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%I #8 Apr 23 2021 14:47:36
%S 3,1,370,123456790,411522633744855967078189300,
%T 137174211248285322359396433470507544581618655692729766803840877914951989026063100
%N Period of the fraction 1/3^n.
%C The length of the period is the number of digits of a(n): 1, 1, 3, 9, 27, 81, ... The terms a(n) are more precisely the integers made from the digits of a period, starting with the first nonzero digit. - _M. F. Hasler_, Apr 23 2021
%F a(n) = floor(10^(3^max(n-2,0)+L(3^n))/3^n) where L(m) = floor(log10(m)). - _M. F. Hasler_, Apr 23 2021
%e 1/3^3 = 0.0370370370..., hence a(3) = 370.
%o (PARI) apply( {A068542(n)=10^(3^max(n-2,0)+logint(3^n,10))\3^n}, [1..6]) \\ _M. F. Hasler_, Apr 23 2021
%K nonn
%O 1,1
%A _Benoit Cloitre_, Mar 22 2002
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