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.

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

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.

Eric W. Weisstein, MathWorld: Riemann Hypothesis

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

Cf. A057641, A000203, A008594, A076633.

Sequence in context: A060500 A187284 A160198 * A131718 A131017 A244227

Adjacent sequences: A207706 A207707 A207708 * A207710 A207711 A207712

Keyword

nonn,changed

Author

Arkadiusz Wesolowski, Feb 19 2012

Status

approved