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A068542
Period of the fraction 1/3^n.
3
3, 1, 370, 123456790, 411522633744855967078189300, 137174211248285322359396433470507544581618655692729766803840877914951989026063100
OFFSET
1,1
COMMENTS
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
LINKS
FORMULA
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
EXAMPLE
1/3^3 = 0.0370370370..., hence a(3) = 370.
MAPLE
f:= proc(n) local k, v;
k:= numtheory:-order(10, 3^n);
v:= (10^k-1)/3^n;
v * 10^(k-ilog10(v)-1)
end proc:
map(f, [$1..8]); # Robert Israel, Jul 23 2025
PROG
(PARI) apply( {A068542(n)=10^(3^max(n-2, 0)+logint(3^n, 10))\3^n}, [1..6]) \\ M. F. Hasler, Apr 23 2021
CROSSREFS
Sequence in context: A266363 A394831 A382849 * A382793 A036112 A266230
KEYWORD
nonn
AUTHOR
Benoit Cloitre, Mar 22 2002
STATUS
approved