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 A334603 Period length 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 (list; graph; refs; listen; history; text; internal format)
 OFFSET 1,1 COMMENTS Conjecture proposed by the authors in References page 205: if p is prime with gcd(p,30) = 1, and if period length of 1/p is m then period length 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. LINKS FORMULA a(n) = 2 * 11^(n-1) [conjectured, see comments]. a(n) = A051626(A001020(n)). EXAMPLE 1/121 = .0082644628099173553719 0082644628099173553719 ...  with a period length {0082644628099173553719} of 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 length 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 Adjacent sequences:  A334600 A334601 A334602 * A334604 A334605 A334606 KEYWORD nonn,base AUTHOR Bernard Schott, May 07 2020 EXTENSIONS More terms from Giovanni Resta, May 07 2020 STATUS approved

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Last modified July 27 12:04 EDT 2021. Contains 346305 sequences. (Running on oeis4.)