If it is a series of rigorous questions, then
a( 1)= 1； a(2)= 3；
a(3)=[-a( 1)-2×a(2)+ 17]/2 = 5；
a(4)=[-a(2)-2×a(3)+ 17]/2 = 2；
a(5)=[-a(3)-2×a(4)+ 17]/2 = 4 .
To sum up, the law is like this
a(n)=[-a(n-2)-2×a(n- 1)+ 17]/2 .
therefore
a(6)=[-a(4)-2×a(5)+ 17]/2 = 7/2 .
Finally found one that is not equal to 6.