# What is the simple dependence of the diagonals (or columns) of the Faulhaber matrix on the first entry (Bernoulli numbers)?

In this presentation by Mathologer you can find the following slide: relating the sum of the $$n$$ first $$k$$ powers of integers, i.e. $$S_k$$ and the Faulhaber matrix diagonals (which correspond to the columns highlighted on the image above. Here is the corresponding Faulhaber matrix:

S 0      1     0     0     0     0     0    0
S 1    1/2   1/2     0     0     0     0    0
S 2    1/6   1/2   1/3     0     0     0    0
S 3      0   1/4   1/2   1/4     0     0    0
S 4  -1/30     0   1/3   1/2   1/5     0    0
S 5      0 -1/12     0  5/12   1/2   1/6    0
S 6   1/42     0  -1/6     0   1/2   1/2   1/7


The patterns along the first four columns are well explained: The harmonic sequence, $$1/2$$'s, $$k/12$$, $$0$$'s, and then the following statement:

Jacob Bernoulli discovered something amazing: The numbers in the different columns always appear to depend in a simple way on numbers at the top of these columns. Unfortunately he doesn't share the simple dependence of the fifth (or higher) column: $$-1/30, -1/12, -1/6, - 7/24, -7/15, -7/10,\dots.$$

or formulate what type of general dependence this is.

• The fifth column starts with the Bernoulli number $B_4=-\frac{1}{30}$. This generalises to higher columns. In general the columns are given by the formula, see here. May 29 at 16:50
• Be cautious with the fifth column... May 29 at 17:27

Let $$a_{k,m}$$ denote the coefficient of the $$k$$th column of $$S_m$$ (corresponding to the monomial $$x^{m-k+2}$$).

The pattern for the first column ($$k=1$$) is $$a_{1,m} = \left(\frac{m}{m+1}\right) a_{1,m-1}.\quad (m\ge 1)$$

The pattern for the second column ($$k=2$$) is $$a_{2,m} = \left(\frac{m}{m}\right) a_{2,m-1}.\quad (m\ge 2)$$

The pattern for the third column ($$k=3$$) is $$a_{3,m} = \left(\frac{m}{m-1}\right) a_{3,m-1}.\quad (m\ge 3)$$

You can see that the pattern for any $$k$$ is $$a_{k,m} = \left(\frac{m}{m-k+2}\right) a_{k,m-1},\quad (m\ge k)$$

where the denominator is easily remembered as the power of $$x$$ attached to the coefficient $$a_{k,m}$$. In other words, to get the coefficient of $$x^r$$ in $$S_m$$, take the coefficient of $$x^{r-1}$$ in $$S_{m-1}$$ and multiply it by $$\frac{m}{r}$$.

• Are these Legendre symbols?
– JAP
May 29 at 16:52
• @JAP These are fractions. The parentheses are just a style choice. May 29 at 16:55
• Makes sense. Sorry for the silly question, but I couldn't actually read the answer, and wanted to catch you still editing to avoid having to wonder later.
– JAP
May 29 at 16:59
• I wonder why you change from $n$ to $x$ since we are considering integers.
– JAP
May 29 at 21:22
• @JAP Good question. I wasn't even consciously aware of changing it, but I suppose it's because $x$ is very well-accepted as a polynomial variable, and because there are already so many integer-valued parameters floating around that keeping it as $n$ felt more cluttered. May 31 at 0:03