p Statistics: Variety of Digits |

by JVSchmidt |

General | |

When testing longer chains we gonna start with a simple question: How many DIFFERENT DIGITS are contained in a chain of length L? Let us call this number the Variety of digits VOD. Here is an example for p with L=5. 14159 -> VOD = 4 26535 -> VOD = 4 89793 -> VOD = 4 23846 -> VOD = 5 26433 -> VOD = 4 83279 -> VOD = 5 50288 -> VOD = 4 41971 -> VOD = 4 69399 -> VOD = 3 37510 -> VOD = 5 The construction law for the VOD-distribution is easy to find. Let w(L,d) be the probability that a chain of length L consists of exactly d different digits. For L=1 we have even one digit: w(1,d)=1 for d=1 w(1,d)=0 for d>1 For L>1 any w(L,d) can be calculated from its predecessor: w(L+1,d) = w(L,d) x d/10 + w(L,d-1) x (10-d)/10 Finding a VOD=1 for L=5 is just the same as searching for a single digit run of length 5 (e.g. 11111). But analyzing single digit runs is limited to L~7 due to the available data. Thus testing variety of digits up to L=40 is a chance to get a better feeling for the regularity of single digit distribution. More than this the VOD can be an indicator for clusters. | |

Result's Overview | |

Digits analyzed: 4.2 * 10 ^{9}Analysis started at digit: 1 Ellapsed computer time for each L: 5 min |

Length of chains L | Chains analyzed | Chi^{2}for the VOD | Number of statistical relevant classes |

10 | 420.000.000 | 6,1977 | 9 |

11 | 381.818.181 | 6,8173 | 9 |

12 | 350.000.000 | 8,2189 | 9 |

15 | 280.000.000 | 14,1435 | 8 |

20 | 210.0000 | 9,3921 | 7 |

40 | 105.0000 | 2,7799 | 5 |

Remarkable detail:

Chain 02220 22202 20202 (L=15,VOD=2) found at position 2.794.419.571

Chain 09905 55595 05009 59509 950 (L=23,VOD=3) found at position 273.876.641

Detailed results for this test you will find here: Details for Variety of Digits (EXCEL file)

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