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For example, in duodecimal, 1 / 2 = 0.6, 1 / 3 = 0.4, 1 / 4 = 0.3 and 1 / 6 = 0.2 all terminate; 1 / 5 = 0. 2497 repeats with period length 4, in contrast with the equivalent decimal expansion of 0.2; 1 / 7 = 0. 186A35 has period 6 in duodecimal, just as it does in decimal. If b is an integer base ...
For any integer coprime to 10, its reciprocal is a repeating decimal without any non-recurring digits. E.g. 1 ⁄ 143 = 0. 006993 006993 006993.... While the expression of a single series with vinculum on top is adequate, the intention of the above expression is to show that the six cyclic permutations of 006993 can be obtained from this repeating decimal if we select six consecutive digits ...
Also the converse is true: The decimal expansion of a rational number is either finite, or endlessly repeating. Finite decimal representations can also be seen as a special case of infinite repeating decimal representations. For example, 36 ⁄ 25 = 1.44 = 1.4400000...; the endlessly repeated sequence is the one-digit sequence "0".
Stylistic impression of the repeating decimal 0.9999..., representing the digit 9 repeating infinitely. In mathematics, 0.999... (also written as 0. 9, 0.., or 0.(9)) is a repeating decimal that is an alternative way of writing the number 1.
1 / 14 = 0.0 714285... 1 / 28 = 0.03 571428... 1 / 35 = 0.0 285714... 1 / 56 = 0.017 857142... 1 / 70 = 0.0 142857... The above decimals follow the 142857 rotational sequence. There are fractions in which the denominator has a factor of 7, such as 1 / 21 and 1 / 42 , that do not follow this ...
A vinculum can indicate a line segment where A and B are the endpoints: ¯. A vinculum can indicate the repetend of a repeating decimal value: . 1 ⁄ 7 = 0. 142857 = 0.1428571428571428571...
[0; 4, 4, 8, 16, 18, 5, 1, 1, 1, 1, 7, 1, 1, 6, 2, 9, 58, 1, 3, 4, …] [OEIS 100] Computed up to 1 011 597 392 terms by E. Weisstein. He also noted that while the Champernowne constant continued fraction contains sporadic large terms, the continued fraction of the Copeland–Erdős Constant do not exhibit this property. [Mw 85]
Let t = 0. Let r = 1. Let n = 0. loop: Let t = t + 1 Let x = r ⋅ b Let d = int(x / p) Let r = x mod p Let n = n ⋅ b + d If r ≠ 1 then repeat the loop. if t = p − 1 then n is a cyclic number. This procedure works by computing the digits of 1/p in base b, by long division. r is the remainder at each step, and d is the digit produced. The ...