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%I #23 Sep 08 2022 08:46:22
%S 70,36,24,18,15,12,11,10,9,8,7,7,6,6,5,5,5,5,4,4,4,4,4,4,4,3,3,3,3,3,
%T 3,3,3,3,3,3,3,3,3,3,3,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,
%U 2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,2,1
%N a(n) = ceiling(log(2)/(log(1+n/100))).
%C a(n) is the least positive integer k such that (1 + n/100)^k >= 2.
%H Wikipedia, <a href="https://en.wikipedia.org/wiki/Rule_of_72">Rule of 72</a>
%F a(n) = 1 for n >= 100.
%e n | a(n) | Rule of 72 (= 72/n)
%e ------+------+---------------------
%e 1 | 70 | 72
%e 2 | 36 | 36
%e 3 | 24 | 24
%e 4 | 18 | 18
%e 5 | 15 | 14.4
%e 6 | 12 | 12
%e 7 | 11 | 10.285714...
%e 8 | 10 | 9
%e 9 | 9 | 8
%e 10 | 8 | 7.2
%e 11 | 7 | 6.545454...
%e 12 | 7 | 6
%p seq(ceil(log(2)/(log(1+n/100))),n=1..100); # _Muniru A Asiru_, Aug 30 2018
%t Table[Ceiling[Log[2]/Log[1 + n/100]], {n, 100}] (* _Vincenzo Librandi_, Sep 01 2018 *)
%o (PARI) {a(n) = ceil(log(2)/(log(1+n/100)))}
%o (Magma) [Ceiling(Log(2)/(Log(1+n/100))): n in [1..100]]; // _Vincenzo Librandi_, Sep 01 2018
%K nonn
%O 1,1
%A _Seiichi Manyama_, Aug 29 2018
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