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On 2015/12/18, I discovered that the Uda equation has the following solution when V =0. where a_{k} is determined by some appropriate condition. I will prove it below.
I want to adjust the value of a_{k} such that all terms other than the term of 2k  j = 1 vanishes. Σ_{k=s}^{2s1} a_{k} b_{k, 2(ks)+1} = 0 (s≧2) Σ_{k=s}^{2s} a_{k} b_{k, 2(ks)} = 0 (s∈N) ∴ a_{2s1} = (1/b_{2s1, 2s1})Σ_{k=s}^{2s2} a_{k} b_{k, 2(ks)+1} (s≧2) a_{2s} = (1/b_{2s, 2s})Σ_{k=s}^{2s1} a_{k} b_{k, 2(ks)} (s∈N) a_{2} = (1/b_{22})a_{1}b_{10} = (1/6)a_{1} ∵ b_{22} = 12, b_{10} = 2, a_{3} = (1/b_{33})a_{2}b_{21} = (1/10)a_{2} ∵ b_{33} = 120, b_{21} = 12 = (1/60)a_{1}, and so on. After the adjustment, the following condition holds. When a_{1} = mα/(i . This is the same as the reduced form of the Uda equation when V = 0. a_{2} = (1/6)a_{1} = mα/(6i a_{3} = (1/60)a_{1} = mα/(60i and so on.  This article is about the problem refered to in 19th @ December 2015 @ News. 

Author Yuichi Uda, Write start at 2016/03/26/15:32JST, Last edit at 2016/04/01/17:19JST  






