Given positive integers $n$ and $d$, where $d\geq 2$, I would like to compute the sum $$\displaystyle\sum_{0\leq i_{1} < i_{2} < ... < i_{d}\leq n} \quad\displaystyle\prod_{1 \leq p < q \leq d}\left(i_{q} - i_{p}\right).$$ Since there are $d\choose 2$ factors in the product, the sum should return a polynomial in $n$ of degree $d(d+1)/2$. Ideally I would like to know all coefficients of the polynomial in $n$. The leading coefficient (i.e., coefficient of $n^{d(d+1)/2}$) is of particular interest.

For example, when $d=2$, our sum becomes $\displaystyle\sum_{i=0}^{n}\displaystyle\sum_{j=i+1}^{n}(j-i) = \frac{1}{6}n(n+1)(n+2)$, and the leading coefficient (of $n^3$) is $1/6$.

When $d=3$, our sum gives $\displaystyle\sum_{i=0}^{n}\displaystyle\sum_{j=i+1}^{n}\displaystyle\sum_{k=j+1}^{n}(k-j)(j-i)(k-i) = \frac{1}{180}(n-1)n(n+1)^{2}(n+2)(n+3)$, and the leading coefficient (of $n^{6}$) is $1/180$.

For $d=4$, WolframAlpha gives $$\displaystyle\sum_{i=0}^{n}\displaystyle\sum_{j=i+1}^{n}\displaystyle\sum_{k=j+1}^{n}\displaystyle\sum_{l=k+1}^{n}(l-k)(l-j)(l-i)(k-j)(k-i)(j-i)\\ = \frac{1}{25200}(n-2)(n-1)n^{2}(n+1)^{2}(n+2)^{2}(n+3)(n+4)$$ and the leading coefficient (of $n^{10}$) is $1/25200$.

I am not sure if this object is well-known, or has a name. Any references will be great too.

  • 3
    $\begingroup$ Extending to $d = 0$ using the empty product gives a sum of $n$, and even with this extension the sequence of ratios $1, \frac{1}{6}, \frac{1}{30}, \frac{1}{140}$ of successive leading coefficients follows the pattern $\frac{n!^2}{(2n+1)!}$. $\endgroup$ – Travis Willse Sep 27 '19 at 21:43
  • 3
    $\begingroup$ And the reciprocals of these ratios, $\frac{(2n+1)!}{n!^2}$, is OEIS A002457 $\endgroup$ – rogerl Oct 2 '19 at 19:02
  • 2
    $\begingroup$ If your primary interest is the leading coefficient, you should be able to replace all the sums by integrals (so that, e.g. the $d=4$ case becomes $$\int_0^1 \int_0^w \int_0^x \int_0^y (w-x)(w-y)(w-z)(x-y)(x-z)(y-z) \, dz \, dy \, dx \, dw=\frac{1}{25200}.$$ Effectively this is the expected product of the pairwise distances of $d$ randomly placed points in the unit interval, and it feels like in this language it should have been studied somewhere (I don't have a reference though). $\endgroup$ – Kevin P. Costello Oct 2 '19 at 19:25
  • 1
    $\begingroup$ @Kevin: I was thinking along similar lines: the Faulhaber's formula gives $\sum_{k=1}^{p}k^{p} = \frac{1}{p+1}n^{p+1} + O(n^{p})$. When we use this in the nested sum, there are many terms contributing to the leading coefficient .... not sure if there is a good way to keep track of them so as to get the leading coefficient as function of $d$. $\endgroup$ – Abhishek Halder Oct 3 '19 at 0:56

In terms of the leading term coefficient the following identities hold true: \begin{eqnarray} c_d&=&\int\limits_{0 \le x_1 \le \cdots x_d \le 1} \prod\limits_{1 \le p < q \le d} (x_p - x_q)\cdot \prod\limits_{p=1}^d dx_p\\ &=&\sum\limits_{\sigma \in \Pi} \mbox{sign($\sigma$)} \frac{1}{\prod\limits_{i=1}^d \sum\limits_{j=1}^i \sigma_j} \quad (1)\\ &=& \int\limits_{[0,1]^d} \left(\prod\limits_{p=1}^d x_p^{\binom{p}{2}+p-1} \right) \cdot \prod\limits_{p=1}^d \prod\limits_{q=p+1}^d \left(1-\prod\limits_{\xi=p}^{q-1} x_\xi\right) \cdot \prod\limits_{p=1}^d d x_p \quad (2) \\ &\underbrace{=}_{?}& \prod\limits_{\xi=1}^{d-1} \frac{(\xi!)^2}{(2 \xi+1)!} \end{eqnarray} where in $(1)$ we expanded the Vandermonde determinant in a sum over permutations $\Pi$ and then integrated term by term and in $(2)$ we we used the trick $1/p = \int\limits_0^1 x^{p-1} dx$ and the definition of the Vandermonde determinant again. The representations above are readily used to compute the result for $d \le 9$. We have:

In[484]:= d =.;
ss = Table[
  Total[Signature[#] Product[1/(Total[Take[#, i]]), {i, 1, d}] & /@ 
    Permutations[Range[1, d]]], {d, 1, 9}]
   Product[x[p]^(Binomial[p, 2] + p - 1), {p, 1, 
      d}] Product[(1 - Product[x[xi], {xi, p, q - 1}]), {p, 1, d}, {q,
       p + 1, d}]] /. x[n_]^p_. :> 1/(p + 1), {d, 1, 9}]
Table[Product[(xi!)^2/(2 xi + 1)!, {xi, 1, d - 1}], {d, 1, 9}]

enter image description here

Update: Let us denote: \begin{eqnarray} {\mathcal S}_d^{(n)} := \sum\limits_{0 \le i_1 < i_2 < \cdots < i_d \le n} \prod\limits_{1 \le p < q \le d} (i_q-i_p) \end{eqnarray} for $n \ge d-1$. Then my conjecture is the following: \begin{eqnarray} {\mathcal S}_d^{(n)} = \left[\prod\limits_{\xi=1}^{d-1} \frac{(\xi!)^2}{(2 \xi+1)!}\right] \cdot \left[\prod\limits_{j=-d+1}^1 (n+j)^{\lceil \frac{j+d-1}{2} \rceil}\right] \cdot (n+2)^{\lfloor \frac{d}{2}\rfloor} \cdot \left[\prod\limits_{j=3}^d (n+j)^{\lceil \frac{d-j+1}{2} \rceil }\right] \end{eqnarray}

I have verified this conjecture for $d \le 6$ using the code below:

d = 2; Clear[a]; Clear[aa]; i[0] = 0;
aa = Table[a[p], {p, 0, d - 1}];
smnD = Product[i[q] - i[p], {p, 1, d}, {q, p + 1, d}];
subst = First@
       smnD - (Sum[Binomial[i[d] - i[d - 1], p] a[p], {p, 0, d - 1}]),
        i[d]] == 0, aa] // Simplify;
(*Sum over i[d] done.*)
S = Sum[Binomial[n + 1 - i[d - 1], p + 1] a[p], {p, 0, d - 1}] /. 
upLim = d - 1;
For[count = 1, count <= d - 1, count++,
  upLim = upLim + d - count;
  aa = Table[a[p], {p, 0, upLim}];
  subst = 
        FunctionExpand@(S - (Sum[
             Binomial[i[d - count] - i[d - count - 1], p] a[p], {p, 0,
               upLim}])), i[d - count]] == 0, aa] // Simplify;
  (*Sum over i[d-count] done.*)
  S = Sum[
     Binomial[n + 1 - i[d - count - 1], p + 1] a[p], {p, 0, 
      upLim}] /. subst;
  Print["count=", count, "done"];

enter image description here

  • $\begingroup$ I think the coefficient of $\mathrm{sign}(\sigma)$, i.e. $\int \prod x_j^{\sigma_j-1} dx$, should be $(\prod_{j=1}^n \sum_{i=1}^j \sigma_i)^{-1}$, not $(\sum \sigma_j)^{-1}$, right? $\endgroup$ – user125932 Oct 4 '19 at 16:19
  • $\begingroup$ Yes, yes, of course, I am sorry for that. I fixed it. $\endgroup$ – Przemo Oct 4 '19 at 17:07
  • $\begingroup$ This looks nice, but my impression is that the polynomial factors completely into linear terms, and that probably cannot be proven by an integral approximation, can it? $\endgroup$ – darij grinberg Oct 5 '19 at 2:58
  • 1
    $\begingroup$ @darij grinberg It is indeed quite remarkable that this factorizes so nicely. I have updated my answer and provided a Mathematica code that computes the sum for higher values of $d$. By the way , why are you interested in this very problem ? $\endgroup$ – Przemo Oct 7 '19 at 11:54
  • 1
    $\begingroup$ I care partly because the sum is the sum of all maximal minors of the Vandermonde-like rectangular matrix $\left(i^{j-1}\right)_{1\leq i\leq n,\ 1\leq j\leq d}$, and it appears that there may be a more general determinant identity behind it. Also, the sum can be reinterpreted as counting something like Gelfand-Tsetlin patterns, since each $i_q - i_p$ factor can be viewed as choosing an integer from the interval $\left[i_p,i_q\right)$. These are not quite Gelfand-Tsetlin patterns, because all these integers in a given addend of the sum are independent, but close enough to be interesting to me. $\endgroup$ – darij grinberg Oct 7 '19 at 13:18

This is a follow up on Przemo's answer in deriving the final expression of the leading coefficient $c_{d}$. Notice that $d! c_{d}$ equals the Vandermonde determinant integrated over $[0,1]^{d}$. In an 1955 paper by De Bruijn (see toward the end of Sec. 9), it is proved that

$$\int_{[0,1]^{d}}\prod_{1\leq i < j \leq d} |x_{i} - x_{j}| \: {\rm{d}}x_{1} ... {\rm{d}}x_{d} = \frac{\{1! \times 2! \times 3! \times ... \times (d-1)!\}^{2} d!}{1!\times 3! \times 5! \times ... \times (2d-1)!}.$$

That proof utilizes the result (also derived in that paper) that integrals of this type are equal to certain Pfaffian form. Equating the above with $d! c_{d}$ recovers the expression conjectured by Przemo:

$$c_{d} = \prod\limits_{\xi=1}^{d-1} \frac{(\xi!)^2}{(2 \xi+1)!}.$$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.