|
%I #31 Nov 19 2023 21:17:31
%S 2,22,242,2662,29282,322102,3543122,38974342,428717762,4715895382,
%T 51874849202,570623341222,6276856753442,69045424287862,
%U 759499667166482,8354496338831302,91899459727144322,1010894056998587542,11119834626984462962,122318180896829092582
%N Period of the fraction 1/11^n for n >= 1.
%C Conjecture proposed by the authors in References page 205: if p is a prime with gcd(p,30) = 1 and if the period of 1/p is m then the period of 1/p^n is m*p^(n-1).
%D J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 346 pp. 50, 204-205, Ellipses, Paris 2004.
%F a(n) = 2 * 11^(n-1) [conjectured, see comments].
%F a(n) = A051626(A001020(n)).
%e 1/121 = 0. 0082644628099173553719 0082644628099173553719 ... with periodic part {0082644628099173553719}, whose length is 22 digits, so a(2) = 22.
%t MultiplicativeOrder[10, 11^#] & /@ Range[20] (* _Giovanni Resta_, May 07 2020 *)
%o (PARI) a(n) = znorder(Mod(10, 11^n)); \\ _Michel Marcus_, May 09 2020
%Y Cf. period of fractions: A051626 (1/n), A133494 (1/3^n), A055272 (1/7^n).
%Y Cf. A001020 (11^n).
%K nonn,base
%O 1,1
%A _Bernard Schott_, May 07 2020
%E More terms from _Giovanni Resta_, May 07 2020
|
