olymmath

Putnam 1966 A4

  • Session A

A-4. Prove that after deleting the perfect squares from the list of positive integers the number we find in the nnth position is equal to n+{n}n+\{\sqrt{ } n\}, where {n}\{\sqrt{ } n\} denotes the integer closest to n\sqrt{ } n.

Dataset text PutnamGAP. Quoted with source.

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