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#30 by Alois P. Heinz at Sat Feb 18 21:19:19 EST 2023
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#29 by Michael B. Porter at Sat Feb 18 21:02:40 EST 2023
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#28 by Luc Rousseau at Wed Feb 08 09:12:52 EST 2023
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#27 by Luc Rousseau at Wed Feb 08 09:12:12 EST 2023
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| COMMENTS
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With n = Product_{i=1..k} p_i the prime factorization of n, if one forgesconstructs for each i a test with a probability of success equal to 1/p_i, and if the tests are independent, then a(n)/n is the probability that at least one of the k tests succeeds. - Luc Rousseau, Jan 14 2023
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proposed
editing
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Discussion
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Wed Feb 08
| 09:12
| Luc Rousseau: Amended
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#26 by Jon E. Schoenfield at Sat Jan 14 16:19:50 EST 2023
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Discussion
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Sat Jan 14
| 16:21
| Luc Rousseau: Thank you
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| 16:34
| Jon E. Schoenfield: You're welcome! :-)
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Tue Feb 07
| 21:45
| Sean A. Irvine: "forges" -> "constructs" ? (For me forges suggests something underhand is going on, which is not the case here)
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#25 by Jon E. Schoenfield at Sat Jan 14 16:19:37 EST 2023
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| COMMENTS
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With n = ProdProduct_{i=1..k} p_i the prime factorization of n, if one forges for each i a test with a probability of success equal to 1/p_i, and if the tests are independent, then a(n)/n is the probability that at least one of the k tests succeeds. - Luc Rousseau, Jan 14 2023
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| STATUS
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proposed
editing
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Discussion
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Sat Jan 14
| 16:19
| Jon E. Schoenfield: (correction as required by the OEIS Style Sheet)
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#24 by Luc Rousseau at Sat Jan 14 12:23:17 EST 2023
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#23 by Luc Rousseau at Sat Jan 14 12:22:51 EST 2023
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| COMMENTS
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With n = Prod_{i=1..k} p_i the prime factorization of n, if one forges for each i a test with a probability of success equal to 1/p_i, and if the tests are independent, then a(n)/n is the probability that at least one of the k tests succeeds. - Luc Rousseau, Jan 14 2023
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| STATUS
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approved
editing
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#22 by Susanna Cuyler at Wed Nov 10 18:25:59 EST 2021
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#21 by Antti Karttunen at Wed Nov 10 17:16:32 EST 2021
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