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Number of trials T of an event that occurs with probability 1-1/n that must be repeated consecutively so that the probability that the event occurs on all T trials (i.e., (1-1/n)^T) is minimally higher than 1/n.
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%I #23 Jul 19 2017 21:33:06

%S 1,3,5,7,10,13,16,19,22,25,29,32,36,39,43,47,51,54,58,62,66,71,75,79,

%T 83,87,92,96,100,105,109,114,118,123,127,132,136,141,146,150,155,160,

%U 165,169,174,179,184,189,194,199,203,208,213,218,223,228,233

%N Number of trials T of an event that occurs with probability 1-1/n that must be repeated consecutively so that the probability that the event occurs on all T trials (i.e., (1-1/n)^T) is minimally higher than 1/n.

%H Charles R Greathouse IV, <a href="/A276498/b276498.txt">Table of n, a(n) for n = 2..10000</a>

%F a(n) = round(log(1/n)/log(1-1/n)).

%F a(n) = n log n - (log n)/2 + O(1). - _Charles R Greathouse IV_, Sep 06 2016

%e a(20) = round(log(1/20)/log(1-1/20)) = round(log(0.05)/log(0.95)) = round(58.4...) = 58.

%e If a phenomenon P occurs with a 95% probability on any given trial, how many trials T must be performed so that the probability that P occurs on all T trials reduces to 5%? 0.95^T = 0.05 => T = log(0.05)/log(0.95) = 58.4... = 58 (rounded).

%t Table[Round[Log[1/n]/Log[1 - 1/n]], {n, 2, 58}] (* _Michael De Vlieger_, Sep 06 2016 *)

%o (PARI) a(n) = round (log(1/n)/log(1-1/n)); \\ _Michel Marcus_, Sep 05 2016

%o (PARI) a(n)=-log(n)\/log1p(-1/n) \\ _Charles R Greathouse IV_, Sep 05 2016

%K nonn

%O 2,2

%A _Talha Ali_, Sep 05 2016

%E Name and Example edited by _Jon E. Schoenfield_, Jul 16 2017