這個(gè)數(shù)列相當(dāng)于a[1] = 7,a[n] = 7^a[n-1] (n > 1).
求a[n]末兩位數(shù),即求a[n]除以100的余數(shù).
首先證明對(duì)任意n,a[n] ≡ -1 (mod 4).
對(duì)n = 1,a[1] = 7 ≡ -1 (mod 4).
而對(duì)n > 1,由a[n-1]為奇數(shù),a[n] = 7^a[n-1] ≡ (-1)^a[n-1] = -1 (mod 4).
因此a[n] ≡ -1 (mod 4)對(duì)任意正整數(shù)n成立,即a[n]-3是4的倍數(shù).
又注意到7^4 = 2401 ≡ 1 (mod 100),于是對(duì)任意正整數(shù)k,有7^(4k) ≡ 1 (mod 100).
因此對(duì)n > 1,a[n] = 7^a[n-1] = 7^(a[n-1]-3)·7³ ≡ 7³ = 343 ≡ 43 (mod 100).
即n > 1時(shí),a[n]的末兩位數(shù)恒為43.