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**Which partial sums of the Taylor series for \(e\) are convergents to \(e\)? (and a link to the primes 2, 5, 13, 37, 463). II.**
*(English)*
Zbl 1227.11031

Amdeberhan, Tewodros (ed.) et al., Gems in experimental mathematics. AMS special session on experimental mathematics, Washington, DC, January 5, 2009. Providence, RI: American Mathematical Society (AMS) (ISBN 978-0-8218-4869-2/pbk). Contemporary Mathematics 517, 349-363 (2010).

Let the \(n\)th partial sum of the Taylor series \(e=\sum_{r=0}^\infty\frac{1}{r!}\) be \(\frac{A_n}{n!}\), and let \(\frac{p_k}{q_k}\) be the \(k\)th convergent of the simple continued fraction for \(e\). The authors prove that given any three consecutive partial sums \(s_n\), \(s_{n+1}\), \(s_{n+2}\) at most two of them are convergents to \(e\). For any positive integer \(k\), there exists a constant \(n(k)\) such that if \(n\geq n(k)\), then among the \(k\) consecutive partial sums \(s_n,s_{n+1},\ldots,s_{n+k-1}\) at most two are convergents to \(e\). Almost all partial sums are not convergents to \(e\). A related result about the denominators \(q_k\) and powers of factorials is proved. There is a connection between the \(A_n\) and the primes \(2,5,13,37,463\).

For Part I see Contemp. Math. 457, 273–284 (2008; Zbl 1159.11004).

For the entire collection see [Zbl 1193.00060].

For Part I see Contemp. Math. 457, 273–284 (2008; Zbl 1159.11004).

For the entire collection see [Zbl 1193.00060].

Reviewer: Florin Nicolae (Berlin)

### MSC:

11A55 | Continued fractions |

11A41 | Primes |

11Y55 | Calculation of integer sequences |

11Y60 | Evaluation of number-theoretic constants |

### Citations:

Zbl 1159.11004### Software:

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\textit{J. Sondow} and \textit{K. Schalm}, Contemp. Math. 517, 349--363 (2010; Zbl 1227.11031)

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### Online Encyclopedia of Integer Sequences:

Total number of ordered k-tuples (k=0..n) of distinct elements from an n-element set: a(n) = Sum_{k=0..n} n!/k!.Decimal expansion of e.

Kempner numbers: smallest positive integer m such that n divides m!.

Numerators of convergents to e.

Denominators of convergents to e.

Numerator of Sum_{k=0..n} 1/k!.

Denominator of Sum_{k=0..n} 1/k!.

Integers n >= 1 such that n divides 0!-1!+2!-3!+4!-...+(-1)^{n-1}(n-1)!.

Primes p such that p divides 0!-1!+2!-3!+...+(-1)^{p-1}(p-1)!.

Cancellation factor in reducing Sum_{k=0...n} 1/k! to lowest terms.

Numbers k such that the denominator of the k-th convergent of the continued fraction expansion of e is prime.

Primes which are the denominators of convergents of the continued fraction expansion of e.

Indices of primes which are denominators of convergents to e.

a(n)! is the smallest factorial divisible by the numerator of Sum_{k=0...n} 1/k!, with a(0) = 1.

Largest prime factor of numerator of Sum_{k=0...n} 1/k!, with a(0) = 1.

Numbers n such that denominator of Sum_{k=0 to n} 1/k! is n!.

Numbers n such that the denominator of Sum_{k=0 to 2n} 1/k! is (2n)!.

Numbers n such that denominator of Sum_{k=0 to 2n+1} 1/k! is (2n+1)!/2.

Numbers n such that denominator of Sum_{k=0..4n+1} 1/k! is (4n+1)!/2.

a(n) = 1/2 times the cancellation factor in reducing Sum_{k=0 to 2n+1} 1/k! to lowest terms.

a(3n) = a(3n-1) + a(3n-2), a(3n+1) = 2n*a(3n) + a(3n-1), a(3n+2) = a(3n+1) + a(3n).

a(3n) = a(3n-1) + a(3n-2), a(3n+1) = 2n*a(3n) + a(3n-1), a(3n+2) = a(3n+1) + a(3n).

a(n) = min{k>0: the n-th convergent to e equals m/k! for some m}.

a(n) = (n+1)!/(d(n)*d(n+1)) where d(n) = cancellation factor in reducing Sum_{k=0...n} 1/k! to lowest terms.

a(n) = (n+3)!/(d(n)*d(n+1)*d(n+2)) where d(n) = cancellation factor in reducing Sum_{k=0...n} 1/k! to lowest terms.

a(n) = (n+3)/gcd(d(n), d(n+2)) where d(n) = cancellation factor in reducing Sum_{k=0..n} 1/k! to lowest terms.

a(n) = gcd(A(n), A(n+2))/gcd(d(n), d(n+2)) where A(n) = Sum_{k=0..n} n!/k! and d(n) = gcd(A(n), n!).

a(n) = gcd(A(n), A(n+2)) where A(n) = A000522(n) = Sum_{k=0..n} n!/k!.

a(n) = gcd(A093101(n), A093101(n+2)) where A093101(n) = gcd(n!, A(n)) and A(n) = A000522(n) = Sum_{k=0..n} n!/k!.

a(n) = (n+3)/gcd(A(n), A(n+2)) where A(n) = A000522(n) = Sum_{k=0..n} n!/k!.

Primes p such that p divides both A061354(p-3) and A061354(p-1).