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In this presentation by Mathologer you can find the following slide:

enter image description here

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.

enter image description here

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.

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  • $\begingroup$ 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. $\endgroup$
    – Dietrich Burde
    May 29 at 16:50
  • $\begingroup$ Be cautious with the fifth column... $\endgroup$
    – Jean Marie
    May 29 at 17:27

1 Answer 1

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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 $\frac{m}{r}$.

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  • $\begingroup$ Are these Legendre symbols? $\endgroup$
    – JAP
    May 29 at 16:52
  • $\begingroup$ @JAP These are fractions. The parentheses are just a style choice. $\endgroup$
    – Erick Wong
    May 29 at 16:55
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    $\begingroup$ 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. $\endgroup$
    – JAP
    May 29 at 16:59
  • $\begingroup$ I wonder why you change from $n$ to $x$ since we are considering integers. $\endgroup$
    – JAP
    May 29 at 21:22
  • $\begingroup$ @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. $\endgroup$
    – Erick Wong
    May 31 at 0:03

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