, combinatorially and otherwise p. 60/103 Twisting forms If f(z) = a(n)q n M k (Γ 0 (N),χ) and ψ is a character modulo M, then (f ψ)(z) := ψ(n)a(n)q n
Modular forms, combinatorially and otherwise p. 60/103 Twisting forms If f(z) = a(n)q n M k (Γ 0 (N),χ) and ψ is a character modulo M, then (f ψ)(z) := ψ(n)a(n)q n M k (Γ 0 (NM 2 ),χψ 2 ).
Modular forms, combinatorially and otherwise p. 61/103 Legendre symbol Given an odd prime p and d Z, ψ p (d) = 1 if d is a nonzero square mod p 1 if d is not a square mod p 0 if p d
d ψ 7 (d) 0 0 1 1 2 1 3 1 4 1 5 1 6 1 Modular forms, combinatorially and otherwise p. 62/103
Modular forms, combinatorially and otherwise p. 63/103 If p l 1, let F l,p,t (z) = (f l,p (z) ± (f l,p ψ p )(z)) E p,t (z)
Modular forms, combinatorially and otherwise p. 63/103 If p l 1, let F l,p,t (z) = (f l,p (z) ± (f l,p ψ p )(z)) E p,t (z) If p l 1, let F l,p,t (z) = (f l,p ψ p )(z) E p,t (z)
Modular forms, combinatorially and otherwise p. 63/103 If p l 1, let F l,p,t (z) = (f l,p (z) ± (f l,p ψ p )(z)) E p,t (z) If p l 1, let F l,p,t (z) = (f l,p ψ p )(z) E p,t (z) E p,t (z) = ( η p3 (z) η(p 3 z) ) 2p t 1 (mod p t )
F l,p,t (z) is a cusp form that for large t vanishes to high enough order at the cusps that on dividing out by the auxiliary eta product, one obtains a modular form Modular forms, combinatorially and otherwise p. 64/103
Durfee squares Modular forms, combinatorially and otherwise p. 65/103
Modular forms, combinatorially and otherwise p. 66/103 Durfee squares Represent the partition 5 + 4 + 4 + 2 + 1 by
Modular forms, combinatorially and otherwise p. 67/103 The Durfee square for this partition is
Modular forms, combinatorially and otherwise p. 68/103 The Durfee square for this partition is
Modular forms, combinatorially and otherwise p. 69/103 The Durfee square for this partition is
Modular forms, combinatorially and otherwise p. 69/103 The Durfee square for this partition is The partition breaks down into the 3 3 Durfee square and two partitions (2 + 1 and 3 + 1) with summands not exceeding 3
Modular forms, combinatorially and otherwise p. 70/103 The generating function for partitions with 3 3 Durfee square is therefore q 3 3 (( 1 ) ( 1 ) ( 1 )) 2 1 q 1 q 2 1 q 3
Modular forms, combinatorially and otherwise p. 70/103 The generating function for partitions with 3 3 Durfee square is therefore q 3 3 (( 1 ) ( 1 ) ( 1 )) 2 1 q 1 q 2 1 q 3 = q 9 (1 q) 2 (1 q 2 ) 2 (1 q 3 ) 2
Modular forms, combinatorially and otherwise p. 71/103 The generating function for partitions with n n Durfee square is q n2 (1 q) 2 (1 q 2 ) 2 (1 q n ) 2
Modular forms, combinatorially and otherwise p. 72/103 Since every partition has a Durfee square of some size, the generating function for p(n) can be written as 1 + n=1 P(q) := n=0 p(n)q n = q n2 (1 q) 2 (1 q 2 ) 2 (1 q n ) 2
Modular forms, combinatorially and otherwise p. 73/103 Modularity q = e 2πiz ρ(z) := q 1 P(q 24 )
Modular forms, combinatorially and otherwise p. 73/103 Modularity q = e 2πiz ρ(z) := q 1 P(q 24 ) ρ( 1/z) = i z ρ(z)
Modular forms, combinatorially and otherwise p. 74/103 Half integer wt modular forms - Theory developed by Shimura in the 1970s
Modular forms, combinatorially and otherwise p. 74/103 Half integer wt modular forms - Theory developed by Shimura in the 1970s - η(24z) S 1/2 (Γ 0 (576),χ 12 )
Modular forms, combinatorially and otherwise p. 74/103 Half integer wt modular forms - Theory developed by Shimura in the 1970s - η(24z) S 1/2 (Γ 0 (576),χ 12 ) - Shimura lift S t,k : S k+ 1 2 (Γ 0(4N),χ) M 2k (Γ 0 (2N),χ 2 ) (t a squarefree positive integer)
Mock theta functions Modular forms, combinatorially and otherwise p. 75/103
Modular forms, combinatorially and otherwise p. 75/103 Mock theta functions Two examples:
Modular forms, combinatorially and otherwise p. 75/103 Mock theta functions Two examples: f(q) := 1 + n=1 q n2 (1 + q) 2 (1 + q 2 ) 2 (1 + q n ) 2
Modular forms, combinatorially and otherwise p. 75/103 Mock theta functions Two examples: f(q) := 1 + n=1 q n2 (1 + q) 2 (1 + q 2 ) 2 (1 + q n ) 2 ω(q) := n=0 q 2n2 +2n (1 q) 2 (1 q 3 ) 2 (1 q 2n+1 ) 2
Φ(z) := q 1/24 f(q) Modular forms, combinatorially and otherwise p. 76/103
Modular forms, combinatorially and otherwise p. 76/103 Φ(z) := q 1/24 f(q) +2 3 i z g f (τ) i(τ + z) dτ
Modular forms, combinatorially and otherwise p. 76/103 Φ(z) := q 1/24 f(q) +2 3 i z g f (τ) i(τ + z) dτ g f (τ) := n= ( ( 1) n n + 1 ) 6 e 3πi (n+ 1 6) 2 τ
Ω(z) := 2q 1/3 ω(q 1/2 ) Modular forms, combinatorially and otherwise p. 77/103
Modular forms, combinatorially and otherwise p. 77/103 Ω(z) := 2q 1/3 ω(q 1/2 ) 2 3 i z g ω (τ) i(τ + z) dτ
Modular forms, combinatorially and otherwise p. 77/103 Ω(z) := 2q 1/3 ω(q 1/2 ) 2 3 i z g ω (τ) i(τ + z) dτ g ω (τ) := n= ( ( 1) n n + 1 ) 3 e 3πi (n+ 1 3) 2 τ
Modular forms, combinatorially and otherwise p. 78/103 Mock theta modularity (S. Zwegers) Φ( 1/z) = iz Ω(z)
Modular forms, combinatorially and otherwise p. 78/103 Mock theta modularity (S. Zwegers) Φ( 1/z) = iz Ω(z) Ω( 1/z) = iz Φ(z)
Modular forms, combinatorially and otherwise p. 78/103 Mock theta modularity (S. Zwegers) (K. Bringmann, K. Ono) Φ( 1/z) = iz Ω(z) Ω( 1/z) = iz Φ(z) Φ(24z) and Ω(6z) are harmonic weak Maass forms of weight 1/2
Modular forms, combinatorially and otherwise p. 79/103 Mock theta congruences ω(q) := n=0 α ω (n)q n = 1 + 2q + 3q 2 + 4q 3 + 6q 4 + 8q 5 + 10q 6 +
Modular forms, combinatorially and otherwise p. 79/103 Mock theta congruences ω(q) := n=0 α ω (n)q n = 1 + 2q + 3q 2 + 4q 3 + 6q 4 + 8q 5 + 10q 6 + (S. Garthwaite-P.) For every prime m 5, there exist positive integers A and B such that for every n 0, α ω (An + B) 0 (mod m).
Ω(6z) = a(n)q n + β n (y)q dn2 Modular forms, combinatorially and otherwise p. 80/103
Modular forms, combinatorially and otherwise p. 81/103 (B. Dandurand-P.) Given a nonnegative integer n, write 6n + 1 = r i=1 p e i i.
Modular forms, combinatorially and otherwise p. 81/103 (B. Dandurand-P.) Given a nonnegative integer n, write 6n + 1 = r i=1 p e i i. For each i with p i 1 (mod 3) write p i = x 2 i + 3y 2 i.
Modular forms, combinatorially and otherwise p. 81/103 (B. Dandurand-P.) Given a nonnegative integer n, write 6n + 1 = r i=1 p e i i. For each i with p i 1 (mod 3) write p i = x 2 i + 3y 2 i. Then b 5 (n) is divisible by 5 if and only if at least one of the following holds:
p i = 5 Modular forms, combinatorially and otherwise p. 82/103
Modular forms, combinatorially and otherwise p. 83/103 p i 2 (mod 3), p i 5 and e i is odd
Modular forms, combinatorially and otherwise p. 84/103 p i 1 (mod 3), 5 x i and e i is odd
p i 1 (mod 3), 5 y i and e i 4 (mod 5) Modular forms, combinatorially and otherwise p. 85/103
p i 1 (mod 3), 5 (x 2 i y2 i ) and e i 2 (mod 3) Modular forms, combinatorially and otherwise p. 86/103
p i 1 (mod 3), 5 x i y i (x 2 i y2 i ) and e i 5 (mod 6) Modular forms, combinatorially and otherwise p. 87/103
Modular forms, combinatorially and otherwise p. 88/103 p(n)x n = n=0 ( 1 ) ( 1 )( 1 ) ( 1 ) 1 x 1 x 2 1 x 3 1 x 4
Modular forms, combinatorially and otherwise p. 88/103 p(n)x n = n=0 ( 1 ) ( 1 )( 1 ) ( 1 ) 1 x 1 x 2 1 x 3 1 x 4 Consider the reciprocal (1 x)(1 x 2 )(1 x 3 )(1 x 4 )
Modular forms, combinatorially and otherwise p. 88/103 p(n)x n = n=0 ( 1 ) ( 1 )( 1 ) ( 1 ) 1 x 1 x 2 1 x 3 1 x 4 Consider the reciprocal (1 x)(1 x 2 )(1 x 3 )(1 x 4 ) = (1 x n ) n=1
Modular forms, combinatorially and otherwise p. 89/103 (1 x n ) n=1
Modular forms, combinatorially and otherwise p. 89/103 (1 x n ) n=1 = 1 x 1 x 2 + x 5 + x 7 x 12 x 15 + x 22 + x 26
Modular forms, combinatorially and otherwise p. 89/103 (1 x n ) n=1 = 1 x 1 x 2 + x 5 + x 7 x 12 x 15 + x 22 + x 26 (1, 2) (5, 7) (12, 15) (22, 26)
Modular forms, combinatorially and otherwise p. 89/103 (1 x n ) n=1 = 1 x 1 x 2 + x 5 + x 7 x 12 x 15 + x 22 + x 26 (1, 2) (5, 7) (12, 15) (22, 26) k 1 1 2 2 3 3 k(3k + 1)/2 1 2 5 7 12 15
Modular forms, combinatorially and otherwise p. 90/103 (1 x n ) 3 n=1
Modular forms, combinatorially and otherwise p. 90/103 (1 x n ) 3 n=1 = 1 3x 1 + 5x 3 7x 6 + 9x 10 11x 15 +
Modular forms, combinatorially and otherwise p. 90/103 (1 x n ) 3 n=1 = 1 3x 1 + 5x 3 7x 6 + 9x 10 11x 15 + (1, 3, 6, 10, 15)
Modular forms, combinatorially and otherwise p. 90/103 (1 x n ) 3 n=1 = 1 3x 1 + 5x 3 7x 6 + 9x 10 11x 15 + (1, 3, 6, 10, 15) l 1 2 3 4 5 2l + 1 3 5 7 9 11 l(l + 1)/2 1 3 6 10 15
Modular forms, combinatorially and otherwise p. 91/103 (1 x n ) 5 n=1
Modular forms, combinatorially and otherwise p. 91/103 (1 x n ) 5 n=1 (1 x n ) 5 = 1 5x n + 10x 2n 10x 3n + 5x 4n x 5n
Modular forms, combinatorially and otherwise p. 91/103 (1 x n ) 5 n=1 (1 x n ) 5 = 1 5x n + 10x 2n 10x 3n + 5x 4n x 5n 1 x 5n (mod 5)
Modular forms, combinatorially and otherwise p. 92/103 (1 x n ) 4 n=1
Modular forms, combinatorially and otherwise p. 93/103 (1 x n ) 4 = n=1 n=1 (1 x n ) 5 1 x n
Modular forms, combinatorially and otherwise p. 93/103 (1 x n ) 4 = n=1 n=1 n=1 (1 x 5n ) 1 x n (1 x n ) 5 1 x n
Modular forms, combinatorially and otherwise p. 93/103 (1 x n ) 4 = n=1 n=1 (1 x 5n ) n=1 n=1 (1 x 5n ) 1 x n n=1 (1 x n ) 5 1 x n ( 1 ) 1 x n
Modular forms, combinatorially and otherwise p. 93/103 (1 x n ) 4 = n=1 n=1 (1 x 5n ) n=1 (1 x 5n ) n=1 n=1 (1 x 5n ) 1 x n n=0 n=1 (1 x n ) 5 1 x n ( 1 ) 1 x n p(n)x n (mod 5)
Modular forms, combinatorially and otherwise p. 94/103 p(m)x m m=0 (1 x m ) 4 m=1 m=1 ( 1 ) 1 x 5m (mod 5)
Modular forms, combinatorially and otherwise p. 94/103 p(m)x m m=0 (1 x m ) 4 m=1 m=1 ( 1 ) 1 x 5m (mod 5) The first Ramanujan congruence is equivalent to showing that for any term of the form cx 5n+4 in the product (1 x m ) 4 m=1 we have c 0 (mod 5). m=1 ( 1 ) 1 x 5m,
Modular forms, combinatorially and otherwise p. 95/103 The first Ramanujan congruence is equivalent to showing that for any term of the form cx 5n+4 in the product (1 x m ) 4 m=1 we have c 0 (mod 5). m=1 ( 1 ) 1 x 5m
Modular forms, combinatorially and otherwise p. 96/103 The first Ramanujan congruence is implied by showing that for any term of the form cx 5n+4 in the product (1 x m ) 4 m=1 we have c 0 (mod 5).
Modular forms, combinatorially and otherwise p. 97/103 The first Ramanujan congruence is implied by showing that for any term of the form cx 5n+4 in the product (1 x m ) m=1 (1 x m ) 3 m=1 we have c 0 (mod 5).
Modular forms, combinatorially and otherwise p. 98/103 (1 x m ) m=1 k(3k + 1) 2
Modular forms, combinatorially and otherwise p. 98/103 (1 x m ) m=1 k(3k + 1) 2 (1 x m ) 3 l(l + 1) 2 m=1
k k(3k + 1)/2 l(l + 1)/2 l 0 0 0 0 1 2 1 1 2 7 3 2 3 15 6 3 4 26 10 4 Modular forms, combinatorially and otherwise p. 99/103
k k(3k + 1)/2 l(l + 1)/2 l 0 0 0 0 1 2 1 1 2 2 3 2 3 0 1 3 4 1 0 4 Modular forms, combinatorially and otherwise p. 100/103
Modular forms, combinatorially and otherwise p. 100/103 k k(3k + 1)/2 l(l + 1)/2 l 0 0 0 0 1 2 1 1 2 2 3 2 3 0 1 3 4 1 0 4 k 4 (mod 5) l 2 (mod 5)
k 4 (mod 5) l 2 (mod 5) Modular forms, combinatorially and otherwise p. 101/103
Modular forms, combinatorially and otherwise p. 101/103 k 4 (mod 5) l 2 (mod 5) l 1 2 3 4 5 2l + 1 3 5 7 9 11 l(l + 1)/2 1 3 6 10 15
Modular forms, combinatorially and otherwise p. 102/103 k 4 (mod 5) l 2 (mod 5) l 1 2 3 4 5 2l + 1 3 5 7 9 11 l(l + 1)/2 1 3 6 10 15
Modular forms, combinatorially and otherwise p. 102/103 k 4 (mod 5) l 2 (mod 5) l 1 2 3 4 5 2l + 1 3 5 7 9 11 l(l + 1)/2 1 3 6 10 15 ±x k(3k+1)/2 (2l + 1)x l(l+1)/2 0 (mod 5)
Modular forms, combinatorially and otherwise p. 103/103 Here is a mock theta function: f(q) := 1 + n=1 q n2 (1 + q) 2 (1 + q 2 ) 2 (1 + q n ) 2
Modular forms, combinatorially and otherwise p. 103/103 Here is a mock theta function: f(q) := 1 + n=1 q n2 (1 + q) 2 (1 + q 2 ) 2 (1 + q n ) 2 = 1 + q (1 + q) 2 + q 4 (1 + q) 2 (1 + q 2 ) 2 +
Modular forms, combinatorially and otherwise p. 103/103 Here is a mock theta function: f(q) := 1 + n=1 q n2 (1 + q) 2 (1 + q 2 ) 2 (1 + q n ) 2 = 1 + q (1 + q) 2 + q 4 (1 + q) 2 (1 + q 2 ) 2 + = 1 + q 2q 2 + 3q 3 3q 4 +