We conjectured in Mathematics of Computation 2003 p.1379-1397 that, essentially,
         (a)~-(a+1/2) for large k (for 0 < a < =1/2)
(in a relative sense, i.e. more precisely  Conjecture II stated [(a)+(a+1/2)]/(a) --> 0 as k --> infinity).  After communication with Mark Coffey, we now generalize this conjecture to, essentially,
        (a)+(a+1/n)+(a+2/n)+...(a+(n-1)/n)~0 (for 0 < a<=1/n)
again in a relative sense.  Here is some numerical evidence, with k=100, a=1/7 and n=7:

a
	3.44786558612769590120072434105326444*1017
	4.945407766645782980895601684909204*1017
	2.7189570309463484160669849229735*1017
1

            Sum of values above is

Similarly, with k=147, a=1/100 and n=15:

a
	1.901777395433241200753844993225364640*1035
	1.6278165124352742710215723103987217260*1035
	1.072391366188489130859935120720069183*1035
	3.31540010358000332847753250067115817484167*1034
	6.90578879662305522263060059022236939142136*1033
	8.4382305905599180362323654059999645854798273*1034
	1.472683557343140394539632981047144255514706*1035
	1.8469036896555535391131402976601917624643*1035

                Sum of the above values is , i.e. zero to 24 digits.

Note that sin(2π/n)+sin(4π/n)+... + sin((n-1)2π/n)=0.  The conjecture above is hence supported by the following figures (i.e.  pictorial evidence), which plot (a) vs a for k=100, 125, and 163, respectively:

We fit a pure sine function to the above data, namely f(a)=5.355*sin(2π(a+.1025)), and superimpose with the graph of (a):

Rick Kreminski, kremin@boisdarc.tamu-commerce.edu    August 8, 2004