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

Chi2-values for the distributions of differences of sums of digits for different length of chains L

 Length of chains L Number of examined pairs ofchains K = N/(2*L) Chi2 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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