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%I #18 Apr 05 2021 20:37:31
%S 2,3,2,8,2,37,2,76,2,217,2,870,2,583,2,5034,2,28494,2,10058,2,187966,
%T 2,383291,2,340992,2
%N Number of different terms in a period of continued fraction for square root of n-th repunit.
%t Do[Print[Length[Union[Last[ContinuedFraction[((-1+10^n)/9)^(1/2)]]]]], {n, 2, 18}]
%o (Python)
%o from sympy.ntheory.continued_fraction import continued_fraction
%o from sympy import sqrt
%o def A096488(n): return len(set(continued_fraction(sqrt((10**n-1)//9))[-1])) # _Chai Wah Wu_, Mar 30 2021
%Y Cf. A002275, A096485, A096487.
%K nonn,more
%O 2,1
%A _Labos Elemer_, Jun 25 2004
%E Name edited by _Michel Marcus_, Aug 22 2019
%E a(19)-a(22) from _Daniel Suteu_, Aug 22 2019
%E a(23)-a(28) from _Chai Wah Wu_, Apr 05 2021
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