Let (a_i)_{1<=i<=n} be a sequence of positive integers and c be a positive integer such that c divides a_1. Suppose that we have this symmetric of continued fraction [a_n, a_n-1,...,a_1]=c^{-1}[a_1, a_2, ...., a_n]. Can we assume that n is even and a_{n-k}=c^(-1)^{k+1}a_{k+1}, for 0<=k<=n-1?
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I think that n is not even necessary – Oussema Oct 02 '19 at 16:04
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We should suppose that i>=2 – Oussema Oct 05 '19 at 10:07
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n shoud be even for example – Oussema Oct 06 '19 at 10:34
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[a_3, a_2, a_1]=1/c[a_1,a_2, a_3]=[c^-1a_1, ca_2, c^-1a_3], so ca_2=a_2 impossible – Oussema Oct 06 '19 at 10:38