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A207709 Floor((H(n) + exp(H(n))*log(H(n)))/sigma(n)), where H(n) is the harmonic number sum_{i=1..n} 1/i. 3
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 1, 1, 2, 1, 2, 1, 2 (list; graph; refs; listen; history; text; internal format)
OFFSET
1,11
COMMENTS
An assertion equivalent to the Riemann hypothesis is: a(n) > 0 for every n >= 1.
a(12*n) = 1 for all 1<=n<=43312.
For n >= 1, a(2^(10^n)) so far appears to equal floor(n*(exp(1)-2/3) - 1/3).
There exist integers n such that (H(n) + exp(H(n))*log(H(n)))/sigma(n) < 1.01 (i.e., n = 100630609505753353981293837481689271234222794240000*1087#). See A215640 for information on how to generate these numbers. - Arkadiusz Wesolowski, Aug 19 2012
LINKS
Arkadiusz Wesolowski, Table of n, a(n) for n = 1..10000
J. C. Lagarias, An elementary problem equivalent to the Riemann hypothesis, Am. Math. Monthly 109 (#6, 2002), 534-543.
Wikipedia, Jeffrey Lagarias
EXAMPLE
a(11) = 2 because (H(11) + exp(1)^H(11)*log(H(11)))/sigma(11) = 2.1387006307....
MATHEMATICA
lst = {}; Do[h = NIntegrate[(1 - x^n)/(1 - x), {x, 0, 1}]; AppendTo[lst, Floor[(h + Exp@h*Log@h)/DivisorSigma[1, n]]], {n, 530}]; lst
CROSSREFS
Sequence in context: A060500 A187284 A160198 * A131718 A131017 A244227
KEYWORD
nonn
AUTHOR
STATUS
approved

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Last modified March 29 01:36 EDT 2024. Contains 371264 sequences. (Running on oeis4.)