It's well known that exp(pi*sqrt(n)) is very close to an integer if the quadratic imaginary field Q(sqrt(-n)) has class number 1. One explanation for this is that the j-invariant of the point (1 + sqrt(-n))/2 must be an integer, and we have the formula j(z) = 1/q + 722 + sum c_n q^n where q = exp(2pi*i*z). However, if the field has class number 2 one can make a slightly more complicated argument (j(z) is no longer an integer, but a quadratic real integer) to show that 1/q must be close to an integer (I don't remember the details, but I'm sure that someone like Noam E. will step in here). The list of discriminants with class number 2 is 5,6,10,13,15,22,35,37,51,58,91,115,123,187,235,267,403,427 Victor On Sun, Mar 2, 2014 at 3:52 PM, Warren D Smith <warren.wds@gmail.com> wrote:
p = 4*a(1) m = 2.0 define v(x) { if(x<0) return(-x) else return(x) } t = 2 for(n=1; n<999999999999; n++){ y = sqrt(n)*p; x = v(s(p*e(y))); if(x<m){ m=x; print n, " ", m, "\n"; } if(n>=t){ t = t*2; print "reached ", n, "\n"; } }
is a bc program which tries to compute the sequence 1, 2, 6, 17, 22, 25, 37, 58, 163, ... aka http://oeis.org/A069014 (You need to preset scale=something large before running it).
This is very slow. Presumably, the next entry in the sequence is of order n=10^12 and would require computing the fractional part of a number having about 10^6 digits before the decimal point. This seems infeasible to reach unless way smarter algorithms can be invented.
-- Warren D. Smith http://RangeVoting.org <-- add your endorsement (by clicking "endorse" as 1st step)
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