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Further corrections from the department of redundancy department.
Thom Brown
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@@ -125,7 +125,7 @@ we get a disk layout like this:
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where the numbers are page numbers *at that level*, starting from 0.
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To find the physical block # corresponding to leaf page n, we need to
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count the number number of leaf and upper-level pages preceding page n.
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count the number of leaf and upper-level pages preceding page n.
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This turns out to be
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y = n + (n / F + 1) + (n / F^2 + 1) + ... + 1
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