olymmath

CZ-SK-MO 1999 Problem 4

    In a certain language there are only two letters, AA and BB . We know that (i) There are no words of length 11 , and the only words of length 22 are ABAB and BBBB . (ii) A segment of length n>2n > 2 is a word if and only if it can be obtained from a word of length less than nn by replacing each letter BB by some (not necessarily the same) word. Prove that the number of words of length nn is equal to 2n+2⋅(−1)n3\frac{2^n +2\cdot (-1)^n}{3}

    Dataset text AI-MO AoPS. Quoted with source.

    Solutions

    No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label.

    Add a solution

    Similar problems

    These problems have the most words in common with this statement.

    Locator cz-sk-mo/1999/p4 · Tag PMQ3CSAZJ4 · Edit this page on GitHub · Keys: j next, k previous