With one-based indexing (u(1)=1, u(2)=2, u(3)=3, ...), I get u(1001)=12336 u(1092)=13533 u(1211)=14894 u(1731)=21870 u(1749)=22151 u(1948)=24851 u(2000)=25511 and for these n <= 2000 I get .5 <= f(n) < .55 (computation of f(n) using PARI, realprecision = 28 significant digits): n u(n) f(n) 7 11 0.5092592592592592592592592592 42 206 0.5370370370370370370370370370 67 400 0.5185185185185185185185185185 123 983 0.5092592592592592592592592598 708 8370 0.5000000000000000000000000000 845 10271 0.5092592592592592592592592630 890 10833 0.5277777777777777777777777828 1092 13533 0.5277777777777777777777777763 1211 14894 0.5370370370370370370370370394 1731 21870 0.5000000000000000000000000129 1749 22151 0.5092592592592592592592592630 1948 24851 0.5092592592592592592592592630 I found no n <= 2000 such that 10.3 <= f(n) < 11.33. Klaus Brockhaus _______________________________________________________________ Dan Hoey wrote:
Jud McCranie <judmccr@bellsouth.net>
I'm getting in on this in the middle, so I don't know what you're discussing, but maybe I could see what I get with my program.
Thanks for the offer. There are two things I want verified. First, is it true that: u(1000)=12336 u(1091)=13533 u(1210)=14894 u(1730)=21870 u(1748)=22151 u(1947)=24851 u(1999)=25511 ? This is with zero-based indexing (u(0)=1, u(1)=2, u(2)=3, ...), so maybe you need to add 1 to the indices.
Second, letting f(n) = (u(n)/21.6) - floor(u(n)/21.6, for what n in 1000...1999 is .5 <= f(n) < .55?
Thanks.
Dan Hoey
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