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A334603
Period of the fraction 1/11^n for n >= 1.
0
2, 22, 242, 2662, 29282, 322102, 3543122, 38974342, 428717762, 4715895382, 51874849202, 570623341222, 6276856753442, 69045424287862, 759499667166482, 8354496338831302, 91899459727144322, 1010894056998587542, 11119834626984462962, 122318180896829092582
OFFSET
1,1
COMMENTS
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).
REFERENCES
J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 346 pp. 50, 204-205, Ellipses, Paris 2004.
FORMULA
a(n) = 2 * 11^(n-1) [conjectured, see comments].
a(n) = A051626(A001020(n)).
EXAMPLE
1/121 = 0. 0082644628099173553719 0082644628099173553719 ... with periodic part {0082644628099173553719}, whose length is 22 digits, so a(2) = 22.
MATHEMATICA
MultiplicativeOrder[10, 11^#] & /@ Range[20] (* Giovanni Resta, May 07 2020 *)
PROG
(PARI) a(n) = znorder(Mod(10, 11^n)); \\ Michel Marcus, May 09 2020
CROSSREFS
Cf. period of fractions: A051626 (1/n), A133494 (1/3^n), A055272 (1/7^n).
Cf. A001020 (11^n).
Sequence in context: A138140 A322283 A151617 * A342232 A082777 A229465
KEYWORD
nonn,base
AUTHOR
Bernard Schott, May 07 2020
EXTENSIONS
More terms from Giovanni Resta, May 07 2020
STATUS
approved