“…(3) Since h(P 5 ) = x 2 (x + 3)h(P 2 ). Then by Lemma 2.4(1),(8) and(10), we have (x + 3)|h(P i ), where i = 5, 11, 17, 23, 29. And, by Lemma 2.2 and h(P n+1 ) = x[h(P n ) + h(P n−1 )], we deduce that h(P 8 ) = h(P 2 )h(P 6 ) + xh(K 1 )h(P 5 ), h(P 14 ) = h(P 2 )h(P 12 ) + xh(K 1 )h(P 11 ), h(P 20 ) = h(P 2 )h(P 18 ) + xh(K 1 )h(P 17 ), h(P 26 ) = h(P 2 )h(P 24 ) + xh(K 1 )h(P 23 ).…”