# Special polynomials and an identity of hypergeometric series

Motivation:

I have a few polynomials and am trying to find a representation for them in terms of special functions. I'm more interested in the techniques here, so I won't give any too particular example, only let's say I find that they look a bit like

$$p_1(x)=a_0+a_1x,~p_2(x)=b_0+b_1x,~p_2(x)=c_0+c_1x+c_2x^2,\dots$$

i.e. they stay the same degree for two steps of $$n,$$ before their degree increases by $$1$$. So $$p_2(x)$$ and $$p_3(x)$$ are quadratics, then $$p_4(x)$$ and $$p_5(x)$$ are cubics, and so on.

A natural first guess is something like

$$p_n(x)=\frac{Q_n(x)}{x^n},$$

where $$Q_n(x)$$ is something like a Hermite or Chebyshev or Legendre polynomial. Basically, if $$Q_n(x)$$ is something like a hypergeometric series that alternates between an odd and even function of $$x,$$ it at least forms that pattern even if the coefficients don't exactly match.

But, when I look at the differential equation polynomials of this type should satisfy, it still has three regular singular points, which could be shifted so that they are at $$x=0,1,\infty,$$ and they generally satisfy some differential equation which looks like

$$x(1+A_nx)\frac{d^2p}{dx^2}+(B_n+C_nx)\frac{dp}{dx}+D_np=0,$$

$$A_n,\dots,D_n$$ being some sequences depending on $$n,$$ and exactly what their values are depending on the special polynomial $$Q_n(x)$$ is.

So after some playing around, I expect that polynomials that look like

$$\frac{{_2}F_1(a,b,c;x)}{x^n}$$

should probably be expressible themselves as hypergeometric functions of some type.

Question:

Is there a natural kind of transformation that would fit the bill here, so that I can write

$${_2}F_1(a',b';c';x')x^n={_2}F_1(a,b;c;x)?$$

Or, is there some 'special form' of the arguments of $${_2}F_1(a,b;c;x)$$ which tends to give polynomials which keep the same degree for two steps?

The differential equation $$x(1+a x)\frac{d^2p}{dx^2}+(b+cx)\frac{dp}{dx}+d\,p=0$$ admits as general solution $$p(x)=C_1 a^{1-b} x^{1-b} \, _2F_1\left(\alpha_1,\alpha_2;2-b;-a x\right)+C_2 \, _2F_1\left(\alpha_3,\alpha_4;b;-a x\right)$$ where $$\alpha_1=\frac{a+c-2ab-\sqrt{(a-c)^2-4 a d}}{2a} \qquad \text{and} \qquad \alpha_2=\frac{a+c-2ab+\sqrt{(a-c)^2-4 a d}}{2 a}$$ $$\alpha_3=\frac{c-a-\sqrt{(a-c)^2-4 a d}}{2 a}\qquad \text{and} \qquad \alpha_4=\frac{c-a+\sqrt{(a-c)^2-4 a d}}{2 a}$$