Divisibility of the 5- and 13-regular partition functions 28 March 2008
Collaborators Joint Work This work was begun as an REU project and is joint with Neil Calkin, Nathan Drake, Philip Lee, Shirley Law, David Penniston and Jeanne Radder.
Thanks to the National Science Foundation. Acknowledgement We have been supported by NSF REU grants DMS 0139569, DMS 0244001, DMS 0552799 since summer 2002. Our current funding is through summer 2010.
Thanks to the National Science Foundation. Acknowledgement We have been supported by NSF REU grants DMS 0139569, DMS 0244001, DMS 0552799 since summer 2002. Our current funding is through summer 2010. Required Disclaimer Any opinions, findings, and conclusions or recommendations expressed in this material are those of the author(s) and do not necessarily reflect the views of the National Science Foundation.
Example of k regular partitions Central Question Generating Function Definition A k-regular partition of a number n is a partition of n in which none of the summands are divisible by k. We denote by b k (n) the number of of n.
Example Clemson Math REU Program Example of k regular partitions Central Question Generating Function Recall the partitions of 4: 4 3 + 1 2 + 2 2 + 1 + 1 1 + 1 + 1 + 1 So, b 3 (4) = 4 because the partition 3 + 1 has a summand divisible by 3.
Number theoretic question Example of k regular partitions Central Question Generating Function Question How often does b k (n) lie in certain arithmetic progressions?
Example of k regular partitions Central Question Generating Function Generating function for b k (n). A useful tool for studying such partition functions is their generating function. Fact n=0 b k (n)q n = k n ( 1 ) 1 q n
Example of k regular partitions Central Question Generating Function Generating function for b k (n). A useful tool for studying such partition functions is their generating function. Fact b k (n)q n = ( ) 1 1 q n n=0 k n ( ) 1 q kn = 1 q n. n=1
Data Ramanujan type congruences The parity of b 5 (n) and b 13 (n). Computational Evidence Computation shows that b 5 (2n) and b 13 (2n) are very frequently divisible by 2.
Data Ramanujan type congruences The parity of b 5 (n) and b 13 (n). Computational Evidence Computation shows that b 5 (2n) and b 13 (2n) are very frequently divisible by 2. Question For what values of n is b 5 (2n) odd?
Data Clemson Math REU Program Data Ramanujan type congruences Values of n for which b 5 (2n) is odd. 2n n 4 2 8 4 20 10 28 14 48 24 60 30 88 44 104 52 140 70
Data Clemson Math REU Program Data Ramanujan type congruences Values of n for which b 5 (2n) is odd. 2n n 12n + 1 4 2 8 4 20 10 28 14 48 24 60 30 88 44 104 52 140 70
Data Clemson Math REU Program Data Ramanujan type congruences Values of n for which b 5 (2n) is odd. 2n n 12n + 1 4 2 25 8 4 20 10 28 14 48 24 60 30 88 44 104 52 140 70
Data Clemson Math REU Program Data Ramanujan type congruences Values of n for which b 5 (2n) is odd. 2n n 12n + 1 4 2 25 8 4 49 20 10 28 14 48 24 60 30 88 44 104 52 140 70
Data Clemson Math REU Program Data Ramanujan type congruences Values of n for which b 5 (2n) is odd. 2n n 12n + 1 4 2 25 8 4 49 20 10 121 28 14 48 24 60 30 88 44 104 52 140 70
Data Clemson Math REU Program Data Ramanujan type congruences Values of n for which b 5 (2n) is odd. 2n n 12n + 1 4 2 25 8 4 49 20 10 121 28 14 169 48 24 289 60 30 361 88 44 529 104 52 625 140 70 841
Data Ramanujan type congruences Theorem b 5 (2n) is odd if and only if n = l(3l + 1) for some l Z.
Data Ramanujan type congruences Theorem b 5 (2n) is odd if and only if n = l(3l + 1) for some l Z. b 13 (2n) is odd if and only if n = l(l + 1) or n = 13l(l + 1) + 3 for some l Z.
Data Ramanujan type congruences Generating Function Version Theorem b 5 (2n)q 2n n=0 q 2n(3n+1) (mod 2)
Data Ramanujan type congruences Generating Function Version Theorem b 5 (2n)q 2n n=0 q 2n(3n+1) (mod 2) (1 q m ) 4 (mod 2) m=1
Data Ramanujan type congruences Generating Function Version Theorem b 5 (2n)q 2n n=0 q 2n(3n+1) (mod 2) (1 q m ) 4 (mod 2) m=1 b 13 (2n)q 2n n=0 (1 + q 4n ) 3 + q 6 (1 + q 52n ) 3 (mod 2). n=1 n=1
Data Ramanujan type congruences Ramanujan type congruences for b 5 (n) Corollary For n 0, b 5 (20n + 5) 0 (mod 2) and b 5 (20n + 13) 0 (mod 2).
Data Ramanujan type congruences Ramanujan type congruences for b 5 (n) Corollary For n 0, b 5 (20n + 5) 0 (mod 2) and b 5 (20n + 13) 0 (mod 2). Theorem For n 0 and 2 l 6 ( b 13 3 l n + 5 ) 3l 1 1 0 (mod 3) 2
Data Ramanujan type congruences Infinitely many Ramanujan congruences? For n 0 and l 2 ( b 13 3 l n + 5 ) 3l 1 1 0 (mod 3) 2
Data Ramanujan type congruences Infinitely many Ramanujan congruences? For n 0 and l 2 ( b 13 3 l n + 5 ) 3l 1 1 0 (mod 3) 2 Theorem Fix l 2. If b 13 (3 l n + 5 3l 1 1 2 0 n 7 3 l 1 3, then b 13 (3 l n + 5 3l 1 1 2 all n. ) 0 (mod 3) for ) 0 (mod 3) for
Modular Forms Clemson Math REU Program Modular Forms Tools of Ramanujan type congruence Definition Let f : H C be holomorphic, let k Z, and let χ be a Dirichlet character modulo N. Then f is said to be a modular form of weight k, level N, and character χ if ( ) az + b f = χ(d)(cz + d) k f (z) cz + d ( ) a b for all Γ c d 0 (N). We denote the complex vector space of such functions by M k (N, χ).
Fourier Expansion Modular Forms Tools of Ramanujan type congruence Note ( ) 1 1 Since Γ 0 1 0 (N), we have that if f (z) M k (N, χ), then f (z + 1) = f (z). Thus f (z) has a Fourier expansion f (z) = a f (n)q n (q = e 2πiz ). n 0
Hecke Operators Modular Forms Tools of Ramanujan type congruence Definition We define linear operators T n on M k (N, χ) as follows. If f (z) = n=0 a(n)qn M k (N, χ), then T p (f )(z) = (a(pn) + χ(p)p k 1 a(n/p))q n M k (N, χ). n=0
Hecke Operators Modular Forms Tools of Ramanujan type congruence Definition We define linear operators T n on M k (N, χ) as follows. If f (z) = n=0 a(n)qn M k (N, χ), then T p (f )(z) = (a(pn) + χ(p)p k 1 a(n/p))q n M k (N, χ). n=0 Note For k > 1, T p (f )(z) a(pn)q n (mod p) n=0
Modular Forms Tools of Ramanujan type congruence Factorization Property of Hecke Operators Fact If f (z) = a(n)q n b(n)q pn M k (N, χ), n 0 n 0 then ( ) T p (f )(z) a(pn)q n b(n)q n (mod p) n 0 n=0
Sturm s Theorem Modular Forms Tools of Ramanujan type congruence Theorem Suppose that 1 f = n 0 a f (n)q n, g = n 0 a g (n)q n M k (N, χ). 2 af (n), a g (n) Z. 3 af (n) a g (n) (mod p) for 0 n k 12 N p N Then a f (n) a g (n) (mod p) n 0. That is, f g (mod p). ( ) 1 + 1 p.
for l = 2 Clemson Math REU Program Modular Forms Tools of Ramanujan type congruence Definition η(z) = q 1 24 (1 q n ). n=1
for l = 2 Clemson Math REU Program Modular Forms Tools of Ramanujan type congruence Definition η(z) = q 1 24 (1 q n ). n=1 Fact g(z) := η(13z) η(z) η 36 (13z) M 18 (13, χ 13 ).
for l = 2 Clemson Math REU Program Modular Forms Tools of Ramanujan type congruence Definition η(z) = q 1 24 (1 q n ). n=1 Fact g(z) := η(13z) η(z) η 36 (13z) M 18 (13, χ 13 ). Note g(z) = b 13 (n)q n+20 ( 1 q 13j ) 36 n=0 j=1
for l = 2 (continued) Modular Forms Tools of Ramanujan type congruence Fact g(z) b 13 (n)q n+20 (1 q 117j ) 4 (mod 3) T 3 (g)(z) T 3 (T 3 (g))(z) n=0 j=1 b 13 (3n + 1)q n+7 (1 q 39j ) 4 (mod 3) n=0 j=1 b 13 (9n + 7)q n+3 (1 q 13j ) 4 (mod 3) n=0 j=1
for l = 2 (continued) Modular Forms Tools of Ramanujan type congruence Since b 13 (9n + 7) 0 (mod 3) for 0 n 18, it follows that the first 21 coefficients of T 3 (T 3 (g))(z) are 0 modulo 3. Thus, by Sturm s theorem b 13 (9n + 7)q n+3 (1 q 13j ) 4 0 (mod 3). n=0 j=1 The theorem follows by noticing that j=1 (1 q13j ) 4 0 (mod 3)