3
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Now let us denote by $\Lambda^{*}(n)$ the algebra of polynomials in $x_{1},\ldots,x_{n}$ that become symmetric in new variables $$ x_{i}'=x_{i}-i+c, \ i \in 1,\ldots,n.$$ Here c is a arbitrary fixed number; note that the definition does not depend on it's choice. We call such polynomials shifted symmetric. The algebra $\Lambda^{*}(n)$ is filtered by degree of polynomials, and the specialization $x_{n+1}=0$ is a morphisms of filtered algebras $$\Lambda^{*}(n+1)\rightarrow \Lambda^{*}(n)$$

Let $$\Lambda^{*} = \lim_{\rightarrow}\Lambda^{*}(n), n\rightarrow \infty $$ be the projective limit in the category of filtered algebras, taken with respect to morpshims. We call $\Lambda^{*}$ the algebra of shifted algebra of shifted symmetric functions.

The most important example of shifted symmetric function is shifted schur function $$s_{\mu}^{*}(x_1,\ldots,x_n)=\frac{det\Big((x_i+n-i|\mu_j+n-j)\Big)}{det\Big((x_i+n-i|n-j)\Big)} $$

Let $s_{\mu}$ denote schur function. Then we have the following generating function $$ \sum_{n\geq 0}s_{(n)}(q_1,q_2,\ldots,q_r)x^n$$ in holonomic function in $x$. where $r$ is fixed and all other $\{q_{s}=0|s>r\}$. There is an explicit expression in terms of exponential. This shows that it's holonomic.

We epxress $s_{(n)}(q_1,q_2,\ldots,q_r)$ in power-sum symmetric function. So $$ \sum_{n\geq 0}s_{(n)}(q_1,q_2,\ldots,q_r)x^n$$ becomes $$exp\Big(\sum_{m=1}^{r}(q_m x^m )/m\Big) .$$

I want to know if the following generating series is holonomic.

$$\sum_{n\geq 0 }s_{(n)}^{*}(q_1,q_2,\ldots,q_r)x^n .$$ is holonomic?

$$\sum_{n\geq 0 }s_{(n)}^{*}(q_1,q_2,\ldots,q_r)(x|n) .$$ is holonomic?

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  • $\begingroup$ Is there a typo? In the second series, the number of $q$-variables grow with n, but in the first, it is fixed. $\endgroup$ – Per Alexandersson Oct 24 '18 at 8:01
  • $\begingroup$ Also, the determinant is of fixed size, so the series should satisfy a linear recurrence if each of the sequences $(x_i+n-i|\mu_j+n-j)$ does. This should be easy to check. $\endgroup$ – Per Alexandersson Oct 24 '18 at 8:03
  • $\begingroup$ It's a typo. I just edit. Thanks a lot. $\endgroup$ – GGT Oct 24 '18 at 10:13
  • $\begingroup$ I forgot to put the denominator for the shifted Schur function. I could not follow the argument you made. Can you please elaborate. $\endgroup$ – GGT Oct 25 '18 at 0:40
  • $\begingroup$ Correct me if I'm wrong, but isn't the set of holonomic sequences closed under addition, and Hadamard multiplication? Then, for the numerator (or denominator) in the definition to be holonomic, it suffices to show that each matrix entry is holonomic. Now, with the denominator that was not there before, it is trickier. You might want to find a Jacobi-Trudy type formula,(without denominator) and try the above argument... $\endgroup$ – Per Alexandersson Oct 25 '18 at 5:46

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