Prime counting sequence

[David Wilson]

Let a(n) be the number of k >= 1 with floor(p(k)/k) = n.

For n = 1, 2, 3, ..., I get a(n) =

3 8 19 41 117 254 616 1642 3766 9461 24183 60252 151368 385600 979844 
2507393 6428977 16513542 42642649

I computed this using a brute force sieve-based prime counting program.  I 
would appreciate verification before submitting.  Presumably a cleverer 
approach using a smart large p(n) algorithm could be used to extend this 
considerably.

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- David Wilson