1
$\begingroup$

By this question I asked before, I know that for any nonnegative integers $a$, $b$, $c$, the coefficient of the monomial $x_1^{a+c+1}x_2^{a+b+1}x_3^{b+c+1}$ in the expansion of $$(x_1-x_2)^{2a+1}(x_2-x_3)^{2b+1}(x_3-x_1)^{2c+1}$$ is always zero.

Then I found for small nonnegative integers $a$, $b$, $c$, $k$, the coefficient of the monomial $x_1^{a+c+k}x_2^{a+b+k}x_3^{b+c+k}$ in the expansion of $$(x_1-x_2)^{2a+2k}(x_2-x_3)^{2b+k}(x_3-x_1)^{2c}$$ is always nonzero.

I want to ask whether it is always right for all nonnegative integers $a$, $b$, $c$, $k$.

Improve this question
$\endgroup$
6
  • $\begingroup$ Is $k$ missing in the exponent of $(x_3-x_1)^{2c}$? $\endgroup$
    – Seva
    Commented Mar 26, 2017 at 9:53
  • $\begingroup$ Hello, Seva, there is no $k$ in the exponent of $(x_3-x_1)^{2c}$, just $2c$. $\endgroup$
    – user173856
    Commented Mar 26, 2017 at 10:58
  • 1
    $\begingroup$ Fwiw, Mathematica says that up to sign it is$$\binom{2 (a+k)}{a-b} \binom{2 c}{-b+c-k} \, _3F_2\left(\begin{array}{ccccc}\smash-a-b-2 k&&-2 b-k&&-b-c-k\\&a-b+1&&-b+c-k+1\end{array}1\right)$$ $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Mar 26, 2017 at 14:59
  • 1
    $\begingroup$ @FedorPetrov Sorry for bad TeXing - 1 means the value at 1 of this hypergeometric series. So the whole is the product of two binomial coefficients and of the value at 1 of the hypergeometric series with three upper parameters $-a-b-2k$, $-2b-k$, $-b-c-k$ and two lower parameters $a-b+1$, $-b+c-k+1$. $\endgroup$
    – მამუკა ჯიბლაძე
    Commented Mar 27, 2017 at 15:12
  • 1
    $\begingroup$ The coefficient is up to a sign equal to $$\sum_{d} (-1)^d\binom{2a+2k}{a+k+d}\binom{2b+k}{b+d}\binom{2c}{c+d}$$ $\endgroup$
    – Markus Sprecher
    Commented Mar 30, 2017 at 20:09

0

Reset to default

You must log in to answer this question.