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India — Team Selection Test, 2013

Combinatorics Difficulty 6.3 National Olympiad Prove it India

For a positive integer nn, a sum-friendly odd partition of nn is a sequence (a1,a2,…,ak)(a_1, a_2, \dots, a_k) of odd positive integers with a1≤a2≤⋯≤aka_1 \le a_2 \le \dots \le a_k and a1+a2+⋯+ak=na_1 + a_2 + \dots + a_k = n such that for all positive integers m≤nm \le n, mm can be uniquely written as a subsum m=ai1+ai2+⋯+airm = a_{i_1} + a_{i_2} + \dots + a_{i_r}. (Two subsums ai1+ai2+⋯+aira_{i_1} + a_{i_2} + \dots + a_{i_r}, and aj1+aj2+⋯+ajsa_{j_1} + a_{j_2} + \dots + a_{j_s} with i1<i2<⋯<iri_1 < i_2 < \dots < i_r and j1<j2<⋯<jsj_1 < j_2 < \dots < j_s are considered the same if r=sr = s and ail=ajla_{i_l} = a_{j_l} for 1≤l≤r1 \le l \le r.) For example, (1,1,3,3)(1, 1, 3, 3) is a sum-friendly odd partition of 88. Find the number of sum-friendly odd partitions of 99999999.

Solution

We consider the sum-friendly odd partitions of a positive integer nn. Clearly (1,1,…,1)(1, 1, \dots, 1) is a sum-friendly odd partition. On the other hand, if ai>1a_i > 1 for some ii, then let rr be the smallest such that ar>1a_r > 1. It follows that ar=ra_r = r and that rr divides aia_i for all i≥ri \ge r. Therefore rr is odd and it divides n+1n+1. Moreover, (arr,ar+1r,…,akr)(\frac{a_r}{r}, \frac{a_{r+1}}{r}, \dots, \frac{a_k}{r}) is a sum-friendly odd partition of (n+1)/r−1(n+1)/r - 1. Thus by induction it follows that the number of sum-friendly odd partitions of nn equals the number of factorisations (n+1)=d1d2…dl(n+1) = d_1d_2 \dots d_l in which d1,d2,…,dl−1d_1, d_2, \dots, d_{l-1} are odd. Hence there are 1616 sum-friendly odd partitions of 99999999. □\square

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