OFFSET
1,2
COMMENTS
a(10) > 99999.
a(10) > 10^7. - Jon E. Schoenfield, Mar 27 2015
At n=163, exp(Pi*sqrt(n)) is remarkably close to its nearest integer, differing from it by less than 7.5*10^-13. Consequently, for values of n=163*k^2 where k is a sufficiently small integer greater than 1, exp(Pi*sqrt(n)) = exp(Pi*sqrt(163*k^2)) = exp(Pi*sqrt(163)*k) = (exp(Pi*sqrt(163)))^k will also be close to an integer. If we exclude numbers of the form 163*k^2 for k=2..4, then the number of values of n < 10^7 at which exp(Pi*sqrt(n)) differs from its nearest integer by less than 10^-6 is 21, which is about what we would expect if we were instead generating random numbers whose fractional parts followed a uniform distribution on the interval [0,1). If the fractional parts continue to behave in this way, then we could expect about a 50% chance of finding a(10) at some value below log(2)/(2u) = 4.62*10^11 where u = abs(t - round(t)) and t = exp(Pi*sqrt(163)). - Jon E. Schoenfield, Mar 27 2015
LINKS
University of Sheffield, Department of Pure Mathematics, Is e^(Pi*Sqrt(163)) an integer?
University of Sheffield, Department of Pure Mathematics, Is e^(Pi*Sqrt(163)) an integer?
MATHEMATICA
s = 1; Do[ t = Abs[ N[ E^(Pi*Sqrt[n]), 10^3] - Round[ E^(Pi*Sqrt[n])]]; If[s > t, s = Abs[t]; Print[n]], {n, 1, 10^4}]
PROG
(PARI) A069014()={default(realprecision, 1000); my(maxx=9999); n=1; minn=1; while (n<maxx, q=abs(exp(Pi*sqrt(n))-round(exp(Pi*sqrt(n)))); if(q<minn, minn=q; print1 (n, ", ")); n+=1); } \\ Bill McEachen, Mar 15 2015
CROSSREFS
KEYWORD
nonn,more
AUTHOR
Robert G. Wilson v, May 24 2002
EXTENSIONS
Name edited by Jon E. Schoenfield, Mar 24 2015
STATUS
approved