Sondow, Jonathan Criteria for irrationality of Euler’s constant. (English) Zbl 1113.11040 Proc. Am. Math. Soc. 131, No. 11, 3335-3344 (2003). By modifying Beukers’ proof of Apéry’s theorem that \(\zeta(3)\) is irrational, the author derives criteria for irrationality of Euler’s constant, \(\gamma\). For \(n>0\), define a double integral \(I_n\) and a positive integer \(S_n\), then it is proved that with \(d_n=\text{lcm}(1,\dots,n)\) the following are equivalent:1) The fractional part of \(\log S_n\) is given by \(\{\log S_n\}=d_{2n}I_n\) for some \(n\). 2) The formula holds for all sufficiently large \(n\).3) Euler’s constant is a rational number. A corollary is that if \(\{\log S_n\}\geq 2^{-n}\) infinitely often, then \(\gamma\) is irrational. Indeed, if the inequality holds for a given \(n\) (numerical evidence for \(1\leq n\leq 2500\) is given by the author) and \(\gamma\) is rational, then its denominator does not divide \(d_{2n}\binom{2n}{n}\). The author also proves a new combinatorial identity in order to show that a certain linear form in logarithms is in fact \(\log S_n\). A by-product is a rapidly converging asymptotic formula for \(\gamma\), used by P. Sebah to compute \(\gamma\) correct to 18063 decimals. Reviewer: Olaf Ninnemann (Berlin) Cited in 3 ReviewsCited in 22 Documents MSC: 11J72 Irrationality; linear independence over a field 05A19 Combinatorial identities, bijective combinatorics Keywords:irrationality; Euler’s constant; Apéry’s theorem; Beukers’ integrals; linear form in logarithms; fractional part; harmonic number; Prime Number Theorem; Laplace’s method; asymptotic formula; combinatorial identity PDFBibTeX XMLCite \textit{J. Sondow}, Proc. Am. Math. Soc. 131, No. 11, 3335--3344 (2003; Zbl 1113.11040) Full Text: DOI arXiv Online Encyclopedia of Integer Sequences: Decimal expansion of Euler’s constant (or the Euler-Mascheroni constant), gamma. Continued fraction for Euler’s constant (or Euler-Mascheroni constant) gamma. Least common multiple (or LCM) of {1, 2, ..., n} for n >= 1, a(0) = 1. a(n) = lcm(1, 2, ..., 2n) / lcm(1, 2, ..., n). Continued fraction for e^gamma. a(n) = lcm{1, 2, ..., 2*n}.