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Let $h(n,t) = \sum\limits_{j = 0}^n {\binom {\lfloor {\frac{n}{2}} \rfloor }{j}\binom {\lfloor {\frac{n+1}{2}}\rfloor }{j}t^j \\ }.$

I am interested in the Hankel determinants $${D_k}(n,t) = \det \left( {h(k + i + j,t)} \right)_{i,j = 0}^{n - 1}.$$ These can easily be computed for $0 \leq k \leq 3.$

It seems that $${D_4}(n,t) = {t^{\lfloor {\frac{{{n^2}}}{4}} \rfloor }}b(n,t)$$ with $b(n,t) = \sum\limits_{j = 0}^{2n} \min({\binom{3+j}{3},\binom{2n+3-j}{3}})t^j.$

In order to prove this, I need the identity $$a{(n,t)^2} = b(n - 1,t)\sum\limits_{j = 0}^{n - 1} {{t^j}} - tb(n - 2,t)\sum\limits_{j = 0}^n {{t^j}} $$ with $a(n,t) = \sum\limits_{j = 0}^{n - 1} {(j + 1){t^j}} .$

Any idea how to prove this identity?

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    $\begingroup$ Please could you include the equivalent form of the identity by equating coefficients of $t^j$ on either side? $\endgroup$ Aug 22, 2021 at 11:04
  • $\begingroup$ @ Mark Wildon: One has to consider many cases which makes the explicit form of the coefficients rather ugly. In the mean time I have seen that Mathematica gives an explicit form of the generating functions. So I now have a proof. But I look for a simple human proof. $\endgroup$ Aug 22, 2021 at 14:25

1 Answer 1

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Denote $a_n=a(n,t)$ and $b_n=b(n,t)$. To help avoiding the min function, write $$b_n=\binom{n+3}3t^n+\sum_{j=0}^{n-1}\binom{3+j}3\left[t^{2n-j}+t^j\right].$$ Notice that $a_n=\frac{nt^{n+1}-(n+1)t^n+1}{(1-t)^2}$ and $\sum_{j=0}^nt^j=\frac{1-t^{n+1}}{1-t}$. Your identity takes the form $$(nt^{n+1}-(n+1)t^n+1)^2=(1-t)^3[(1-t^n)b_{n-1}-t(1-t^{n+1})b_{n-2}].$$ Now, as Mark Widon mentioned, try to read-off the coefficients of $t^k$.

UPDATE. Resorting back the original formulation of the claim $$a_n^2=b_{n-1}\sum_{j=0}^{n-1}t^j-t\,b_{n-2}\sum_{j=0}^nt^j,$$ I was able (after lots of routine algebraic simplification and reorganization) to rewrite the right-hand side as \begin{align*} \left(\sum_{j=0}^{n-1}(j+1)t^j\right)^2 &=\sum_{j=0}^{n-1}\binom{3+j}3t^j+\sum_{j=0}^{n-2}\left[\beta_n-\beta_{j+1}-\binom{n-j}3\right]\,t^{n+j} \\ &=\sum_{j=0}^{n-1}\binom{3+j}3t^j+\sum_{j=0}^{n-2} \frac{(n - j - 1)(j^2 + 4jn + n^2 + 5j + 7n + 6)}6\, t^{n+j} \end{align*} where $\beta_k=\frac{k(k+1)(2k+1)}6$ (the sum of squares function).

Once we got this far, the next step is to compare the coefficients of $t^k$.

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  • $\begingroup$ it looks that $(1-t)^4b_n$ is a fewnomial (have only bounded number oа non-zero coefficients), so this actually an identity with fewnomials $\endgroup$ Aug 24, 2021 at 19:44
  • $\begingroup$ @FedorPetrov: True. But, the OP did not like the case-by-case analysis and I want to keep it that way. The current form I have requires only two cases $j<n$ and $j>n$. $\endgroup$ Aug 24, 2021 at 20:05
  • $\begingroup$ I want to thank T. Amdeberhan and Fedor Petrov for their answers and remarks which led me to the following proof: If we write $a_n=\frac{nt^{n+1}-(n+1)t^n+1}{(1-t)^2}$ and $b_n=\binom{n+3}{3}t^n+\sum_{j=0}^{n-1}\binom{3+j}{3}\left[t^{2n-j}+t^j\right]=\frac{1-(n+2)^2 t^{n+1}+2(n^2+4n+3)t^{n+2}-(n+2)^2 t^{n+3}+t^{2n+4}}{(1-t)^4},$ then the identity is a simple verification. $\endgroup$ Aug 25, 2021 at 15:39

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