Let $(a_1,a_2\ldots,a_n)$ and $(b_1,b_2,\ldots,b_n)$ be two permutations of arithmetic progressions of natural numbers. For which $n$ is it possible that $(a_1b_1,a_2b_2,\dots,a_nb_n)$ is an arithmetic progression?

The sequence is (trivially) an arithmetic progression when $n=1$ or $2$. We can notice that $(a_1,a_2,\ldots,a_n)$ and $(b_1,b_2,\ldots,b_n)$ cannot be in identical order, since that would mean they both have to be increasing or decreasing, but then the differences of the resulting product terms cannot be constant.


1 Answer 1


My computer found these for 4, 5 and 6: $$A: 1×10, 11×4, 6×13, 16×7\\ B: 8×4, 6×9, 4×19, 7×14, 5×24\\ C: 7×35, 31×11,19×23,13×41,37×17,25×29$$

  • $\begingroup$ $4×41,6×31,8×26,5×46,7×36$ $\endgroup$
    – Empy2
    Sep 12, 2018 at 10:25

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