Consider mod 5. Now the condition equivalent to a_{2n+1} = - (a_{2n-1} + a_{n}), a_{2n} = - (a_{2n-2} + 2a_{n-1}) (1) Claim: a_{4k+3} + a_{4k+5} = a_{4k+2} + a_{4k+4} Proof: By (1), it equivalent to a_{2k+2} = 2a_{2k+1}, which is obviously be the determination Back to the problem: We proceed by induction. Obviously it true for n=11, assume it true for 8k-5, consider 8k+3. By (1), we have a_{2n+1} = - a_{2n-1} - a_{n} = a_{2n-3} + a_{n-1} - a_{n} = - a_{2n-5} - a_{n-2} - a_{n} + a_{n-1} = a_{2n-7} - a_{n} + a_{n-1} - a_{n-2} + a_{n-3}. Plugin n=4k+1, k>0 by claim we have the conclusion: