|
%I #27 Jan 04 2017 07:23:35
%S 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,
%T 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,
%U 0,0,1,0,2,0,2,1,1,0,1,0,1,0,1,1,1,0,1
%N a(n) = floor((k/phi(k) - (e^gamma)*loglog(k))*sqrt(log(k))) where k = A100966(n).
%C 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.
%C 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).
%H Peter J. C. Moses, <a href="/A279291/b279291.txt">Table of n, a(n) for n = 1..5000</a>
%H J.-L. Nicolas, <a href="http://dx.doi.org/10.4064/aa155-3-7">Small values of the Euler function and the Riemann hypothesis</a>, Acta Arithmetica, 155(2012), 311-321.
%e 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.
%Y Cf. A000010, A001620, A279161, A100966.
%K nonn
%O 1,4
%A _Vladimir Shevelev_, Dec 09 2016
%E More terms from _Peter J. C. Moses_, Dec 09 2016
|
