A279291 a(n) = floor((k/phi(k) - (e^gamma)*loglog(k))*sqrt(log(k))) where k = A100966(n).

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

Offset: 1

Author

Vladimir Shevelev, Dec 09 2016

Keywords

• nonn

Comments

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).

Examples

			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.
		

Links

• Peter J. C. Moses, Table of n, a(n) for n = 1..5000
• J.-L. Nicolas, Small values of the Euler function and the Riemann hypothesis, Acta Arithmetica, 155(2012), 311-321.

Crossrefs

Cf. A000010, A001620, A279161, A100966.

Extensions

More terms from Peter J. C. Moses, Dec 09 2016