olymmath

Putnam 2007 B3

  • Session B

Let x0=1x_0 = 1 and for n≥0n \geq 0, let xn+1=3xn+⌊xn5⌋x_{n+1} = 3x_n + \lfloor x_n \sqrt{5} \rfloor. In particular, x1=5x_1 = 5, x2=26x_2 = 26, x3=136x_3 = 136, x4=712x_4 = 712. Find a closed-form expression for x2007x_{2007}. (⌊a⌋\lfloor a \rfloor means the largest integer ≤a\leq a.)

Curated text Kedlaya Putnam archive. Quoted with source.

Solutions

Human solution Solution 1 by Manjul Bhargava, Kiran Kedlaya and Lenny Ng

The terms of the source let olymmath link to this solution but not copy it. Read the solution (Kedlaya Putnam archive).

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