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A279291
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a(n) = floor((k/phi(k) - (e^gamma)*loglog(k))*sqrt(log(k))) where k = A100966(n).
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1
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1, 1, 0, 2, 1, 0, 1, 2, 0, 0, 0, 1, 0, 0, 1, 0, 0, 2, 1, 0, 2, 1, 0, 1, 2, 1, 0, 0, 1, 1, 2, 0, 0, 0, 0, 2, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 0, 0, 0, 0, 3, 0, 0, 0, 0, 1, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 2, 0, 0, 1, 0, 2, 0, 2, 1, 1, 0, 1, 0, 1, 0, 1, 1, 1, 0, 1
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OFFSET
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1,4
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COMMENTS
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Assuming the Riemann hypothesis, no term exceeds 4. Indeed, let c(n) = (n/phi(n) - (e^gamma)*loglog(n))*sqrt(log(n)). Then, by [Nicolas], the Riemann hypothesis is equivalent to the inequality: for n>=2, c(n)<=c(N), where N is the product of the first 66 primes such that c(N)=4.0628356921... . Since for n in [or "not in", the grammar of the original was ambiguous here - N. J. A. Sloane, Jan 04 2017] A100966, we have c(n)<=0, for those n c(n)<=c(N). Thus assuming the R. H. we see that a(n)<=4.
On the other hand, we conjecture that a(n)<=4 should be true independent of the R. H. If so, then the statement that the R. H. is false would be equivalent to the existence of n for which c(n) is in interval (c(N),5).
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LINKS
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EXAMPLE
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The first term in A100966 is k=3. So a(1) = {floor((3/phi(3) - (e^gamma)*loglog(3))*sqrt(log(3)))} = floor((3/2 - 1.78...*0.094...)*1.048...) = 1.
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CROSSREFS
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KEYWORD
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nonn
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AUTHOR
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EXTENSIONS
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STATUS
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approved
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