>>279
　b[n] = a[2n+1] −a[n+1]^2 −(a[n+2]-a[n+1])^2 −(a[n+1]-a[n])^2,
とおくと
　b[1]=0, b[2]=0, b[3]=0,
また漸化式より
　b[n+3] - 5b[n+2] + 6b[n+1] - a[n] = a[2n+7] - 5a[2n+5] + 6a[2n+3] - a[2n+1]
　=0,　　　>>280
ゆえ
　b[n] =0.