OFFSET
1,1
COMMENTS
Let X_1, X_2, ... be a sequence of independent Bernoulli trials with probability of success 1/n^3. Let Y be the position of the last success in the sequence. 1.428189... is the expected value of Y. - Geoffrey Critzer, Aug 19 2019
If m tends to infinity, Product_{k>=1} (1 + m/k^3) ~ exp(2*Pi*m^(1/3)/sqrt(3)) / (2^(3/2)*Pi^(3/2)*sqrt(m)). - Vaclav Kotesovec, Aug 30 2024
LINKS
G. C. Greubel, Table of n, a(n) for n = 1..25000
Simon Plouffe, product(1+1/n**3,n=1..infinity).
S. Ramanujan, Notebook entry.
FORMULA
Equals cosh(sqrt(3) * Pi / 2)/Pi.
Equals exp(Sum_{j>=1} (-(-1)^j*zeta(3*j)/j)). - Vaclav Kotesovec, Mar 28 2019
Equals Product_{n>=1} (1 + 1/(n^2 + n)). - Amiram Eldar, Sep 01 2020
Equals 3*Product_{n >= 2} (1-n^(-3)) = 3*A109219. - Robert FERREOL, Oct 06 2021
EXAMPLE
2.42818979209887032873604143617914635811836294478339049763...
MATHEMATICA
RealDigits[ Cosh[Sqrt[3]*Pi/2]/Pi, 10, 105][[1]] (* Jean-François Alcover, Nov 18 2015 *)
PROG
(Magma)
m:= 510; R:=RealField(m); SetDefaultRealField(R);
Prune(Reverse(IntegerToSequence(Floor(( Cosh(Sqrt(3)*Pi(R)/2)/Pi(R) )*10^(Floor(m/2)) )))); // G. C. Greubel, Sep 09 2025
(SageMath)
numerical_approx( cosh(sqrt(3)*pi/2)/pi , digits=265 ) # G. C. Greubel, Sep 09 2025
CROSSREFS
KEYWORD
cons,nonn
AUTHOR
Robert G. Wilson v, Aug 03 2002
STATUS
approved
