bump this

Do some ugly calculations just to get that; $N_m=N_{100}\prod_{A=99}^{m-2}[1+\sum_{\substack{a\neq b\\1\le a,b\le A}}^{}N_aN_b]$ This yields that for all $m>100$ $N_m$ is a multiple of $N_{100}$ since the product is an odd number we can say that $2^{2018}|N_{100}$ and$N_{100} |N_{161}$ and $N_{161}|N_{1440}$

wizixez wrote: Do some ugly calculations just to get that; $N_m=N_{100}\prod_{A=99}^{m-2}[1+\sum_{\substack{a\neq b\\1\le a,b\le A}}^{}]$ This yields that for all $m>100$ $N_m$ is a multiple of $N_{100}$ since the product is an odd number we can say that $2^{2018}|N_{100}$ and$N_{100} |N_{161}$ and $N_{161}|N_{1440}$ Reason behind the product being an odd number is that the number of pairs of odd numbers[ which equals to $43.42.41$ is even

Folks hello again, $N_{102}=\sum_{\substack{i\neq j\neq v\\1\le i,j,v\le 101}}^{}N_iN_jN_v=\sum_{\substack{i\neq j\neq v\\ 1\le i,j,v\le 100}}^{}N_iN_jN_v+\sum_{\substack{j\neq v\\ 1\le j,v\le 100}}^{}N_{101}N_jN_v=N_{101}[1+\sum_{\substack{i\neq j\\ 1\le i,j,v\le 100}}^{}N_iN_j]$ I called it "ugly" but it's actually beautiful.From here on,it is your job to prove the bizarre divisibility.

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