Part II. The power of q. Michael D. Hirschhorn. A course of lectures presented at Wits, July 2014.

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2 n 1(1 q n ) 3 = ( 1) n (2n+1)q (n2 +n)/2. Note that the power on q, (n 2 +n)/2 0, 1 or 3 mod 5. And when (n 2 +n)/2 3 (mod 5), n 2 (mod 5), and then the coefficient, (2n+1) 0 (mod 5). So we can write (1 q n ) 3 J 0 +J 1 (mod 5), n 1 most where J 0 consists of those terms in which the power of q is congruent to 0 modulo 5, and J 1 those terms in which the power of q is congruent to 1 modulo 5. Also, (1 q) 5 1 q 5 (mod 5).

3 So we can write p(n)q n = 1 1 q n n 1 n 1 = (1 qn ) 9 n 1 (1 qn ) 10 ( ) 3 n 1 (1 qn ) 3 = ( n 1 (1 qn ) 5 ) 2 most (J 0 +J 1 ) 3 ( (1 q 5n )) 2 = J3 0 +3J2 0 J 1 +3J 0 J 2 1 +J3 1 ( n 1 (1 q5n )) 2. There are no terms on the right in which the power of q is 4 modulo 5, so p(5n+4)q n 0 (mod 5).

4 n 1(1 q n ) 3 = ( 1) n (2n+1)q (n2 +n)/2. The power, (n 2 +n)/2 0, 1, 3 or 6 (mod 7), and when (n 2 +n)/2 6 (mod 7), n 3 (mod 7), and then the coefficient (2n+1) 0 (mod 7). So we can write (1 q n ) 3 J 0 +J 1 +J 3, where J i consists of those terms in which the power of q is i (mod 7), i = 0, 1, 3. Also, (1 q) 7 1 q 7 (mod 7). most

5 So p(n)q n = 1 1 q n n 1 ( ) 2 n 1 (1 qn ) 3 = n 1 (1 qn ) 7 (J 0 +J 1 +J 3 ) 2 n 1 (1 q7n ) J2 0 +2J 0J 1 +J1 2 +2J 0J 3 +2J 1 J 3 +J 3 2. n 1 (1 q7n ) most There are no terms in which the power of q is 5 (mod 7), so p(7n+5)q n 0 (mod 7).

6 In the same way, we can write (1 q n ) 3 J 0 +J 1 +J 3 +J 6 +J 10 (mod 11), n 1 where J i consists of those terms in which the power of q is i (mod 11). p(n)q n = ( n 1 (1 qn ) 3 ) 7 ( n 1 (1 qn ) 11 ) 2 (J 0 +J 1 +J 3 +J 6 +J 10 ) 7 ( n 1 (1 q11n )) 2. most It follows that, modulo 11, p(11n +6)q n P 6 n 1 (1 qn ) 2 where P is a polynomial in the J i homogeneous of degree 7 in the J i, of 30 terms. Not obviously congruent to 0 mod 11.

7 P 6 = 7J0 6 J 6 +10J0 5 J2 3 +J4 0 J 1J 6 J J6 5 J2 10. Here is the trick I discovered last year. (J 0 +J 1 +J 3 +J 6 +J 10 ) 4 n 1(1 q n ) 3 = n 1(1 q n ) = n 1(1 q n ) (1 q n ) n 1 n 1(1 q 11n ) ( 1) n q (3n2 +n)/2 4 most

8 (J 0 +J 1 +J 3 +J 6 +J 10 ) 4 n 1(1 q 11n )(E 0 +E 1 +E 2 +E 4 +E 5 +E 7 ). where each E i consists of powers i (mod 11). Comparing terms in which the power of q is 3, 6, 8, 9 and 10 (mod 11) gives five quartics in the J i that are 0 (mod 11) For example,q 3 = J 3 0J 3 +J 0 J J 0 J 1 J 3 J 10 +7J 2 1J J 3 J 2 6J 10 +J 6 J most If we can express P 6 as a linear combination of these five quartics (using cubics as coefficients) then we will have succeeded in proving that p(11n +6)q n 0 (mod 11).

9 And, indeed we can! P 6 C 3 Q 3 +C 0 Q 6 +C 9 Q 8 +C 8 Q 9 +C 7 Q 10, where, for example, C 3 = 7J 9 J 3 J 10 +5J 2 0J 3 +7J 3 1, and the other C i are of similar form. See my paper in the Journal of Number Theory 139 (2014), , for details. most

10 Some notation and some simple results We can say a lot more about the preceding results, but it may help to introduce some more notation. (a;q) = k 0(1 aq k ), (a 1,a 2, a n ;q) = (a 1 ;q) (a 2 ;q) (a n ;q), ( ) a1,a 2,,a n ;q = (a 1,a 2,,a n ;q). b 1,b 2,,b m (b 1,b 2,,b m ;q) Euler s product has many notations, among them most E(q) = n 1(1 q n ) = (q;q). Bruce Berndt persuaded me to use (q k ;q k ) in preference to (q), so I often use that in my papers..

11 Ramanujan defined = φ(q) = q n2 = (1+q 2n 1 ) 2 (1 q 2n ) n 1 (q 2 ;q 2 ) 5 (q;q) 2 (q4 ;q 4 ) 2 ψ(q) = = φ( q) = E(q)2 E(q 2 ), E(q 2 ) 5 E(q) 2 E(q 4 ) 2, q (n2 +n)/2 = E(q2 ) 2 E(q), most ψ( q) = E(q)E(q4 ) E(q 2 ) 2

12 Jacobi s triple product identity in one form becomes ( a 1 q, aq,q 2 ;q 2 ) = in another, (a 1 q,aq 2,q 3 ;q 3 ) = and yet another, a n q n2, ( 1) k a k q (3k2 +k)/2, most (a 1 q,aq,q;q) = 1+ ( a k +a k 1 + +a k) q (k2 +k)/2. k 1

13 most We saw that, modulo 5, We can do much better. We had E(q) = E 0 +E 1 +E 2. (a 1 q,aq,q;q) = 1+ k 1 ( a k + +a k) q (k2 +k)/2. Let η = exp{ 2πi 5 }. We set a = η in the above, and use the fact that 1 if k 0 (mod 5), (η 2 +η 2 ) if k 1 (mod 5), η k + +η k = 0 if k 2 (mod 5), η 2 +η 2 if k 3 (mod 5), 1 if k 4 (mod 5). most

14 We now split the sum in five, according to the residue modulo 5 of k For k 0, write k = 5m, m 1, for k 1 write k = 5m +1, m 0, for k 2, write k = 5m+2, m 0, for k 3 write k = 5m 2, m 1 and for k 4, write k = 5m 1, m 1. We obtain = (η 1 q,ηq,q;q) q (25m2 +5m)/2 +(η 2 +η 2 )q q (25m2 +5m)/2 = (q 10,q 15,q 25 ;q 25 ) +(η 2 +η 2 )q(q 5,q 20,q 25 ;q 25 ), most where we have used JTP to sum the series.

15 We found that (ηq,η 1 q,q;q) = (q 10,q 15,q 25 ;q 25 ) +(η 2 +η 2 )q(q 5,q 20,q 25 ;q 25 ). Taking a = η 2 gives (η 2 q,η 2 q,q;q) = (q 10,q 15,q 25 ;q 25 ) +(η +η 1 )q(q 5,q 20,q 25 ;q 25 ). If we multiply these two equations, we find most (q;q) (q 5 ;q 5 ) = = (q 10,q 15,q 25 ;q 25 ) 2 q(q5 ;q 5 ) (q 25 ;q 25 ) q 2 (q 5,q 20,q 25 ;q 25 ) 2.

16 If we divide by (q 5 ;q 5 ), we find (q;q) = (q 25 ;q 25 ) (( q 10,q 15 ) ( q 5,q20 ;q25 q q 2 q 5,q 20 ) ) q 10,q15 q25. We now define and then we have ( ) q,q 4 R = q 2,q 3;q5, E(q) = E(q 25 ) ( R(q 5 ) 1 q q 2 R(q 5 ) ), most which is the refined form of E = E 0 +E 1 +E 2.

17 We have p(n)q n = 1 E = E(ηq)E(η2 q)e(η 3 q)e(η 4 q) E(q)E(ηq)E(η 2 q)e(η 3 q)e(η 4 q). We can write the denominator in two different ways. E(q)E(ηq)E(η 2 q)e(η 3 q)e(η 4 q) = E(q 25 ) 5( R(q 5 ) 1 q q 2 R(q 5 ) ) ( R 1 (q 5 ) ηq η 2 q 2 R(q 5 ) ) ( R(q 5 ) η 4 q η 3 q 2 R(q 5 ) ) most = E(q 25 ) 5( R(q 5 ) 5 11q 5 q 10 R(q 5 ) 5).

18 Alternatively, E(q)E(ηq)E(η 2 q)e(η 3 q)e(η 4 q) = n 1(1 q n )(1 η n q n )(1 η 2n q n )(1 η 3n q n )(1 η 4n q n ) = 5 n (1 q ) 5n (1 q 5n ) 5 n = (1 q 5n ) 6 / (1 q 25n ) = E(q5 ) 6 E(q 25 ). n 1 n 1 And the numerator is most E(ηq)E(η 2 q)e(η 3 q)e(η 4 q) = E(q 25 ) 4 ( R(q 5 ) 4 +qr(q 5 ) 3 +2q 2 R(q 5 ) 2 +3q 3 R(q 5 ) 1 +5q 4 3q 5 R(q 5 )+2q 6 R(q 5 ) 2 q 7 R(q 5 ) 3 +q 8 R(q 5 ) 4).

19 It follows that p(n)q n = E(q25 ) 5 E(q 5 ) 6 ( R(q 5 ) 4 +qr(q 5 ) 3 +2q 2 R(q 5 ) 2 +3q 3 R(q 5 ) 1 +5q 4 3q 5 R(q 5 )+2q 6 R(q 5 ) 2 q 7 R(q 5 ) 3 +q 8 R(q 5 ) 4). And therefore Fanfare! p(5n+4)q n = 5 E(q5 ) 5 E(q) 6. Hardy thought this perhaps the best, in some sense (beauty, simplicity, elegance) of work. [ The corrsponding formula for the prime 7 is p(7n+5)q n = 7 E(q7 ) 3 E(q) 4 +49qE(q7 ) 7 E(q) 8. most

20 for powers of 5 We saw that and that p(5n+4)q n = 5 E(q5 ) 5 E(q) 6 1 E(q) = E(q25 ) 5 E(q 5 ) 6 ( R(q 5 ) 4 +qr(q 5 ) 3 +2q 2 R(q 5 ) 2 +3q 3 R(q 5 ) 1 +5q 4 3q 5 R(q 5 )+2q 6 R(q 5 ) 2 q 7 R(q 5 ) 3 +q 8 R(q 5 ) 4). most If we substitute the second equation into the first, we find

21 p(5n+4)q n = 5 E(q25 ) 30 E(q 5 ) 31 ( R(q 5 ) 4 +qr(q 5 ) 3 +2q 2 R(q 5 ) 2 +3q 3 R(q 5 ) 1 +5q 4 3q 5 R(q 5 )+2q 6 R(q 5 ) 2 q 7 R(q 5 ) 3 +q 8 R(q 5 ) 4) 6 = 5 E(q25 ) 30 E(q 5 ) 31 ( R(q 5 ) 24 +6qR(q 5 ) q 48 R(q 5 ) 24). most

22 If we now extract those terms in which the power of q is congruent to 4 modulo 5, divide by q 4 and replace q 5 by q, we find p(25n+24)q n = 5 E(q5 ) 30 E(q) 31 ( 315R(q) qR(q) q 2 R(q) q 3 R(q) q q 5 R(q) q 6 R(q) q 7 R(q) q 8 R(q) 20). Now, we saw before that most E(q 25 ) 5( R(q 5 ) 5 11q 5 q 10 R(q 5 ) 5) = E(q5 ) 6 E(q 25 ), so R(q) 5 11q q 2 R(q) 5 = E(q)6 E(q 5 ) 6.

23 Using the facts that (1 11x x 2 ) 2 = 1 22x +119x 2 +22x 3 +x 4, (1 11x x 2 ) 3 = 1 33x +360x x 3 360x 4 33x 5 x 6 (1 11x x 2 ) 4 = 1 44x +722x x x 4 It is routine to show that p(25n+24)q n +5192x x 6 +44x 7 +x 8 = 25 E(q5 ) 30 E(q) 31 ( 63 ( R(q) 5 11q R(q) 5) q ( R(q) 5 11q q 2 R(q) 5) q 2( R(q) 5 11q q 2 R(q) 5) q 3( R(q) 5 11q q 2 R(q) 5) q 4). most

24 That is, p(25n+24)q n = 25 E(q5 ) 30 E(q) 31 ( ( ) E(q) 6 4 ) E(q) q( E(q 5 ) 6 E(q 5 ) 6 ( ) E(q) q ( ) E(q) E(q 5 ) q 3 6 )+5 10 E(q 5 ) 6 q 4 = 25 (63 E(q5 ) 6 ) 12 E(q) qE(q5 E(q) q 2E(q5 ) 18 E(q) q 3E(q5 ) 24 E(q) q 4E(q5 ) 30 ) E(q) 31. most

25 This process can be repeated indefinitely, and a general theorem obtained. p(5 α n+δ α )q n i 1 x α,iq i 1E(q5 ) 6i 1 E(q) 6i, = i 1 x α,iq i 1 E(q5 ) 6i E(q) 6i+1, α odd α even most where ν 5 (x α,i ) { α+ 1 2 (5i 5) α α+ 1 2 (5i 4) α odd, even

26 and δ α = { 1 24 (19 5α +1) α odd, 1 24 (23 5α +1) α even. Note that 1 δ α < 5 α and 24δ α 1 (mod 5 α ). Details may be found in M. D. and D. C. Hunt, A simple proof of the Ramanujan conjectures for powers of 5, J. Reine Angew. Math. 326(1981), You may also like to consult F. G. Garvan, A simple proof of Watson s for powers of 7, J. Austral. Math. Soc. (Ser. A) 36(1984), most

27 Returning to the basic, there is more we can say. Modulo 5, E(q) 3 = J 0 +J 1 +J 3, J 0 = 1+9q 10 11q 15 19q q q 10 q 15 +q 45 +q 55 = ( 1) n q (25n2 5n)/2 = (q 10,q 15,q 25 ;q 25 ), J 1 = 3q 7q 6 13q q 36 23q 66 27q q +3q 6 +3q 21 3q 36 3q 66 +3q 91 + = 3q ( 1) n q (25n2 15n)/2 most = 3q(q 5,q 20,q 25 ;q 25 ), J 3 = 5q 3 15q q 78 35q = 5q 3( 1 3q 25 +5q 75 ) = 5q 3 E(q 25 ) 3.

28 p(n)q n (J 0 +J 1 ) 3 E(q 5 ) 2 (q10,q 15,q 25 ;q 25 ) 3q(q 5,q 20,q 25 ;q 25 ) ) 3 E(q 5 ) 2 It follows that p(5n)q n (q2,q 2,q 2,q 3,q 3,q 3,q 5,q 5,q 5 ;q 5 ) E(q) 2, p(5n+1)q n (q,q2,q 2,q 3,q 3,q 4,q 5,q 5,q 5 ;q 5 ) E(q) 2, p(5n+2)q n 2 (q,q,q2,q 3,q 4,q 4,q 5,q 5,q 5 ;q 5 ) E(q) 2, p(5n+3)q n 3 (q,q,q,q4,q 4,q 4,q 5,q 5,q 5 ;q 5 ) E(q) 2. most

29 Similarly,modulo 7, p(7n)q n (q3,q 3,q 4,q 4,q 7,q 7 ;q 7 ), E(q) p(7n+1)q n (q2,q 3,q 4,q 5,q 7,q 7 ;q 7 ), E(q) p(7n+2)q n 2 (q2,q 2,q 5,q 5,q 7,q 7 ;q 7 ), E(q) p(7n+3)q n 3 (q,q3,q 4,q 6,q 7,q 7 ;q 7 ), E(q) p(7n+4)q n 5 (q,q2,q 5,q 6,q 7,q 7 ;q 7 ), E(q) p(7n+6)q n 11 (q,q,q6,q 6,q 7,q 7 ;q 7 ). E(q) most

30 There are similar simple relations modulo 11 for p(11n +r)qn. For example, p(11n)q n (q2,q 3,q 4,q 5,q 6,q 7,q 8,q 9,q 11,q 11 ;q 11 ). E(q) My proofs follow from a formula for E(q) 10, which itself follows from Winquist s identity. I refer you to M.D., A generalisation of Winquist s identity and a conjecture of Ramanujan, J. Indian Math. Soc., 51(1987), 49 55, and to M. D., Winquist and the for modulus 11, Australas. J. Combin., 22(2000), most