Timeline for More questions involving characteristic 2 theta series identities
Current License: CC BY-SA 3.0
7 events
| when toggle format | what | by | license | comment | |
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| Sep 29, 2011 at 1:54 | history | edited | paul Monsky | CC BY-SA 3.0 |
Conjectured equations made specific. Quadratic forms introduced.
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| Sep 23, 2011 at 3:02 | answer | added | paul Monsky | timeline score: 2 | |
| Sep 9, 2011 at 21:42 | history | edited | paul Monsky | CC BY-SA 3.0 |
added 1633 characters in body; edited tags
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| Sep 6, 2011 at 23:54 | comment | added | paul Monsky | @ARupinski---Quadratic forms arguments(Jacobi's 4 square theorem is the key!) will give (2) above. When l=31, similar arguments show that C(3,3,2,3)+C(2,3,7) is equal to C(1,1,2,5)+C(1,5,6). But I can't use these arguments to prove (3); the trouble is that 31=15 mod 16. | |
| Sep 5, 2011 at 12:18 | comment | added | paul Monsky | In that case, it also could be twice the l-value; multiplication by 12 takes(2,5,8) to(-7,-2,3). So C(2,5,8)=C(2,3,7) when l=31. But I have no idea what happens in general. When l is 7 mod 16, I think I see how to write C as a sum of various C(r1,r2,r3) and various C(s1,s2,s3,s4). But my argument fails when l is 15 mod 16. | |
| Sep 5, 2011 at 5:06 | comment | added | ARupinski | Extending your comment that $3^2+3^2+2^2+5^2 = 47$ etc., is there any reason you can see as to why the sum of the squares of the arguments of the second function is always twice the I-value except in case (3) where it is triple the I-value? | |
| Sep 4, 2011 at 2:53 | history | asked | paul Monsky | CC BY-SA 3.0 |