A polynomial sequence

Question

I have a sequence of polynomials $Q_k(x, y)$, $k\geq 1$ defined recursively as follows:

1. $Q_1=x$.

2. There is a sequence of polynomials $p_j(y)$ of degree $j$ such that $Q_{2m}$ is of the form \begin{eqnarray}\frac{p_{0}(y)}{(2m)!}x^{2m}+\frac{p_{1}(y)}{(2m-2)!}x^{2m-2}+\cdots+\frac{p_{m-1}(y)}{2!}x^2+p_{m}(y)\end{eqnarray} and $Q_{2m+1}$ \begin{eqnarray} \frac{p_{0}(y)}{(2m+1)!}x^{2m+1}+\frac{p_{1}(y)}{(2m-1)!}x^{2m-1}+\cdots+\frac{p_{m-1}(y)}{6}x^3+p_{m}(y)x \end{eqnarray}

3. $Q_k(k+1-2i, k+1)$ as a polynomial in $i$ has roots $1, 2, \cdots, k$.

Find a general formula of $Q_k(x, y)$. The following are the first 4 polynomials in the sequence: \begin{align*} Q_1&=x\\ Q_2&=\frac{x^2}{2}-\frac{y}{6}\\ Q_3&=\frac{x^3-xy}{6}\\ Q_4&=\frac{x^4-2x^2y}{24}+\frac{y(5y+2)}{360} \end{align*}

In fact, it suffices to find the sequence of polynomials $p_j(y)$. Here's what I've got so far. By condition (2) and (3), we have \begin{eqnarray} \sum_{j=0}^m \frac{p_j(2m+1)}{(2m-2j)!}(2i-2m-1)^{2m-2j}=\frac{2^{2m}}{(2m)!}(i-1)(i-2)\cdots(i-2m) \end{eqnarray} If we let $m=j+k$, differentiate both sides with respect to $i$ $2k$ times and put $\displaystyle i=\frac{2j+2k+1}{2}$, we have \begin{eqnarray} p_j(2j+2k+1)=\left.\frac{d^{2k}}{di^{2k}}\right|_{i=\frac{2j+2k+1}{2}}\frac{2^{2j}}{(2(j+k))!}(i-1)(i-2)\cdots(i-2(j+k)) \end{eqnarray} Letting $k=0, 1, \cdots, j$, we get the $j+1$ values taken by $p_j$ at $2j+1, 2j+3, \cdots, 4j+1$. $p_j$ then can be computed using Lagrangian interpolation.

It seems to me that, though the above algorithm can be implemented on a computer to get a few polynomials in the sequence, it does not yield directly a general formula I want. Is there another better way to go about getting a general formula?

Edit: The first 6 members in the polynomial sequence $p_j(y)$ are the following: \begin{align*} p_0(y)&=1\\ p_1(y)&=-\frac{y}{6}\\ p_2(y)&=\frac{y(5y+2)}{360}\\ p_3(y)&=-\frac{y(35y^2+42y+16)}{45360}\\ p_4(y)&=\frac{y(5y+4)(35y^2+56y+36)}{5443200}\\ p_5(y)&=-\frac{y(385y^4+1540y^3+2684y^2+2288y+768)}{359251200} \end{align*}

It is noteworthy that the denominators happen to be the first 6 numbers in this sequence, as pointed out by Solomonoff's Secret.

Answer 1

We are essentially looking for the coefficients of falling factorials. I am not a fan of the first kind of Stirling numbers, because they are (irreversibly?) recursive. (See the more explicit Bernoulli numbers for comparison). Never the less, begin with condition (3): $$ 1 \le m\implies\ Q_m(m+1-2i,m+1) = (-2)^m\binom{i-1}{m} $$ (Corrected by San). When we use the translation $z=m+1-2i,\ $ we can apply the structure of condition (2): $$ Q_m(z,m+1) = \sum_{k=0}^{\lfloor\frac{m}{2}\rfloor}p_k(m+1)\frac{z^{m-2k}}{(m-2k)!} = (-2)^m\binom{\frac{m-1-z}{2}}{m} $$ Notice the top of the binomial expands to $(z+1-m)(z+3-m)\cdots (z-3+m)(z-1+m),\ $ and so the parity of the powers of $z$ match the parity of $m$. This reassures us that condition (2) is well defined (for all complex $z$) even though the sum omits half of the powers of $z$. We also require roots of unity, the exact choice of roots is irrelevant as long as it is consistent: $\mu_N \in\mathbb{C}$ is a $N$th root unity when $k\in N\mathbb{Z}\Leftrightarrow \mu_N^k = 1.\ $ When $m$ is odd and $m < 2(m-2N),\ \ (m-2N)\mathbb{Z} \cap \{m,m-2,\ldots,3,1\} = \{m-2N\};\ $ which makes the discrete fourier transform of $z^{m-2k}$ in $Q_m$ useful: $$\mbox{when } m \mbox{ is odd, and } \mu_B := \exp(\frac{2\pi i}{B}): $$ $$ 4N < m\implies\ p_N(m+1)\frac{m-2N}{(m-2N)!} = (-2)^m\sum_{j=1}^{m-2N} \binom{(m-1-\mu_{m-2N}^j)/2}{m} $$ We do not see an obvious way to show that $p_N$ is a degree $N$ polynomial, other than to tediously expand the powers of $z$ in the right hand binomial. We assume that everyone is convinced that $p_N$ is polynomial. Now we too resort to Lagrange interpolation: $$\mbox{using }\ m+1=4N+2+2h,\ \ h=0\ldots N$$ $$ p_N(y) = 4^{2N+1}(-1)^{N+1}(N+1)\binom{\frac{y}{2}-2N-1}{N+1} $$$$ \sum_{h=0}^N \sum_{j=1}^{2N+1+2h} \frac{(2N+2h)!(-4)^h}{y-4N-2-2h}\binom{N}{h} \binom{2N+h-\frac{\mu_{2N+1+2h}^j}{2}}{4N+1+2h} $$

This is a finite formula, that takes no limits. It is not suitable for computation. (The case $N=4$ evaluates a degree 25 binomial). Now that Winther corrected the fourier transform, we can numerically evaluate the first few polynomials. Here is the output of a test program (by gcc):

This machine uses 128 bit long doubles:

Will claims 'No_way's polynomial, p1(y), is about:
-1.0000y^1   -0.0000y^0
over 1!6^1  =  6
with relative error > 7.12e-16   making coefficent errors > 4.27e-15

Will claims 'No_way's polynomial, p2(y), is about:
+1.0000y^2   +0.4000y^1   -0.0000y^0
over 2!6^2  =  72
with relative error > 3.20e-14   making coefficent errors > 2.31e-12

Will claims 'No_way's polynomial, p3(y), is about:
-1.0000y^3   -1.2001y^2   -0.4561y^1   -0.0053y^0
over 3!6^3  =  1296
with relative error > 7.44e-11   making coefficent errors > 1.16e-07

Will claims 'No_way's polynomial, p4(y), is about:
+0.8094y^4   +18.4170y^3   -500.4859y^2   +6988.6559y^1   -36277.1494y^0
over 4!6^4  =  31104
with relative error > 1.75e-06   making coefficent errors > 1.97e+03

The alternative recurrence relation to this "formula" would be to interpolate the correct combination of stirling numbers. If you construct a generating function, you might need to integrate something like $\ln(z)/(z^2+y^2)$, but this is a wild guess from past experience.

Answer 2

Using Wolfram Alpha, I find that the first 6 members $p_j(x)$, $0\leq j\leq 5$, of the polynomial sequence happen to be the first 6 non-zero coefficients of the Maclaurin series of \begin{eqnarray}\left(\frac{t}{\sinh(t)}\right)^x.\end{eqnarray} I conjecture that \begin{eqnarray}\left(\frac{t}{\sinh(t)}\right)^x=\sum_{j=0}^\infty p_j(x)t^{2j}.\end{eqnarray}