# Some more binomial coefficient determinants

The setup is similar to this question, but generalizes the size of the Hankel matrix. We'll define

$$d(n,k,r):=\det\left(\binom{2i+2j+k+r}{i+j}\right)_{i,j=0}^{kn-1}.$$ Edit: Thanks to Johann Cigler for checking, which made me discover that the originally given formula $d(n,k,r):=\det\left(\binom{2i+2j+k}{i+j}\right)_{i,j=0}^{kn-1+r}$ was not the one I used in my implementation. Sorry for that! So in fact my generalization only concerns matrix sizes that are multiples of $k$. Now I am not sure if the picture becomes more coherent when including other matrix sizes as well, but it may be wotrh giving it a look.

The original question is the special case when $k$ is odd and $r=0$ and conjectures $d(n,k,0)=(2n+1)^\frac{k-1}{2}$. (For even $k$, we have $d(n,k,0)=(-1)^{ n\frac k2}$ according to Krattenthaler.)
Computer experiments suggest that in the more general case, there is still a certain proportion of those determinants which are, up to sign, exact $\bigl[\frac{k+r-1}2\bigr]^\text{th}$ powers (including $\pm1$), while the others are not only no powers but usually have quite large prime factors (e.g. see the two grey entries in the $k=4$ column just below). In the following I will ignore those other values and only consider those which are powers. It may not be a surprise that they all come in intriguing patterns.

E.g. for $r=1$, the table of $d(n,k,1)$ starts with $$\begin{matrix} n \backslash k &2&3&4&5&6&7&8&9&10&11 \\ \hline \ 0 \quad| &1 &1 &1 &1 &1 &1 &1 &1 &1 &1\\ \ 1 \quad|&-4 & 8 & \color{grey} {\scriptstyle {(-61)}} & & & & & \\ \ 2 \quad| &3& -16& -8^2 & -24^2 & & & & \\ \ 3 \quad|&\color{red}5 &1 & \color{grey}{\scriptstyle {(5\cdot139)}} & -36^2 & 12^3 & -48^3 & & \\ \ 4 \quad|&\color{red}{-12} & 1 & 7^2 & & & 64^3 & 16^4 & 80^4 \\ \ 5 \quad|&\color{red}7 &32 & \color{blue}{9^2} & 1 & & & & 100^4& -20^5 & 120^5 \\ \ 6\quad|&9 &-40 & & -1 & 11^3 & & & & & -144^5 \\ \ 7 \quad|&-20 &1 & \color{blue}{-24^2} & & \color{red}{13^3} & 1 & & \\ \ 8 \quad|&11 &1 & & 84^2 & & 1 & 15^4 & \\ \ 9 \quad|&13 & 56 & \color{blue}{15^2} & 96^2 & & & 17^4 & 1 \\ 10\quad|&-28 & -64& 17^2& & \color{red}{36^3} & & & -1 & 19^5 \\ 11\quad|&15 & 1 & & & & -160^3 & & & 21^5 & 1 & \\ 12 \quad|&17 & 1 & -40^2 & & & 176^3 & & & & 1 & \\ 13 \quad|&-36 & 80 & & & \color{red}{23^3} & & 48^4 & \\ \end{matrix}$$

These first entries largely suffice to conjecture the general patterns of $d(n,k,1)$ for even and odd $k$.
For $r=2,3,...$ the patterns are not too different from the ones encountered for $r=1$. (See at the end.)

Instead of $d(n,k,r)$, wherever it is up to sign such a $\bigl[\frac{k+r-1}2\bigr]^\text{th}$ power, let us consider the root $e(n,k,r)$ defined by $$e(n,k,r):=\sqrt[{\bigl[\frac{k+r-1}2 \bigr]}]{|d(n,k,r)|} \in\mathbb N$$

If $r$ and $k$ are fixed, the $e(n+\lambda(k+r),k,r)_{\lambda\in\mathbb N_0}$ always seem to form an arithmetic sequence. For $k+r$ even, its difference is either $2k(r+k)$ or $0$ (in the latter case, we have the constant sequence $1,1,...$), i.e. the first conjecture will be

$$e(n+k+r,k,r)-e(n,k,r)=\begin{cases}2k(k+r)\\ 0\end{cases}.$$

It seems much harder to formalize where the $\pm1$'s occur (and where the others), at least for odd $r\ge3$. OTOH, the pattern seems clear if $r$ is even:

For even $r$, we have $|d(n,k,r)|=1$ iff $k$ is also even and $n=\lambda\frac{k+r}{\gcd(k,r)}$, where $\lambda \in\mathbb N_0$. In this case, more precisely, $$d(n,k,r)= (-1)^{ n\frac k2(\frac r2-1)}.$$

Further, if $k$ and $r$ are coprime and moreover $(k+r)$ is odd, the $k^\text{th}$ column consists of equidistant triples (three of them are colored in the above table) such that the middle value of the root is the sum of the two others, e.g. from $d(10,4,1)$ and $d(12,4,1)$ we can guess $d(14,4,1)=23^2$. BTW, the middle value is always $\equiv0\pmod4$. So for these columns, the roots come in exactly $3$ arithmetic sequences.

There must be a deeper reason for this "sum behavior". Any idea what is happening here?

The density (i.e. the proportion of $\bigl[\frac{k+r-1}2\bigr]^\text{th}$ powers among all values) of a given column is for coprime $k,r$ $$\begin{cases}\frac3{k+r} & \text{if}\ k\not\equiv r\pmod 2\\ \frac4{k+r} & \text{if} \ k\equiv r\pmod 2\text{ i.e. both odd}\end{cases}$$ and a certain multiple of that if $k,r$ are not coprime.

The first entries of $e(n,k,r)$ for $r=2,...,8$ are displayed below, with a "$-$" sign meaning that the corresponding power $d(n,k,r)$ bears a negative sign. (So in fact, compared with the $d(n,k,1)$ table above I have just omitted the exponents $\bigl[\frac{k+r-1}2\bigr]$ for better lisibility.) Sadly I couldn't get the entries of the columns to flush right, and I did not want to do that manually.

r=2:
n\k 0   1   2   3   4   5   6   7   8   9   10
0   1   1   1   1   1   1   1   1   1   1   1
1   1   1   -8  -4  12
2   1   -4  1   3           32  8   40
3   1   3   -16     1   5                   -72
4   1   3   1       36      1   7
5   1   -8  -24 7       16          1   9
6   1   5   1   -16 1       80              1
7   1   5   -32 9   60  11          120
8   1   -12 1               1           -28
9   1   7   -40     1           15          -192
10  1   7   1   13  84  15  128     1
11  1   -16 -48 -28             36      19
12  1   9   1   15  1   36  1       200     1
13  1   9   -56     108         21

r=3:
n\k 0   1   2   3   4   5   6   7   8   9
0   1   1   1   1   1   1   1   1   1   1
1   1   1       -12 4   -16
2   1   -8      -1  3                   -48
3   1   8   3   24          5   1
4   1   1   -8  1       48              1
5   1   1   5   -36     1       -80
6   1   -16     -1          9           -120
7   1   16      48  9               11
8   1   1   7   1   20  1       120     1
9   1   1   -16 -60 11  -96 13      -28
10  1   -24 9   -1              -1      -192
11  1   24      72                  17
12  1   1       1       128 17          1
13  1   1   11  -84     1       -1
14  1   -32 -24 -1  17                  -264
15  1   32  13  96  36      21  220

r=4:
n\k 0   1   2   3   4   5   6   7   8   9
0   1   1   1   1   1   1   1   1   1   1
1   1   1           16  4   -20
2   1       12      1   3
3   1   -4  -1      32              1   5
4   1               1           -12
5   1   3   -24 5   48      -1
6   1   3   1   12  1       80      1
7   1           7   64
8   1   -8  36      1            11     -24
9   1       -1      80  11          1
10  1   5           1   24  1
11  1   5   -48     96  13  -140 15
12  1       1   11  1               1
13  1   -12     24  112                 19

r=5:
n\k 0   1   2   3   4   5   6   7   8   9
0   1   1   1   1   1   1   1   1   1   1
1   1   1               20  -4  24
2   1       4           -1  3
3   1   -12     1       -40
4   1   12  3   32      1               84
5   1                   60          7
6   1   -1              -1      -96
7   1   -1  5   -48 7   -80     1       -140
8   1           1   16  1
9   1   24  12      9   100         -24
10  1   -24             -1
11  1       7   1       -120 13         -1
12  1   1       80      1   -28 1
13  1   1               140  15 192 17

r=6:
n\k 0   1   2   3   4   5   6   7   8   9   10
0   1   1   1   1   1   1   1   1   1   1   1
1   1   1                   -24 -4  28
2   1       16              1   3
3   1           3           -48
4   1   4   1       40      1               96
5   1               1       -72         7
6   1       32  5           1
7   1   3                   -96     1
8   1   3   1               1       140     1
9   1           7   80  9   -120
10  1       48      1   -20 1           13
11  1   8               11  -144
12  1       1   9           1               256
13  1                       -168 15

r=7:
n\k 0   1   2   3   4   5   6   7   8   9   10  11
0   1   1   1   1   1   1   1   1   1   1   1   1
1   1   1                       -28 4   -32
2   1           -20             -1  3
3   1               3           56
4   1   16                      1
5   1   -16 3   -40     1       -84             -1
6   1                   -72     -1          16
7   1       8   1   -12         112
8   1   1                       1       160
9   1   1   5                   -140    1       -216
10  1           -1              -1
11  1               9   120 11  168
12  1   32      80      1   -24 1           15
13  1   -32                 13  -196

r=8:
n\k 0   1   2   3   4   5   6   7   8   9   10  11  12
0   1   1   1   1   1   1   1   1   1   1   1   1   1
1   1   1                           32  4   -36
2   1           -4                  1   3
3   1       -20     1               64
4   1           3       -8          1
5   1   4   -1                      96              1
6   1               1       84      1
7   1                       -1      128         9
8   1       40          7           1
9   1   3           1               160     -1
10  1   3   1                       1       216     1
11  1           7                   192
12  1               1               1
13  1       -60 -16     11  -168 13 224         -32

• Perhaps I have misunderstood something, but I do not understand your table of $d(n,k,1)$. For example my computations with Mathematica give $d(n,k,0)=d(n,k,1)$. – Johann Cigler Apr 23 '18 at 11:38
• @JohannCigler That is funny. What do you get for $d(n,k,2)$? Actually I stumbled on this question when I wanted to check the original question with GP-Pari. As Pari does not accept zero indices, I had to shift them and did a mistake in this, which made me come up with the current generalization. – Wolfgang Apr 23 '18 at 19:43
• Note that $r$ may also be negative, with similar outcomes for the columns where $k\ge 1-r$. – Wolfgang Apr 23 '18 at 20:28
• For $r=2$ I could not find a formula. I have only a conjecture for $s(n,k):=d(n,k,-1)+d(n,k,2)$ for $n>1:$ $$s(n,2k+1)=(2-\binom{2k+2}{2})(2n+1)^k$$ and $$s(n,2k)=(-1)^{kn}(\binom{2k+1}{2}-2)$$. – Johann Cigler Apr 24 '18 at 8:22
• @JohannCigler Oh sorry. My Pari code is $d(n,k,r)=matdet(matrix(N=k*n,N,i,j,binomial(2*i+2*j+k-4+r,i+j-2)))$ and I just realized that I added the r in the "wrong" place. So my generalization does imply changing the size of the matrix, but only certain sizes (in fact, only multiples of k). You have (correctly) assumed what corresponds to $d(n,k,r)=matdet(matrix(N=k*n+r,N,i,j,binomial(2*i+2*j+k-4,i+j-2))).$ I have edited the main formula so my claims should be correct now. – Wolfgang Apr 24 '18 at 14:04

Let $d_r(N)=\det\left({2i+2j+r\choose i+j}\right)_{i,j=0}^{N-1}$. We show that for $k,n\ge 1$, \begin{align} &d_{2k+1}((2k+1)n)=d_{2k+1}((2k+1)n+1)=(2n+1)^k,\\ &d_{2k+1}((2k+1)n+k+1)=(-1)^{k+1\choose 2}4^k(n+1)^k,\\ &d_{2k}(2kn)=d_{2k}(2kn+1)=(-1)^{kn},\\ &d_{2k}(2kn+k)=-d_{2k}(2kn+k+1)=(-1)^{kn+{k\choose 2}}4^{k-1}(n+1)^{k-1}. \end{align}
We do not prove the formulas for $d_r(rn-1)+d_r(rn+2)$. Possibly these could be proven by modifying the first or last step of the inductions in Lemmas 7.2 or 7.5.
• Very nice! I see that your notation has only 2 variables but covers all matrix sizes and is better than mine. What I call $d(n,k,r)$ is $d_{k+r}(kn)$ in yours. In my notation, nothing hinders $n$ to be non integer, so it can capture it all... but yours is of course more straightforward, needing only integers. - On the other hand, it seems to me that you haven't gained any insights into that curious "sum behavior", have you? – Wolfgang Jul 24 '18 at 18:49
• Indeed we haven't. I suspect that for values of $N$ close to $rn$ (or any family with a nice formula) you can find $w$ vectors that almost satisfy the hypotheses of 7.2 and 7.5, save for an extra nonzero entry or two. This will only come into play in the last step of the induction, so there will be only a couple extra terms in the final formula. Why these cancel across different values of $N$, I don't know. – MTyson Jul 24 '18 at 19:32