4 ms·
The researcher's answer to their variant of "Year: 2016 ID: A1" in the appendix is wrong. The solution (sum of 1,2,5,6,9,10,13,14, ...) has an alternating patt
by lupire 2y ago
The researcher's answer to their variant of "Year: 2016 ID: A1" in the appendix is wrong.
The solution (sum of 1,2,5,6,9,10,13,14, ...) has an alternating pattern, so has to be two piecewise interleaved polynomials, which cannot be expressed as a single polyomial.
Their answer works for k=1,2, but not k=3.
https://openreview.net/pdf?id=YXnwlZe0yf https://openreview.net/pdf?id=YXnwlZe0yf
This does not give me confidence in the results of their paper.
- Chinjut 2y agoYou are correct. Their answer is instead the sum of the first k terms of 1, 2, 6, 10, 14, 18, ..., for positive k.
- zeroonetwothree 2y agoVery astute. Did you communicate this to the authors?
- pfedak 2y agoYou're misreading the solution, the first part reads n=1, a trivial special case, not n congruent to 1 mod 4. The statement doesn't hold for e.g. n=5. Taking m=2 gives the permutation (1 2 4 3), which is odd, and thus cannot have a square root.