|
|
A365797
|
|
Decimal expansion of smallest positive number x such that Gamma(x) = 2.
|
|
0
|
|
|
4, 4, 2, 8, 7, 7, 3, 9, 6, 4, 8, 4, 7, 2, 7, 4, 3, 7, 4, 5, 2, 0, 3, 2, 5, 1, 6, 5, 2, 0, 6, 0, 5, 6, 7, 1, 7, 1, 0, 3, 6, 4, 5, 3, 8, 0, 6, 6, 3, 6, 6, 4, 0, 2, 9, 9, 1, 2, 3, 0, 7, 1, 9, 8, 9, 5, 8, 5, 2, 4, 8, 2, 2, 8, 4, 1, 7, 4, 0, 8, 0, 4, 0, 7, 7, 0, 0, 9, 3, 7, 7, 2, 9, 8, 4, 4, 8, 2, 2, 1, 0, 8, 3, 6, 3, 4
(list;
constant;
graph;
refs;
listen;
history;
text;
internal format)
|
|
|
OFFSET
|
0,1
|
|
COMMENTS
|
Second branch (i.e., the first after the principal branch) of the inverse gamma function Gamma(y) = x at x=2. See for instance Uchiyama.
Since 1 - x = 0.55712260351... (approximately equal to A249649), we can obtain the interesting approximation Gamma(zeta(2) - zeta(3)) ≈ 2.000001... - David Ulgenes, Feb 19 2024
|
|
LINKS
|
K. Amenyo Folitse, David J. Jeffrey, and Robert M. Corless, Properties and Computation of the Functional Inverse of Gamma, 2017 19th International Symposium on Symbolic and Numeric Algorithms for Scientific Computing (SYNASC). International Symposium on Symbolic and Numeric Algorithms for Scientific Computing (SYNASC). p. 65.
|
|
FORMULA
|
Equals ((((1/2)!/2)!/2)!/2)!/2...
Proof: Since y = y! / x we substitute the expression into itself to obtain an iterative scheme for the inverse gamma function.
Equals (1/(2*Pi))*Integral_{x=-oo..oo} log((2-Gamma(i*x))/(2-Gamma(1+i*x))) dx. Proof: Follows from writing the inverse gamma function using the Lagrange inversion theorem together with Cauchy's formula for differentiation. - David Ulgenes, Feb 11 2024
|
|
EXAMPLE
|
0.4428773964847274374520325165206056717103645380663664...
|
|
MAPLE
|
Digits:= 140:
with(RootFinding):
NextZero(x -> (x - 1)! - 2, 0);
|
|
MATHEMATICA
|
FindRoot[-2 + (-1 + x)! == 0, {x, 0, 1}, WorkingPrecision -> 15]
|
|
PROG
|
|
|
CROSSREFS
|
|
|
KEYWORD
|
|
|
AUTHOR
|
|
|
STATUS
|
approved
|
|
|
|