On Tuesday 12 April 2005 17:58, R. William Gosper wrote:
Numerical experiments suggest that, for m = 1..9, ..., f(m) :=
inf ==== m k \ %pi (k 2 - 5) n (- 1) > tan(--------------) (- 1) inf / n + m ==== ==== 2 5 m - 1 \ n = 1 3 2 > --------------------------------------- / m ==== m 3 10 %pi (k 2 - 5) k = - inf (k 2 - 5) csc(-----------------) 2 m 2 ------------------------------------------------------------ 3 5 %pi %pi csc(-----) m 2
= 2, 8, 32 - sqrt(2), 113, 382, 833, 1822, 3713, 7582, ...
Screwy indeed. I can (numerically) confirm that the recurrence that holds for the last few of those continues at least through m=14, not that there can have been much doubt about that. Write f(m) = g(2^m) in the obvious way. Then the following values of g, at least, seem to be (very close to) integers: 2 : 1 0 0 : 2 4 : 2 0 0 : 8 6 : 1 1 0 : 20 10 : 1 0 1 : 50 12 : 2 1 0 : 71 15 : 0 1 1 : 100 16 : 4 0 0 : 113 20 : 2 0 1 : 175 24 : 3 1 0 : 244 30 : 1 1 1 : 350 32 : 5 0 0 : 382 40 : 3 0 1 : 500 48 : 3 1 0 : 611 60 : 2 1 1 : 775 64 : 6 0 0 : 833 80 : 4 0 1 : 1075 96 : 5 1 0 : 1324 120 : 3 1 1 : 1700 128 : 7 0 0 : 1822 160 : 5 0 1 : 2300 192 : 6 1 0 : 2771 240 : 4 1 1 : 3475 256 : 8 0 0 : 3713 320 : 6 0 1 : 4675 384 : 7 1 0 : 5644 480 : 5 1 1 : 7100 512 : 9 0 0 : 7582 640 : 7 0 1 : 9500 768 : 8 1 0 : 11411 960 : 6 1 1 : 14275 1024 : 10 0 0 : 15233 1280 : 8 0 1 : 19075 1536 : 9 1 0 : 22924 1920 : 7 1 1 : 28700 The numbers in the middle are the number of 2,3,5 factors in m. It seems that we get integer values for m = 2^i3^j5^k with i,j,k integers and j,k 0 or 1 -- except for a few oddities at small m, such as g(3)=1/4 and g(5)=oo. I suppressed a couple of entries my program puts in the table above, because they don't come out so close to integers and I'm suspicious. g(691) ~= 10261 g(1382) ~= 20609 Hmm, 691. Haven't I seen that number somewhere before? :-) Anyway, I am fairly convinced after a little more computation that g(691) and g(1382) are not actually integers, but that they *are* very close to integers. (g(691) is about 10261 - 1/3301, and g(1382) is about 20609 + 1/3323. I wouldn't make any large bets on the last couple of digits of 3301 and 3323.) There are some provocative things in the table, to those of us who are unfortunate enough to feel most comfortable in base 10. And some provocative whatever base one works in. u(r) = 1*2^r satisfies u(2r)-2u(r) = 156,69,156,69,... after a while u(r) = 3*2^r satisfies u(2r)-2u(r) = 132,102,123,102,... after a while u(r) = 5*2^r satisfies u(2r)-2u(r) = 150,75,150,75,... u(r) = 15*2^r satisfies u(2r)-2u(r) = 150,75,150,75,... Similar things seem to happen with u(3r)-3u(r), and probably (though I'm too lazy to check) with u(5r)-5u(r). Enough numerology! Surely there must be some real mathematics to be done here. (Where did the definition of f(m) come from, anyway?) -- g