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2^f(n) divides sum of binomials

Source: Vietnam TST 1996 for the 37th IMO, problem 2

June 26, 2005
floor functionnumber theory unsolvednumber theory

Problem Statement

For each positive integer nn, let f(n)f(n) be the maximal natural number such that: 2f(n)2^{f(n)} divides ∑i=0⌊n−12⌋(n2⋅i+1)3i\sum^{\left\lfloor \frac{n - 1}{2}\right\rfloor}_{i=0} \binom{n}{2 \cdot i + 1} 3^i. Find all nn such that f(n)=1996.f(n) = 1996. [hide="old version"]For each positive integer nn, let f(n)f(n) be the maximal natural number such that: 2f(n)2^{f(n)} divides ∑i=1n+1/2(2⋅i+1n)\sum^{n + 1/2}_{i=1} \binom{2 \cdot i + 1}{n}. Find all nn such that f(n)=1996.f(n) = 1996.
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