Last Update 2011/12/11
JVSchmidt 2011

Analyzing Decimal Digits of π


Sum of Digits

The idea of this test is to divide data into substrings of length L, building the sum of digits of each substring and calculating the Chi2-value for the sum's distribution.
Here is an example for the Pi digits with L=5.

First sequence 14159 -> SUM = 20
Second sequence 26535 -> SUM = 21
Third sequence 89793 -> SUM = 36

For single digits (L=1) the sum is equal to the digits value and thus we have a probability of 1/10 for any sum of digits from 0 to 9. Let w(L,S) be the probability that a chain of length L has a sum of digits equal to S, then w(1,y)=1/10 with y=0,1,..,9 and w(1,y)=0 when y>9..

Any further distribution can be calculated recursively:
w(L,y) = w(L-1,y) + 1/10 * sum (for all i=0..y) w(L-1,y-i)

For any L the sum of digits is located between S=0 (all digits=0) and S=9*L (all digits=9). When going to longer and longer chains the min and max sums became extremely improbable because the likelihood for a single digit long run falls like 10-L.