5 P=1
10 for K=2 to 40
20 P=P*(K^3-1)/(K^3+1)
30 print K,P
40 next
finds the partial products up to various k:
max partial product
k
2 0.7777777777777777777
3 0.7222222222222222221
4 0.7
5 0.6888888888888888887
6 0.6825396825396825395
7 0.6785714285714285712
8 0.6759259259259259257
9 0.6740740740740740738
10 0.672727272727272727
11 0.6717171717171717168
12 0.6709401709401709398
13 0.6703296703296703292
14 0.6698412698412698407
15 0.6694444444444444439
16 0.6691176470588235288
17 0.6688453159041394329
18 0.668615984405458089
19 0.6684210526315789467
20 0.6682539682539682532
21 0.6681096681096681089
22 0.6679841897233201573
23 0.6678743961352656997
24 0.6677777777777777769
25 0.6676923076923076915
26 0.6676163342830009488
27 0.6675485008818342143
28 0.6674876847290640385
29 0.6674329501915708803
30 0.6673835125448028664
31 0.6673387096774193539
32 0.667297979797979797
33 0.6672608437314319658
34 0.66722689075630252
35 0.6671957671957671947
36 0.6671671671671671661
37 0.6671408250355618766
38 0.6671165092217723785
39 0.6670940170940170928
40 0.6670731707317073159
and, when extended there seems to be a pattern in the partial products of max k = 4 x 10^i (more precision was used here than shown, to allow for losses in rounding at each step):
40 0.6670731707317073170731707
400 0.6666708229426433915211970
4000 0.6666667083229192701824543
40000 0.6666666670833229169270768
400000 0.6666666666708333229166927
4000000 0.6666666666667083333229166
40000000 0.6666666666666670833333229
In any event, the limit seems to be 2/3.
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Posted by Charlie
on 2013-06-24 12:29:29 |
computer exploration (spoiler)
