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

w h a t a r e t h e n e x t t w o n u m b e r

o f t h i s s e r i e s 4, 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

w a w \dots .

Commented by MJS last updated on 27/Apr/20

you know this is only one possibility \dots the

next two numbers can be any z_{1}, z_{2} \in C

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