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A257821
Decimal expansion of the unique real number a>0 such that the real part of li(-a) is zero.
4
2, 4, 6, 6, 4, 0, 8, 2, 6, 2, 4, 1, 2, 6, 7, 8, 0, 7, 5, 1, 9, 7, 1, 0, 3, 3, 5, 0, 7, 7, 5, 9, 3, 2, 9, 5, 0, 2, 9, 0, 7, 8, 0, 8, 7, 8, 2, 7, 7, 4, 0, 9, 9, 8, 2, 3, 7, 8, 6, 0, 8, 9, 8, 8, 1, 6, 1, 2, 2, 4, 0, 9, 4, 1, 5, 0, 0, 9, 1, 5, 0, 7, 1, 7, 1, 6, 2, 7, 8, 1, 5, 8, 0, 4, 6, 5, 5, 8, 4, 7, 2, 9, 3, 3, 6
OFFSET
1,1
COMMENTS
As discussed in A257819, the real part of li(z) is a well behaved function for any real z, except for the singularity at z=+1. It has three roots: z=A070769 (the Soldner's constant), z=0, and z=-a. However, unlike in the other two cases, the imaginary part of li(-a) is not infinitesimal in the neighborhood of this root; it is described in A257822.
LINKS
Eric Weisstein's World of Mathematics, Logarithmic Integral
FORMULA
Satisfies real(li(-a)) = 0.
EXAMPLE
2.4664082624126780751971033507759329502907808782774099823786...
MATHEMATICA
RealDigits[a/.FindRoot[Re[LogIntegral[-a]]==0, {a, 2}, WorkingPrecision->120]][[1]] (* Vaclav Kotesovec, May 11 2015 *)
PROG
(PARI) li(z) = {my(c=z+0.0*I); \\ If z is real, convert it to complex
if(imag(c)<0, return(-Pi*I-eint1(-log(c))),
return(+Pi*I-eint1(-log(c)))); }
a=-solve(x=-3, -1, real(li(x))) \\ Better use excess realprecision
CROSSREFS
KEYWORD
nonn,cons
AUTHOR
Stanislav Sykora, May 11 2015
STATUS
approved