A092605 Decimal expansion of e^(-1/2) or 1/sqrt(e).
6, 0, 6, 5, 3, 0, 6, 5, 9, 7, 1, 2, 6, 3, 3, 4, 2, 3, 6, 0, 3, 7, 9, 9, 5, 3, 4, 9, 9, 1, 1, 8, 0, 4, 5, 3, 4, 4, 1, 9, 1, 8, 1, 3, 5, 4, 8, 7, 1, 8, 6, 9, 5, 5, 6, 8, 2, 8, 9, 2, 1, 5, 8, 7, 3, 5, 0, 5, 6, 5, 1, 9, 4, 1, 3, 7, 4, 8, 4, 2, 3, 9, 9, 8, 6, 4, 7, 6, 1, 1, 5, 0, 7, 9, 8, 9, 4, 5, 6, 0, 2, 6, 4, 2, 3
Offset: 0
Examples
0.6065306597126334...
References
- Paulo Ribenboim, The Little Book of Bigger Primes, Springer-Verlag NY 2004. See p. 225.
- C. L. Siegel, Zu zwei Bemerkungen Kummers. Nachr. Akad. d. Wiss. Göttingen, Math. Phys. Kl., II, 1964, 51-62. Reprinted in Gesammelte Abhandlungen (edited by K. Chandrasekharan and H. Maas), Vol. III, 436-442. Springer-Verlag, Berlin, 1966.
Links
- Michael I. Shamos, A catalog of the real numbers, (2007). See p. 502.
- Index entries for transcendental numbers.
Programs
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Mathematica
RealDigits[E^-(1/2),10,120][[1]] (* Harvey P. Dale, Jul 23 2012 *)
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PARI
exp(-.5) \\ Charles R Greathouse IV, Oct 02 2022
Formula
Equals Sum_{k>=0} (-1)^k/(2^k * k!) = Sum_{k>=0} (-1)^k/A000165(k). - Amiram Eldar, Aug 15 2020
From Peter Bala, Jan 16 2022; (Start)
Equals 16*Sum_{n >= 0} (-1)^n*n^2/((4*n^2 - 1)*(4*n^2 - 9)*(2^n)*n!).
Equals 8*Sum_{n >= 0} (-1)^n/(p(n)*p(n+1)*(2^n)*n!), where p(n) = 4*n^2 + 8*n + 1.
Equals 48*Sum_{n >= 0} (-1)^n/(q(n)*q(n+1)*(2^n)*n!), where q(n) = 8*n^3 + 36*n^2 + 34*n + 1. (End)
Equals i^(i/Pi), where i denotes the imaginary unit. - Stefano Spezia, Mar 01 2025
Equals 1 - A290506. - Amiram Eldar, Apr 22 2025
Comments