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A345208 Decimal expansion of log(2*Pi) - gamma - 1, where gamma is Euler's constant (A001620). 1
2, 6, 0, 6, 6, 1, 4, 0, 1, 5, 0, 7, 8, 1, 2, 6, 2, 2, 9, 5, 4, 1, 4, 7, 3, 8, 2, 7, 2, 8, 8, 3, 2, 8, 4, 8, 6, 8, 0, 6, 3, 5, 6, 1, 1, 3, 3, 5, 6, 4, 3, 2, 2, 6, 8, 2, 8, 5, 3, 5, 8, 4, 6, 0, 8, 0, 6, 6, 3, 6, 6, 5, 0, 7, 6, 8, 5, 6, 1, 2, 4, 4, 5, 2, 5, 3, 9 (list; constant; graph; refs; listen; history; text; internal format)
OFFSET
0,1
COMMENTS
The first two formulae (in the Formula section) are similar to the sum and integral lim_{n->oo} (1/n) * Sum_{k=1..n} frac(n/k) = Integral_{x=0..1} frac(1/x) dx = 1 - gamma (A153810).
The second raw moment of the distribution of the fractional part of 1/x, where x is chosen uniformly at random from (0, 1]. Since the expected value is 1 - gamma, the second central moment, or variance, is log(2*Pi) - gamma - 1 - (1 - gamma)^2 = log(2*Pi) - gamma^2 + gamma - 2 = 0.081914807503... and the standard deviation is sqrt(log(2*Pi) - gamma^2 + gamma - 2) = 0.2862076300...
REFERENCES
Ovidiu Furdui, Limits, Series, and Fractional Part Integrals: Problems in Mathematical Analysis, New York: Springer, 2013. See Problem 3.42, pages 145 and 195.
LINKS
Ovidiu Furdui, Solution to Problem U27, Mathematical Reflections, Vol. 6 (2006), pp. 27-28.
Ovidiu Furdui, Exotic fractional part integrals and Euler's constant, Analysis, Vol. 31, No. 3 (2011), pp. 249-257.
Mircea Ivan and Alexandru Lupaş, Problem 11206, The American Mathematical Monthly, Vol. 113, No. 2 (2006), p. 180; A Limit Involving Euler's Constant, Solution to problem 11206 by Richard A. Stong, ibid., Vol. 114, No. 10 (2007), pp. 928-929.
Albert F. S. Wong, Problem 1845, Mathematics Magazine, Vol. 83, No. 2 (2010), p. 150; Integrating a square-fractional-reciprocal function, Solution to problem 1845 by Allen Stenger, ibid., Vol. 84, No. 2 (2011), pp. 155-156.
FORMULA
Equals lim_{n->oo} (1/n) * Sum_{k=1..n} frac(n/k)^2, where frac(x) = x - floor(x) is the fractional part of x.
Equals Integral_{x=0..1} frac(1/x)^2 dx.
Equals 2 * Sum_{k>=2} (zeta(k)-1)/(k*(k+1)).
Equals A061444 - A001620 - 1.
Equals -2 * Sum_{k>=1} (H(k) - log(k) - gamma - 1/(2*k)), where H(k) = A001008(k)/A002805(k) is the k-th harmonic number (Furdui, 2013). - Amiram Eldar, Mar 26 2022
EXAMPLE
0.26066140150781262295414738272883284868063561133564...
MATHEMATICA
RealDigits[Log[2*Pi] - EulerGamma - 1, 10, 100][[1]]
CROSSREFS
Sequence in context: A336082 A327280 A350256 * A241810 A156991 A229586
KEYWORD
nonn,cons
AUTHOR
Amiram Eldar, Jun 10 2021
STATUS
approved

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Last modified April 25 16:23 EDT 2024. Contains 371989 sequences. (Running on oeis4.)