http://www.siam.org/students/siuro/vol1issue1/S01009.pdf
bendford'law
e=log10(1+l/n)
o=freq of first digit / total
x2= N*sum(power((o-e),2)/e)
1-9的乘法表中的数字算出来的chisquare test is 4.881
其第一个数字出现频率和benford标准值放大一百倍为：
| Digit | Benford Probability | Observed Probability |
| 1 | 30.10 | 22.22 |
| 2 | 17.61 | 18.52 |
| 3 | 12.49 | 13.58 |
| 4 | 9.69 | 14.81 |
| 5 | 7.92 | 7.41 |
| 6 | 6.69 | 8.64 |
| 7 | 5.80 | 4.94 |
| 8 | 5.12 | 6.17 |
| 9 | 4.58 | 3.70 |
(表格中因为都放大了100倍，所以下面的公式中需要除以100.)
x2= N/100*sum(power((o-e),2)/e)