## n_1 X n_2 X n_3 Dots_Puzzle: A Method to Improve the Current Upper Bound

**Authors:** Valerio Bencini

The aim of this paper is to lower down the current upper bound for the Ripà's n_1 X n_2 X n_3 Dots Problem, with n_1>n_2>=n_3, using the same method Ripà and I used for the case n_1=n_2=n_3:=n. The new value is now 1/2*floor(1/(3*n_1-3*n_2-3*n_3+5))*((n_1-n_2-n_3)*(n_1-n_2-n_3+1)-2*floor(1/2*(-n_1+n_2+n_3)))+2*n_2*n_3-1.
At the end of the article, I also extend this result, and that I previously found with Ripà for n_1=n_2=n_3=:n, to the k-dimensional problem n_1 X n_2 X n_3 X n_4 X ... X n_k, using the equation found by Ripà in 2013.

**Comments:** 8 Pages.

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### Submission history

[v1] 2019-06-07 13:21:19

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