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A218245 Nicolas's sequence, whose nonnegativity is equivalent to the Riemann hypothesis. 2
2, 1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,1
COMMENTS
a(n) = floor(p(n)#/phi(p(n)#) - log(log(p(n)#))*exp(gamma)), where p(n)# is the n-th primorial, phi is Euler's totient function, and gamma is Euler's constant.
J.-L. Nicolas proved that all terms are >= 0 if and only if the Riemann hypothesis (RH) is true. In fact, results in his 2012 paper imply that RH is equivalent to a(n) = 0 for n > 6. Nicolas's refinement of this result is in A233825.
He also proved that if RH is false, then infinitely many terms are >= 0 and infinitely many terms are < 0.
See Nicolas's sequence A216868 for references, links, and additional cross-refs.
LINKS
FORMULA
a(n) = [p(n)#/phi(p(n)#) - log(log(p(n)#))*exp(gamma)].
a(n) = [A002110(n)/A005867(n) - log(log(A002110(n)))*e^gamma].
EXAMPLE
p(2)# = 2*3 = 6 and phi(6) = 2, so a(2) = [6/2 - log(log(6))*e^gamma] = [3-0.58319...*1.78107...] = [3-1.038...] = 1.
MATHEMATICA
primorial[n_] := Product[Prime[k], {k, n}]; Table[ With[{p = primorial[n]}, Floor[N[p/EulerPhi[p] - Log[Log[p]]*Exp[EulerGamma]]]], {n, 1, 100}]
CROSSREFS
Sequence in context: A165105 A325674 A055641 * A352516 A086075 A316865
KEYWORD
nonn
AUTHOR
Jonathan Sondow, Oct 24 2012
STATUS
approved

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Last modified April 23 03:30 EDT 2024. Contains 371906 sequences. (Running on oeis4.)