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Number of the fraction which has (the digits of) n as repeating period in its decimal expansion, according to the canonical enumeration A038566/A038567.
1

%I #6 Oct 22 2021 23:44:04

%S 23,24,3,25,26,4,27,28,1,2951,23,327,2952,2953,328,2954,2955,34,2956,

%T 2957,329,24,2958,330,2959,2960,35,2961,2962,331,2963,2964,3,2965,

%U 2966,36,2967,2968,332,2969,2970,333,2971,25,37,2972,2973,334,2974,2975,335,2976,2977,38,26

%N Number of the fraction which has (the digits of) n as repeating period in its decimal expansion, according to the canonical enumeration A038566/A038567.

%C The fraction f = n/(10^L-1), where L is the number of decimal digits of n, has the infinite decimal expansion 0.{n}{n}{n}... (with special cases n = 9, 99 etc., where f = 0.999... = 1). This sequence lists the index of this fraction, for given n, corresponding to the "canonical enumeration of positive fractions <= 1", i.e., m such that f = A038566(m) / A038567(m-1).

%C Inspired by Angelini's blog post, which however takes a different approach to encoding fractions (in a more "decimal" way) and to deal with n's made of digits 9 or with repetitions, such as 11, 111, or 1010, etc.

%H Eric Angelini, <a href="http://cinquantesignes.blogspot.com/2021/10/cantorisation.html">Cantorisation</a>, personal blog CinquanteSignes.blogspot.com, Oct 17 2021

%F a(n) = m such that A038566(m)/A038567(m-1) = n/A002283(A055642(n)), where A002283(1, 2, 3, ...) = (9, 99, 999, ...) and A055642(n) = number of digits of n.

%e a(n = 1) = 23 because A038566(23)/A038567(22) = 1/9, the unique fraction (in lowest terms) whose decimal expansion is 0.111..., i.e., period = (1), repeated.

%e a(n = 2) = 24 because A038566(24)/A038567(23) = 2/9, the unique fraction (in lowest terms) whose decimal expansion is 0.222..., i.e., period = (2), repeated.

%e a(n = 3) = 3 because A038566(3)/A038567(2) = 1/3, the unique fraction (in lowest terms) whose decimal expansion is 0.333..., i.e., period = (3), repeated.

%e a(n = 9) = 1 because A038566(1)/A038567(0) = 1/1, the unique fraction (in lowest terms) equal to 0.999..., i.e., period = (9), repeated.

%e a(n = 10) = 2951 because A038566(2951)/A038567(2950) = 10/99, the unique fraction (in lowest terms) whose decimal expansion is 0.1010..., i.e., period = (10), repeated.

%e a(11) = a(1) because the unique fraction that has decimal expansion 0.1111..., i.e., period (11) repeated, is 1/9, the same as for 0.111..., i.e., period (1), repeated.

%o (PARI) apply( {A343634(n, d=10^(logint(n,10)+1)-1, g=gcd(n,d)) = A002088(d\g-1) - sum(k=1, n\=g, gcd(k, d) > 1) + n}, [1..55])

%Y Cf. A038566/A038567, A002283, A055642, A002088.

%K nonn,base

%O 1,1

%A _M. F. Hasler_, Oct 18 2021