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A257822
Decimal expansion of the absolute value of the imaginary part of li(-A257821).
3
3, 8, 7, 4, 5, 0, 1, 0, 4, 9, 3, 1, 2, 8, 7, 3, 6, 2, 2, 3, 7, 0, 9, 6, 9, 7, 1, 3, 5, 0, 6, 3, 3, 9, 0, 1, 2, 3, 8, 4, 0, 5, 8, 0, 4, 0, 5, 4, 5, 0, 4, 8, 4, 6, 3, 7, 7, 3, 4, 0, 2, 1, 4, 5, 6, 4, 6, 0, 3, 2, 4, 7, 8, 2, 1, 6, 8, 6, 5, 4, 3, 7, 2, 6, 5, 3, 3, 8, 6, 7, 8, 2, 3, 8, 9, 5, 3, 1, 1, 4, 8, 4, 6, 1, 2
OFFSET
1,1
COMMENTS
As discussed in A257820, the absolute value of the imaginary part is continuous and its value is a well behaved function of any real argument, excepting z=+1. The above value corresponds to |imag(li(z))| at z=-A257821, the unique point in the real interval (-infinity,+1) where the corresponding real part is zero.
LINKS
Eric Weisstein's World of Mathematics, Logarithmic Integral
EXAMPLE
3.87450104931287362237096971350633901238405804054504846377340...
MATHEMATICA
RealDigits[Im[LogIntegral[-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)))); }
root=solve(x=-3, -1, real(li(x))); \\ Better use excess realprecision
a=imag(li(root))
CROSSREFS
Sequence in context: A371527 A177346 A357319 * A201293 A335810 A225016
KEYWORD
nonn,cons
AUTHOR
Stanislav Sykora, May 11 2015
STATUS
approved