A334603 Period of the fraction 1/11^n for n >= 1.

2, 22, 242, 2662, 29282, 322102, 3543122, 38974342, 428717762, 4715895382, 51874849202, 570623341222, 6276856753442, 69045424287862, 759499667166482, 8354496338831302, 91899459727144322, 1010894056998587542, 11119834626984462962, 122318180896829092582

Offset: 1

Author

Bernard Schott, May 07 2020

Keywords

• nonn
• base

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).

Examples

			1/121 = 0. 0082644628099173553719 0082644628099173553719 ... with periodic part {0082644628099173553719}, whose length is 22 digits, so a(2) = 22.
		

References

• J.-M. De Koninck & A. Mercier, 1001 Problèmes en Théorie Classique des Nombres, Problème 346 pp. 50, 204-205, Ellipses, Paris 2004.

Crossrefs

Cf. period of fractions: A051626 (1/n), A133494 (1/3^n), A055272 (1/7^n).

Cf. A001020 (11^n).

Programs

Mathematica

MultiplicativeOrder[10, 11^#] & /@ Range[20] (* Giovanni Resta, May 07 2020 *)

PARI

a(n) = znorder(Mod(10, 11^n)); \\ Michel Marcus, May 09 2020

Formula

a(n) = 2 * 11^(n-1) [conjectured, see comments].

a(n) = A051626(A001020(n)).

Extensions

More terms from Giovanni Resta, May 07 2020