MathDB

6

Part of 2000 German National Olympiad

Problems(1)

a_{2^m} = 1/m, a_{2n-1}a_{2n} = a_n, a_{2n}a_{2n+1} = a_{2^m+n}

Source: Germany 2000 p6

2/23/2020
A sequence (ana_n) satisfies the following conditions: (i) For each m∈Nm \in N it holds that a2m=1/ma_{2^m} = 1/m. (ii) For each natural n≥2n \ge 2 it holds that a2n−1a2n=ana_{2n-1}a_{2n} = a_n. (iii) For all integers m,nm,n with 2m>n≥12m > n \ge 1 it holds that a2na2n+1=a2m+na_{2n}a_{2n+1} = a_{2^m+n}. Determine a2000a_{2000}. You may assume that such a sequence exists.
Sequencerecurrence relationalgebra