My first thought would be the log function (log(1) = 0, and log(1e9) = 20.7232658369464). Otherwise, you have 1000 ranges of size n .. n + ( 999_999 ), so you could just divide by 1_000_000.

Hope that helps.

Update: 2015-02-09
Thanks to choroba for catching my error in linking to the log function documentation.

In reply to Re: Encoding function needed by atcroft
in thread Encoding function needed by baxy77bax

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