A318571 - OEIS

%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