Maths Olympiad Prep

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Combinatorics Difficulty 5.1 AIME, harder Prove it United States

Problem:

An infinite table of nonnegative integers is constructed as follows: in the top row, some number is 11 and all other numbers are 00's; in each subsequent row, every number is the sum of some two of the three closest numbers in the preceding row. An example of such a table is shown below.

000010000000011000000112100001133200012446320. \begin{array}{ccccccccccc} \cdots & 0 & 0 & 0 & 0 & 1 & 0 & 0 & 0 & 0 & \cdots \\ \cdots & 0 & 0 & 0 & 0 & 1 & 1 & 0 & 0 & 0 & \cdots \\ \cdots & 0 & 0 & 0 & 1 & 1 & 2 & 1 & 0 & 0 & \cdots \\ \cdots & 0 & 0 & 1 & 1 & 3 & 3 & 2 & 0 & 0 & \cdots \\ \cdots & 0 & 1 & 2 & 4 & 4 & 6 & 3 & 2 & 0 & \cdots \\ . \cdot & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots \end{array}

The top row (with the one 11) is called row 00; the next row is row 11; the next row is row 22, and so forth.

Note that the following problems require you to prove the statements for every table that can be constructed by the process described above, not just for the example shown.

1. Show that any number in row nn (for n>0n>0) is at most 2n12^{n-1}.

Solution

Solution:

We use induction on nn. It is clear that any number in row 11 is at most 1=201 = 2^{0}. Now, if every number in row nn is at most 2n12^{n-1}, then every number in row n+1n+1 is the sum of two numbers in row nn and so is at most 2n1+2n1=2n2^{n-1} + 2^{n-1} = 2^{n}. This gives the induction step, and the result follows.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.