olymmath

BALKAN 2004 Problem 1

  • Paper

The sequence {an}n≥0\{a_n\}_{n\geq 0} of real numbers satisfies the relation: am+n+am−n−m+n−1=12(a2m+a2n)a_{m+n} + a_{m-n} - m + n -1 = \frac12 (a_{2m} + a_{2n}) for all non-negative integers mm and nn , m≥nm \ge n . If a1=3a_1 = 3 find a2004a_{2004} .

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