HMMT 二月 2026 · 冲刺赛 · 第 8 题

HMMT February 2026 — Guts Round — Problem 8

[6] Let a , a , . . . be a sequence of positive integers such that a = 2 and for all n ≥ 2 , it holds that a is 1 2 1 n the sum of a and the largest prime divisor of a . Compute the smallest integer greater than 2026 n − 1 n − 1 that appears in this sequence. ©2026 HMMT . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . HMMT February 2026, February 14, 2026 — GUTS ROUND Organization Team Team ID#

解析

[6] Let a , a , . . . be a sequence of positive integers such that a = 2 and for all n ≥ 2 , it holds that a 1 2 1 n is the sum of a and the largest prime divisor of a . Compute the smallest integer greater than n − 1 n − 1 2026 that appears in this sequence. Proposed by: Matthew Qian Answer: 2068 Solution: The first few terms of the sequence are 2 , 4 , 6 , 9 , 12 , 15 , 20 , 25 , 30 , 35 , 42 , 49 , ... Notice that 2 · 3 = 6 , 3 · 5 = 15 , and 5 · 7 = 35 all appear in the sequence. This suggests the following more general fact. Claim 1. Let p , p , . . . be the list of primes in increasing order. Then, for all i ≥ 1 , the value p p 1 2 i i +1 appears in the sequence a , a , . . . . 1 2 Proof. We proceed by induction on i . For the base case i = 1 , we know p p = 6 appears in the 1 2 sequence, as desired. Now assume a = p p for some k . The largest prime factor of a is p , so we have a = k i i +1 k i +1 k +1 ( p + 1) p . Continuing this reasoning, we get that i i +1 a = ( p + 1) p , a = ( p + 2) p , . . . , k +1 i i +1 k +2 i i +1 and this will continue until a term has a prime factor greater than p . But this occurs precisely when i +1 the term becomes p p , so the induction is complete. i +2 i +1 Using our claim, we know that the term 43 · 47 = 2021 appears in the sequence. Therefore, the smallest term greater than 2026 is the following term, which is 2021 + 47 = 2068 .