Bill Gosper wrote:

George Andrews kindly informs me that any fixed power of 2 divides almost all Q(n) (usually written q(n), to confuse you with the nome).

I looked at Q(n) mod small powers of 2 a few years ago. I found that the value of Q(n) mod 64 is determined by representations of 24n+1 by the binary quadratic form x^2 + 24 y^2. Specifically: Let f(x,y) be determined by the values of x mod 192 and y mod 2: x mod 192 1 5 7 11 13 17 19 23 25 29 31 35 37 41 43 47 f(x,even) 1 51 55 29 5 15 27 25 41 11 31 53 45 39 3 49 f(x,odd) 7 5 1 11 3 9 29 15 31 13 25 19 27 17 21 23 x mod 192 49 53 55 59 61 65 67 71 73 77 79 83 85 89 91 95 f(x,even) 17 35 7 13 21 63 43 9 57 59 47 37 61 23 19 33 f(x,odd) 23 21 17 27 19 25 13 31 15 29 9 3 11 1 5 7 x mod 192 97 101 103 107 109 113 115 119 121 125 127 131 133 137 139 143 f(x,even) 33 19 23 61 37 47 59 57 9 43 63 21 13 7 35 17 f(x,odd) 7 5 1 11 3 9 29 15 31 13 25 19 27 17 21 23 x mod 192 145 149 151 155 157 161 163 167 169 173 175 179 181 185 187 191 f(x,even) 49 3 39 45 53 31 11 41 25 27 15 5 29 55 51 1 f(x,odd) 23 21 17 27 19 25 13 31 15 29 9 3 11 1 5 7 Let b(n) be the sum of f(x,y) over all solutions of 24n+1 = x^2 + 24 y^2, x>0. (0) Then Q(n) == b(n) (mod 64). (1) For example, for n=5 there are 3 solutions of (0): (x,y) = (11,0) or (5,2) or (5,-2). So Q(5) == f(11,0) + f(5,2) + f(5,-2) = 29 + 51 + 51 = 131 (mod 64); in fact Q(n) = 3. For most values of n, there are no solutions of (0), so Q(n) is divisible by 64. The values of f(x,odd) only range from 1 to 31, while those of f(x,even) range from 1 to 63. That's because, if (x,y) is a solution of (0) in which y is odd, then (x,-y) is another solution. Together these contribute 2 f(x,odd) to the value of b(n), so only the value of f(x,odd) mod 32 matters. This doubling also happens if y is even and nonzero. The only time that we need the value of f(x,y) mod 64 rather than just mod 32 is when y=0; i.e. when 24n+1 is a square; i.e. when n is pentagonal. If we're willing to ignore such values of n, then we can simplify the table above by reducing f(x,even) mod 32, which implies that we only need the value of x mod 96 rather than mod 192, since f(x+96,0) == f(x,0) (mod 32) for all x. There's no hope of modifying f(x,y) to give Q(n) mod 128: For n=20, there are no solutions of (0), so b(n) would be 0 no matter how we define f(x,y). Yet Q(20) = 64 is not divisible by 128. (I think I showed that Q(n) mod 128 was determined by representations of 24n+1 by (0) and by the quadratic form x^2 + 48 y^2, but I don't see the details in my notes.) I don't know if the value of Q(n) mod 2^k can be expressed in terms of binary quadratic forms for larger values of k, but it's not hard to show that it can be expressed in terms of representations of 24n+1 by quadratic forms in at most k variables. Outline of proof: The generating function for Q(n) can be written as J[1]/J[1,2] (2) where n n n(3n+1)/2 J[1] = PRODUCT (1 - q ) = SUM (-1) q (3) n>=1 n and 2n-1 2 2n J[1,2] = PRODUCT (1 - q ) (1 - q ) = 1 + 2A (4) n>=1 where 2 n n A = SUM (-1) q . (5) n>=1 Thus J[1]/J[1,2] = J[1]/(1 + 2A) 2 3 4 = J[1] (1 + 2 A + 4 A + 8 A + 16 A + ...). (6) Working mod 2^k, we only need the terms on the right up to A^(k-1). And the coefficient of n in J[1] A^i is determined by representations of 24n+1 by the quadratic form 2 2 2 r + 24 (s + ... + s ). (7) 1 i Dean Hickerson dean@math.ucdavis.edu