# The Josephus Problem: Filling the Gaps for Odds

As was promised earlier, in this post we're going to review The Josephus Problem
for an **odd number of people**. This time we assume the last person is assigned
the number `2n + 1`

(since this expression can represent any odd integer, for an
arbitrary integer `n`

).

**Three People**

**Five People**

**Seven People**

**Generalized**

Once again, here's a table that summarizes how "re-numbering" can be performed
and verifies that the formula `2n + 1`

calculates initial person's number, when
`n`

is the person's number after the 1st round.

Initial Person's Number | After the 1st Round | Verify 2n + 1 |
---|---|---|

1 | Dead | - |

2 | Dead | - |

3 | 1 | 2 x 1 + 1 = 3 |

4 | Dead | - |

5 | 2 | 2 x 2 + 1 = 5 |

6 | Dead | - |

7 | 3 | 2 x 3 + 1 = 7 |

… | … | … |

2n - 3 | n - 2 | 2(n - 2) + 1 = 2n - 3 |

2n - 2 | Dead | - |

2n - 1 | n - 1 | 2(n - 1) + 1 = 2n - 1 |

2n | Dead | - |

2n + 1 | n | 2(n + 1) + 1 = 2n |

These considerations help us to establish, that in a group of the odd number of
people `2n + 1`

, the survived person number is the same as the survived person
number in a group of `n`

people, but doubled and increased by `1`

.

The next post is devoted to the proof of the *Josephus problem* closed form.