In different bases:

Binary (base 2) | \(100010111000001001_2\) |

Ternary (base 3) | \(21020222000_3\) |

Quaternary (base 4) | \(202320021_4\) |

Quinary (base 5) | \(14032412_5\) |

Senary (base 6) | \(3021213_6\) |

Octal (base 8) | \(427011_8\) |

Decimal (base 10) | \(142857_{10}\) |

Duodecimal (base 12) | \(\rm{6A809_{12}}\) |

Hexadecimal (base 16) | \(\rm{22E09_{16}}\) |

Vigesimal (base 20) | \(\rm{HH2H_{20}}\) |

Base 36 | \(\rm{3289_{36}}\) |

Other miscellaneous information:

Prime factorization | \(3^3 × 11 × 13 × 37\) |

Divisors | \(1, 3, 9, 11, 13, 27, 33, 37, 39, 99, 111, 117, 143, 297, 333, 351, 407,\\ 429, 481, 999, 1221, 1287, 1443, 3663, 3861, 4329, 5291, 10989, 12987,\\ 15873, 47619, 142857\) |

Roman numeral | \(\rm{\overline{CXL}MMDCCCLVII}\) (vinculum system) |

If it is multiplied by 1, 2, 3, 4, 5, or 6, the answer will be a cyclic permutation of itself.

\(142857 × 1 = 142857\)

\(142857 × 2 = 285714\)

\(142857 × 3 = 428571\)

\(142857 × 4 = 571428\)

\(142857 × 5 = 714285\)

\(142857 × 6 = 857142\)

\(142857 × 7 = 999999 \space\longleftarrow\) Doesn’t work.

Those numbers are the six repeating digits of \(\frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}\), and \(\frac{6}{7}\).

\(1/7 = 0.\overline{142857}\)

\(2/7 = 0.\overline{285714}\)

\(3/7 = 0.\overline{428571}\)

\(4/7 = 0.\overline{571428}\)

\(5/7 = 0.\overline{714285}\)

\(6/7 = 0.\overline{857142}\)

\(7/7 = 0.\overline{999999} = 1\)

\(8/7 = 1.\overline{142857} \space\longleftarrow\) The cycle starts repeating.

\(9/7 = 1.\overline{285714}\)

\(…\)

If you square the last three digits and subtract the square of the first three digits, you also get back a cyclic permutation of the number.

\(857^2 = 734449\)

\(142^2 = 20164\)

\(857^2 − 142^2 = 714285\)

\(734449 − 20164 = 714285\)