p Statistics: Difference of Sums |

by JVSchmidt |

General | |

Compared with the SUM OF DIGITS we go one step further. Taking two consecutive substrings of length L we can calculate the difference of these sums to proof if there would be any correlation between these two values that is a dependence of the 2*L digits. Here is an example for p with L=5. First sequence X=14159 -> SUM (X)=20 First sequence Y=26535 -> SUM (Y)=21 Difference of sums = SUM(X)-SUM(Y) = 20-21 = -1 We have a simple recursive law for the expected distribution. Let w(L,d) be the probability that two chains of length L have a difference of sums d. Then w(L+1,d) = w(L,d) / 10 + sum w(L,d-i) x (10-i)/10 + sum w(L,d+i) x (10-i)/10 where sums taken for i=1 to 9. Starting condition is: w(1,d) = (10-abs(d))/100 for d= -9,-8,..+8,+9. For any L the difference of sums of digits is located between -9*L (X=000..0,Y=999..9) and S=9*L (X=999..9, Y=000..0). Again, when going to longer and longer chains the min and max difference became extremely improbable because the likelihood for a single digit long run falls like 10 ^{-L}. | |

Result's Overview | |

Digits analyzed: 4.2 * 10 ^{9}Analysis started at digit: 1 Ellapsed computer time for one class: 3 min 40 sec |

Length of chains L | Number of examined pairs of chains K = N/(2*L) | Chi^{2}
| Number of statistical relevant classes |

2 | 1.050.000.000 | 40,5520 | 37 |

5 | 420.000.000 | 86,4518 | 85 |

10 | 210.000.000 | 134,8602 | 127 |

20 | 105.000.000 | 181,1919 | 179 |

40 | 52.500.0000 | 246,2498 | 247 |

70 | 30.000.0000 | 291,6570 | 319 |

80 | 26.250.000 | 319,3401 | 339 |

Detailed results for this test you will find here: Details for Sum's Differences (EXCEL file)

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