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Question Number 90801 by john santu last updated on 26/Apr/20

$$whatarethenexttwonumber$$$$ofthisseries4,7,9,2,5,6,3,8?$$

Commented by MJS last updated on 26/Apr/20

$$a_n=-\frac{17}{630}{n}^7+\frac{643}{720}{n}^6-\frac{8683}{720}{n}^5+\frac{12287}{144}{n}^4-\frac{242917}{720}{n}^3+\frac{263981}{360}{n}^2-\frac{27791}{35}n+328$$$$a_9=-122$$$$a_{10}=-1163$$

Commented by john santu last updated on 26/Apr/20

$$waw....$$

Commented by MJS last updated on 27/Apr/20

$$\mathrm{you}\mathrm{know}\mathrm{this}\mathrm{is}\mathrm{only}\mathrm{one}\mathrm{possibility}...\mathrm{the}$$$$\mathrm{next}\mathrm{two}\mathrm{numbers}\mathrm{can}\mathrm{be}\mathrm{any}{z}_1,z_2\in \mathbb{C}$$

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