Prime counting sequence

[Victor S. Miller]

>>>>> "David" == David Wilson <davidwwilson at comcast.net> writes:

David> Let a(n) be the number of k >= 1 with floor(p(k)/k) = n.  For n
David> = 1, 2, 3, ..., I get a(n) =

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

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

I wouldn't say "considerably" since p(n) is approximately n log n, so
that the n-th term of your sequence is around exp(n).
But, fast prime-counting techniques could probably get this up to
about 20 terms or so.

Victor