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Old 11-13-2008   #42 (permalink)
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CraigD
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Post Failure at the 18th prime

Using exact arithmetic starting with Don’s estimated value of

A = \frac{2566543832171388844467529}{1000000000000000000000000}

, I get a slightly different, but similar behavior for the generating function

A_n = \frac1{\frac{A_{n-1}}{\lfloor A_{n-1} \rfloor}-1}

to what Hal calculated using a spreadsheet:
Code:
Prime     A
  2       2.5665438321713888444675290
  3       3.530176989721365539402422
  5       5.658487746849688216649061
  7       7.593155717658844724384335
 11      11.801285550493624044601748
 13      13.72794004986556753523763
 17      17.85861349763730877250927
 19      19.79936263147479092664292
 23      23.76893696537426002453788
 29      29.91142451944074443786296
 31      31.81832327464107973441092
 37      37.88233936471712759962716
 41      41.93397855694896393020755
 43      43.89822410263364625518520
 47      47.87224020589233116148280
 53      53.88423932134315471134298
 59      59.93852424419815301864134
 61      62.86465199458734434295017
 67      71.70514887852601938925004
 71     100.6879570572559038006540
 73     145.3579099818760320973203
 79     405.129801745018365638482
 83    3120.14295295257072632134
 89   21825.36242793846740266
 97   60218.86748657202264851
101   69416.63645535693254109
with failure at the 18th prime, 61.

I’m guessing, however, that Don believe that a higher precision estimate may increase the number of correct primes the algorithm generates. Unfortunately, I don’t have such an estimate handy. When I do, I’ll retry the algorithm, and see if does, though based on trying it with +/- an additional digit of precision, I don’t think it will.


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Last edited by CraigD; 11-13-2008 at 12:44 AM..
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