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