Looking for a closed form solution $P_{n+2}(x)=x^2 P'_{n+1}(x)+(n+1)xP_{n+1}(x)-(n+1)P'_n(x) $

Question

$$P_{n+2}(x)=x^2 P'_{n+1}(x)+(n+1)xP_{n+1}(x)-(n+1)P'_n(x) \tag{1}$$

where $P_{0}(x)=x$ , $P_{1}(x)=x^2$ are the given initial conditions.

I calculated few terms: $$P_{2}(x)=3x^3-1$$ $$P_{3}(x)=3.5x^4-6x$$ $$P_{4}(x)=3.5.7x^5-51x^2$$ $$P_{5}(x)=3.5.7.9x^6-546x^3+24$$

We can estimate the $P_n(x)$ form from a few terms ,

$P_{n}(x)=a_{n}x^{n+1}+b_{n}x^{n-2}+c_{n}x^{n-5}+d_{n}x^{n-8}+\cdots.$ If we put these terms in Equation 1, We get; $$a_{n+2}x^{n+3}+b_{n+2}x^{n}+c_{n+2}x^{n-3}+....= a_{n+1}(2n+3)x^{n+3}+[b_{n+1}2n-a_{n}(n+1)^2]x^{n}+[c_{n+1}(2n-3)-b_{n}(n+1)(n-2)]x^{n-3}+...... $$

We can write that $$a_{n+2}=(2n+3)a_{n+1}$$ $$b_{n+2}=2nb_{n+1}-a_{n}(n+1)(n+1)$$ $$c_{n+2}=(2n-3)c_{n+1}-b_{n}(n+1)(n-2)$$

Next term can be gotten as $$d_{n+2}=(2n-6)d_{n+1}-c_{n}(n+1)(n-5)$$

$a_{n}$ can be expressed as for $n>0$

$a_{n}=1.3.5.7.....(2n-1)=\frac{(2n-1)!}{2^{n-1}(n-1)!}$

I am looking for a closed form $P_n(x)$ or a generating function. Please help me to find a similar Rodrigues formula of Legendre Polynomials for the defined $P_{n}(x)$ above (if it is possible)?

Note: Legendre Polynomials, Rodrigues formula is $$\frac{1}{2^nn!}\frac{d^n}{dx^n}(x^2-1)^n$$

Edit: $P_n(x)$ satisfy the relation

$$\frac{x^2-y}{1-xy}=\cfrac{\sum_{n=0}^\infty \frac{y^{n}P_{n+1}(x)}{n!}}{\sum_{n=0}^\infty \frac{y^{n}P'_{n}(x)}{n!}} $$

$$\sum_{n=0}^\infty \frac{x^{2n}P_{n+1}(x)}{n!}=0 $$

My atempt to find generating function:

$$G(x,t)=\sum_{n=0}^\infty P_{n}(x) t^n =P_0(x)+P_1(x)t+P_2(x)t^2+P_3(x)t^3+..... \tag {2}$$

$$xt\frac {\partial G(x,t)}{\partial t}=xP_1(x)t+2xP_2(x)t^2+3xP_3(x)t^3+.....\tag {3}$$

$$x^2\frac {\partial G(x,t)}{\partial x}=x^2P'_0(x)+x^2P'_1(x)t+x^2P'_2(x)t^2+x^3P'_3(x)t^3+.....\tag {4}$$

$$t\frac {\partial G(x,t)}{\partial x}+t^2\frac {\partial^2 G(x,t)}{\partial x \partial t}=P'_0(x)t+2P'_1(x)t^2+3P'_2(x)t^3+4P'_3(x)t^4+.....\tag {5}$$

$$xt\frac {\partial G(x,t)}{\partial t}+x^2\frac {\partial G(x,t)}{\partial x}-t\frac {\partial G(x,t)}{\partial x}-t^2\frac {\partial^2 G(x,t)}{\partial x \partial t}=x^2P'_0(x)+(x^2P'_1(x)+xP_1(x)-P'_0(x))t+(x^2P'_2(x)+2xP_2(x)-2P'_1(x))t^2+(x^2P'_3(x)+3xP_3(x)-3P'_2(x))t^3+....\tag {6}$$

$$xt\frac {\partial G(x,t)}{\partial t}+x^2\frac {\partial G(x,t)}{\partial x}-t\frac {\partial G(x,t)}{\partial x}-t^2\frac {\partial^2 G(x,t)}{\partial x \partial t}=x^2+P_2(x)t+P_3(x)t^2+P_4(x)t^3+....\tag {7}$$

$$xt^2\frac {\partial G(x,t)}{\partial t}+x^2t\frac {\partial G(x,t)}{\partial x}-t^2\frac {\partial G(x,t)}{\partial x}-t^3\frac {\partial^2 G(x,t)}{\partial x \partial t}=x^2t+P_2(x)t^2+P_3(x)t^3+P_4(x)t^4+....\tag {8}$$

$$xt^2\frac {\partial G(x,t)}{\partial t}+x^2t\frac {\partial G(x,t)}{\partial x}-t^2\frac {\partial G(x,t)}{\partial x}-t^3\frac {\partial^2 G(x,t)}{\partial x \partial t}=x^2t+G(x,t)-x-x^2t\tag {9}$$

$$G(x,t)-xt^2\frac {\partial G(x,t)}{\partial t}+(t^2-x^2t)\frac {\partial G(x,t)}{\partial x}+t^3\frac {\partial^2 G(x,t)}{\partial x \partial t}=x\tag {10}$$

I could not get progress after here. Please help me how to get the generating function after this step.

UPDATE (03/23/2018): Thanks a lot for all answers. All answers are very helpful to solve the problem. I would like to write other property of the $P_{n}(x)$. I believe that it will be helpful to find the closed form of $P_n(x)$.

$$U(x,y)=\sum_{n=0}^\infty \frac{(xy)^{n}P_{n}(x+y)}{n!} $$

$$\frac {\partial U(x,y)}{\partial x}=\sum_{n=0}^\infty \frac{(xy)^{n}P'_{n}(x+y)}{n!}+y \sum_{n=0}^\infty \frac{(xy)^{n}P_{n+1}(x+y)}{n!} $$

$$\frac {\partial U(x,y)}{\partial y}=\sum_{n=0}^\infty \frac{(xy)^{n}P'_{n}(x+y)}{n!}+x \sum_{n=0}^\infty \frac{(xy)^{n}P_{n+1}(x+y)}{n!} $$

We know that

$$\frac{x^2-y}{1-xy}=\cfrac{\sum_{n=0}^\infty \frac{y^{n}P_{n+1}(x)}{n!}}{\sum_{n=0}^\infty \frac{y^{n}P'_{n}(x)}{n!}} $$

Thus We can rewrite it as

$$\frac{(x+y)^2-xy}{1-xy(x+y)}=\frac{x^2+xy+y^2}{1-xy(x+y)}=\cfrac{\sum_{n=0}^\infty \frac{(xy)^{n}P_{n+1}(x+y)}{n!}}{\sum_{n=0}^\infty \frac{(xy)^{n}P'_{n}(x+y)}{n!}} $$

$$\frac {\partial U(x,y)}{\partial x}=\sum_{n=0}^\infty \frac{(xy)^{n}P'_{n}(x+y)}{n!}+y \frac{x^2+xy+y^2}{1-xy(x+y)}\sum_{n=0}^\infty \frac{(xy)^{n}P'_{n}(x+y)}{n!} $$ $$\frac {\partial U(x,y)}{\partial y}=\sum_{n=0}^\infty \frac{(xy)^{n}P'_{n}(x+y)}{n!}+x \frac{x^2+xy+y^2}{1-xy(x+y)}\sum_{n=0}^\infty \frac{(xy)^{n}P'_{n}(x+y)}{n!} $$

$$\frac {\partial U(x,y)}{\partial x}/\frac {\partial U(x,y)}{\partial y}=\frac {1-xy(x+y)+y(x^2+xy+y^2)}{1-xy(x+y)+x(x^2+xy+y^2)}$$

$$(1+x^3)\frac {\partial U(x,y)}{\partial x}=(1+y^3)\frac {\partial U(x,y)}{\partial y}$$

$$\int_0^{U(x,y)} F(z)\mathrm dz=\int_0^{x} \frac{\;\mathrm dz}{1+z^3}+\int_0^{y} \frac{\;\mathrm dz}{1+z^3}$$

If we derivative both side for x

$$\frac{\partial U(x,y)}{\partial x} F(U(x,y))=\frac{1}{1+x^3}$$

$$\frac{1}{F(U(x,y))}=(1+x^3)\frac{\partial U(x,y)}{\partial x}$$ If we derivative both side for y $$\int_0^{U(x,y)} F(z)\mathrm dz=\int_0^{x} \frac{\;\mathrm dz}{1+z^3}+\int_0^{y} \frac{\;\mathrm dz}{1+z^3}$$

$$\frac{\partial U(x,y)}{\partial y} F(U(x,y))=\frac{1}{1+y^3}$$

$$\frac{1}{F(U(x,y))}=(1+y^3)\frac{\partial U(x,y)}{\partial y}$$

To find $F(z)$,We can put $y=0$

$$\int_0^{U(x,0)} F(z)\mathrm dz=\int_0^{x} \frac{\;\mathrm dz}{1+z^3}$$

$$U(x,0)=P_0(x)=x $$

$$\int_0^{x} F(z)\mathrm dz=\int_0^{x} \frac{\;\mathrm dz}{1+z^3}$$ $$F(z)=\frac{1}{1+z^3}$$ Thus the solution of $U(x,y)$ is $$\int_0^{U(x,y)} \frac{\;\mathrm dz}{1+z^3}=\int_0^{x} \frac{\;\mathrm dz}{1+z^3}+\int_0^{y} \frac{\;\mathrm dz}{1+z^3}$$

Let's define $g(x)$ , $g(0)=0$; $$x=\int_0^{g(x)} \frac{\;\mathrm dz}{1+z^3}$$

$$y=\int_0^{g(y)} \frac{\;\mathrm dz}{1+z^3}$$

$$x+y=\int_0^{g(x+y)} \frac{\;\mathrm dz}{1+z^3}$$

$$\int_0^{g(x+y)} \frac{\;\mathrm dz}{1+z^3}=\int_0^{g(x)} \frac{\;\mathrm dz}{1+z^3}+\int_0^{g(y)} \frac{\;\mathrm dz}{1+z^3}$$

$$g(x+y)=U(g(x),g(y))$$

$$\int_0^{U(g(x),g(y))} \frac{\;\mathrm dz}{1+z^3}=\int_0^{g(x)} \frac{\;\mathrm dz}{1+z^3}+\int_0^{g(y)} \frac{\;\mathrm dz}{1+z^3}$$

$$x=\int_0^{g(x)} \frac{\;\mathrm dz}{1+z^3}$$

$$g^{-1}(x)=\int_0^{x} \frac{\;\mathrm dz}{1+z^3}$$

$$g'(x)=1+g^3(x)$$

So we can write that $$g(g^{-1}(x)+g^{-1}(y))=U(x,y)=\sum_{n=0}^\infty \frac{(xy)^{n}P_{n}(x+y)}{n!} $$

Addition formula of $g(x)$ can be written as:

$$g(x+y)=\sum_{n=0}^\infty \frac{g^n(x)g^n(y)P_{n}(g(x)+g(y))}{n!} $$

$$\int_0^{g(x+y)} \frac{\;\mathrm dz}{1+z^3}=\int_0^{g(x)} \frac{\;\mathrm dz}{1+z^3}+\int_0^{g(y)} \frac{\;\mathrm dz}{1+z^3}$$

If a closed form of $P_n(x)$ can be found, We can write a closed form addition formula for $g(x)$ like $\tan(x)$

Answer 1

You can first calculate $ b_{n+2}$ from $ b_{n+2}=2nb_{n+1}-a_{n}(n+1)(n+1) $. Let $r_{n+1} = -a_{n}(n+1)(n+1) = -(n+1)^2 \frac{(2n-1)!}{2^{n-1}(n-1)!}$ then, by recursively applying the equation,

$$ b_{n+2}= r_{n+1} + 2nb_{n+1} = r_{n+1} + 2n (r_{n} + 2(n-1) ( \cdots ( r_{2} +2b_{2}) \cdots ) \\ = r_{n+1} + 2n r_{n} + 2^2 n(n-1) r_{n-1} + \cdots + 2^{n-1}n(n-1)\cdots(3)(2)(r_{2} +2b_{2})\\ = 2^{n}n!b_{2} + r_{n+1} + \sum_{k=2}^{n} r_k 2^{n+1-k}(k)(k+1)\cdots(n) \\ = 2^{n}n!b_{2} + r_{n+1} + \sum_{k=2}^{n} r_k 2^{n+1-k}\frac{n!}{(k-1)!} \\ = - 2^{n}n! - \sum_{k=2}^{n+1} 2^{n+1-k}\frac{n!}{(k-1)!} k^2 \frac{(2k-3)!}{2^{k-2}(k-2)!} = \\ - \frac{ (n^2 + 4 n + 5) (2 n + 1)!}{5 n! 2^n } $$ where the last step was done with the help of Wolfram Alpha.

Now, having obtained a closed formula for the $b_n$, you can use the very same recursive application method with the equation $c_{n+2}=(2n-3)c_{n+1}-b_{n}(n+1)(n-2)$. This will give you a closed formula for the $c_n$. Continue in this way for $d_n$ etc. Hence you obtain your full $P_n$.

Note that this indeed works since all recursive equations are of the same type $a^{k+1}_{n+2}=(2n-3(k-1))a^{k+1}_{n+1}-a^{k}_{n}(n+1)(n+1-3(k-1))$ where $a^{k}$ stands for the $k$-th letter of the alphabet.

I conjecture that the result will be, with polynomials $p^{k}(n) $, $$ a^{k}_{n+2} = p^{k}(n) \frac{(2(n+k-1)-1)!}{2^{n-2+k}(n-2+k)!} $$ This holds with $p^{1}(n) =1$ and $p^{2}(n) = - \frac{n^2 + 4 n + 5}{5}$

Answer 2

So we are given that $$ \left\{ \matrix{ P_{\,0} (x) = x,\quad P_{\,1} (x) = x^{\,2} \hfill \cr P_{\,n + 2} (x) = x^{\,2} P\,'_{\,n + 1} (x) + \left( {n + 1} \right)x\,P_{\,n + 1} (x) - \left( {n + 1} \right)P\,'_{\,n} (x) \hfill \cr} \right. $$

We rewrite it, so as to inglobate the initial conditons into the recurrence $$ P_{\,n} (x) = x^{\,2} P\,'_{\,n - 1} (x) + \left( {n - 1} \right)x\,P_{\,n - 1} (x) - \left( {n - 1} \right)P\,'_{\,n - 2} (x) + x\left[ {n = 0} \right] $$ where $[P]$ denotes the Iverson bracket.
We can clearly deduce that $P_n (x)$ is polynomial of degree $n+1$.

Putting $$ \left\{ \matrix{ P_{\,n} (x) = \sum\limits_{0\, \le \,k\,\, \left( { \le \,n + 1} \right)} {a_{\,n,\,k} x^{\,k} } \hfill \cr Q(x,y) = \sum\limits_{0\, \le \,n} {P_{\,n} (x)y^{\,n} } = \sum\limits_{0\, \le \,n} {\sum\limits_{0\, \le \,k\,\left( { \le \,n + 1} \right)} {a_{\,n,\,k} x^{\,k} y^{\,n} } } \hfill \cr} \right. $$ then we get $$ \bbox[lightyellow] { y^{\,3} Q_{\,x\,y} + y\left( {y - x^{\,2} } \right)Q_{\,x} - x\,y^{\,2} Q_{\,y} + Q = x } \tag{1}$$ which confirms the PDE you already found

On the other hand, putting $$ R(x,y) = \sum\limits_{0\, \le \,n} {\,P_{\,n} (x){{y^{\,n} } \over {n!}}} $$ then the identity you give in the edit to your post becomes $$ \eqalign{ & \left( {x^{\,2} - y} \right)\sum\limits_{0\, \le \,n} {\,P\,'_{\,n} (x){{y^{\,n} } \over {n!}}} = \left( {x^{\,2} - y} \right)R_{\,x} = \cr & = \left( {1 - x\,y} \right)\sum\limits_{0\, \le \,n} {\,P_{\,n + 1} (x){{y^{\,n} } \over {n!}}} = \left( {1 - x\,y} \right){d \over {dy}}\sum\limits_{0\, \le \,n} {\,P_{\,n + 1} (x){{y^{\,n + 1} } \over {\left( {n + 1} \right)!}}} = \cr & = \left( {1 - x\,y} \right){d \over {dy}}\sum\limits_{1\, \le \,n} {\,P_{\,n} (x){{y^{\,n} } \over {n!}}} = \left( {1 - x\,y} \right){d \over {dy}}\sum\limits_{0\, \le \,n} {\,P_{\,n} (x){{y^{\,n} } \over {n!}}} = \cr & = \left( {1 - x\,y} \right)R_{\,y} \cr} $$ i.e., the 1st order linear homogeneous PDE $$ \bbox[lightyellow] { \left( {x^{\,2} - y} \right)R_{\,x} - \left( {1 - x\,y} \right)R_{\,y} = 0 } \tag{2}$$

Unfortunately this PDE is not readily solvable.

However, being $$ R(x,y) = \sum\limits_{0\, \le \,n} {\,P_{\,n} (x){{y^{\,n} } \over {n!}}} = \sum\limits_{\left\{ {\matrix{ {0\, \le \,m\,\,\left( { \le \,n + 1} \right)} \cr {0\, \le \,n} \cr } } \right.} {r_{\,n,\,m} \,x^{\,m} y^{\,n} /n!} $$ we get the following recurrence for the $r$ coefficients $$ \bbox[lightyellow] { r_{\,n,\,m} = \left( {m + n - 2} \right)r_{\,n - 1,\,m - 1} - \left( {n - 1} \right)\left( {m + 1} \right)r_{\,n - 2,\,m + 1} + \left[ {0 = n} \right]\left[ {1 = m} \right] } \tag{3}$$ where it is understood that whenever one of the indices is negative the coefficient $r$ is null.

The recurrence becomes more readable if we write it by diagonals. That is, putting $n=m+d$, with $-1 \le d$, we get $$ \bbox[lightyellow] { r_{\,m + d,\,m} = \left( {d + 2m - 2} \right)r_{\,m - 1 + d,\,m - 1} - \left( {m + d - 1} \right)\left( {m + 1} \right)r_{\,m + 1 + d - 3,\,m + 1} + \left[ { - 1 = d} \right]\left[ {1 = m} \right] } \tag{4}$$ which tells that the matrix of the coefficients
- is lower triangular, except for the upper diagonal with $d=-1$;
- that only the diagonals $d \equiv 2\;\left( {\bmod 3} \right)$ are not null;
- that the upper diagonal is the double factorial $(2m-3)!!$, taken as $1$ for $m=1$,
i.e. sequence A001147;
- and that for the following diagonals we can apply the above recurrence.

From this we get $$ \bbox[lightyellow] { \eqalign{ & R(x,y) = \sum\limits_{0\, \le \,n} {\,P_{\,n} (x){{y^{\,n} } \over {n!}}} = \sum\limits_{\left\{ {\matrix{ {0\, \le \,m\,\,\left( { \le \,n + 1} \right)} \cr {0\, \le \,n} \cr } } \right.} {r_{\,n,\,m} \,x^{\,m} y^{\,n} /n!} = \cr & = \sum\limits_{\left\{ {\matrix{ { - 1\, \le \,d} \cr {0\, \le \,m,\;m + d} \cr } } \right.} {r_{\,m + d,\,m} \,{{x^{\,m} y^{\,m + d} } \over {\left( {m + d} \right)!}}} = \sum\limits_{1\, \le \,m} {r_{\,m - 1,\,m} \,{{x^{\,m} y^{\,m - 1} } \over {\left( {m - 1} \right)!}}} + \sum\limits_{\left\{ {\matrix{ {0\, \le \,d} \cr {0\, \le \,m} \cr } } \right.} {r_{\,m + d,\,m} \,{{x^{\,m} y^{\,m + d} } \over {\left( {m + d} \right)!}}} = \cr & = x\sum\limits_{0\, \le \,m} {r_{\,m,\,m + 1} \,{{x^{\,m} y^{\,m} } \over {m!}}} + \sum\limits_{\left\{ {\matrix{ {0\, \le \,j} \cr {0\, \le \,m} \cr } } \right.} {r_{\,m + 3j + 2,\,m} \,{{x^{\,m} y^{\,m + 3j + 2} } \over {\left( {m + 3j + 2} \right)!}}} = \cr & = x\sum\limits_{\left\{ {\matrix{ {0\, \le \,m} \cr {\left( {\left( { - 1} \right)!! = 1} \right)} \cr } } \right.} {\left( {2m - 1} \right)!!\,{{\left( {x\,y} \right)^{\,m} } \over {m!}}} + y^{\,2} \sum\limits_{\left\{ {\matrix{ {0\, \le \,j} \cr {0\, \le \,m} \cr } } \right.} {q_{j,\,m} \,{{\left( {x\,y} \right)^{\,m} \left( {y^{\,3} } \right)^{\,j} } \over {\left( {m + 3j + 2} \right)!}}} = \cr & = x\sum\limits_{0\, \le \,m} {\left( \matrix{ 2m \cr m \cr} \right)\left( {x\,y/2} \right)^{\,m} } + y^{\,2} \sum\limits_{\left\{ {\matrix{ {0\, \le \,j} \cr {0\, \le \,m} \cr } } \right.} {q_{j,\,m} \,{{\left( {x\,y} \right)^{\,m} \left( {y^{\,3} } \right)^{\,j} } \over {\left( {m + 3j + 2} \right)!}}} = \cr & = {x \over {\sqrt {1 - 2x\,y} }} + y^{\,2} \sum\limits_{\left\{ {\matrix{ {0\, \le \,j} \cr {0\, \le \,m} \cr } } \right.} {q_{j,\,m} \,{{\left( {x\,y} \right)^{\,m} \left( {y^{\,3} } \right)^{\,j} } \over {\left( {m + 3j + 2} \right)!}}} = \cr & = {x \over {\sqrt {1 - 2x\,y} }} + y^{\,2} Q\left( {xy,y^{\,3} } \right) \cr} } \tag{5.a}$$ which gives an idea of the structure of $R(x,y)$, and where $$ \bbox[lightyellow] { \left\{ \matrix{ q_{\,n,\,\,m} = 0\quad \left| {\;n < 0\; \vee \;m < 0} \right. \hfill \cr q_{\,n,\,\,m} = r_{\,m + 3n + 2,\,\,m} \hfill \cr q_{\,n,\,\,m} = \left( {3n + 2m} \right)q_{\,n,\,\,m - 1} - \left( {m + 3n + 1} \right)\left( {m + 1} \right)q_{\,n - 1,\,\,m + 1} - \left[ {0 = n} \right]\left( {m + 1} \right)^{\,2} {{\left( {2m} \right)!} \over {2^{\,m} m!}} \hfill \cr} \right. } \tag{5.b}$$ ---- update 24 . 3 -------

The neat PDE you have found $$ \left\{ \matrix{ U(x,y) = \sum\limits_{0\, \le \,n} {P_{\,n} (x + y){{\left( {xy} \right)^{\,n} } \over {n!}}} \hfill \cr \left( {1 + x^{\,3} } \right)U_x - \left( {1 + y^{\,3} } \right)U_y = 0 \hfill \cr} \right. $$ is solvable. In fact, as per this eqworld link, putting $$ I(x,y) = \int {{{dx} \over {\left( {1 + x^{\,3} } \right)}}} + \int {{{dy} \over {\left( {1 + y^{\,3} } \right)}}} $$ which results in a complicated expression in terms of $\ln$ and $\arctan$, then any function of $I(x,y)$ $$ U(x,y) = \Psi \left( {I(x,y)} \right) $$ will be a solution, as is readily verified.
The problem is that unfortunately, being $I(x,y)$ quite complex, it is not easy to find a solution that verifies the starting condition $U(x,0)=P_0(x)=x$.
Actually, I am convinced that the kernel of the problem lies in that the formal power series above is not convergent, so that $U(x,y)$ does not have a closed form, not even in a small domain around the origin. A look at (4) and (5.b) supports the convinction.

Answer 3

This question reminds me of some problem I have been struggling with for a long time. Andreas above has clearly explained that it is fairly easy to find closed forms for the consecutive coefficients by going top -down in the order of the coefficient. As a matter of fact it didn't take me long by using Andreas' approach to find the coefficients: \begin{eqnarray} a_n &=& 2^{-n} (n+1) \frac{\Gamma(2n+1)}{\Gamma(n+2)} & \mbox{if $n\ge 0$}\\ b_n&=& -\frac{2^{2-n}}{5} (1+n^2) \cdot \frac{\Gamma(2n-2)}{\Gamma(n-1)} & \mbox{if $n\ge 2$}\\ c_n&=& \frac{ 2^{1-n}}{25} \left(10+13 n-7 n^2+4 n^3 \right) \cdot \frac{\Gamma(2n-5)}{\Gamma(n-4)} & \mbox{if $n\ge 5$}\\ d_n&=& -\frac{2^{3-n}}{4125} \left(420+97 n+466 n^2-187 n^3+44 n^4\right) \cdot \frac{\Gamma(2n-8)}{\Gamma(n-7)} & \mbox{if $n\ge 8$}\\ e_n&=&\frac{2^{1-n}}{20625} \left(8580+8839 n-4140 n^2+5025 n^3-1320 n^4+176 n^5 \right)\cdot \frac{\Gamma(2n-11)}{\Gamma(n-10)} & \mbox{if $n\ge 11$}\\ f_n&=&-\frac{2^{3-n}}{8765625} \left(930930+417143 n+756683 n^2-404005 n^3+192525 n^4-34408 n^5+2992 n^6\right) \cdot \frac{\Gamma(2n-14)}{\Gamma(n-13)} & \mbox{if $n\ge 14$}\\ g_n&=&\frac{2^{2-n}}{1446328125}\left(313331760+272072562 n-63004435 n^2+149739602 n^3-60171925 n^4+16703588 n^5-2139280 n^6+131648 n^7\right)\cdot \frac{\Gamma(2n-17)}{\Gamma(n-16)} & \mbox{if $n\ge 17$} \end{eqnarray} As Andreas pointed out we can clearly see a pattern in here; the coefficient involves a ratio of gamma functions, a power of two and a polynomial of known order. However the frustrating thing about this question is that it is not quite clear how to find the polynomials for generic order. Maybe inserting the conjectured form into the recurrence relations and solving those will help us achieve this goal.

Update: Clearly we have: \begin{equation} a^{(k)}_n = 2^{-n} \cdot \left( \sum\limits_{l=0}^k \theta^{(k)}_l n^l\right) \cdot \frac{\Gamma(2 n+4-3 k)}{\Gamma(n+5-3 k)} \end{equation} where $k=1,2,\cdots$ and $a^{(k)}_\cdot$ denotes the $k$th letter in the alphabet. Now, the coefficients $\left\{\theta^{(k)}_l\right\}_{l=0}^k$ in question satisfy the following recurrence relations: \begin{eqnarray} &&(3-3 k) \theta^{(k)}_l -2 \sum\limits_{l_1=l}^k \binom{l_1}{l-1} \theta^{(k)}_{l_1} (-1)^{l_1-l+1}-(8-6 k) \sum\limits_{l_1=l}^k \binom{l_1}{l} \theta^{(k)}_{l_1} (-1)^{l_1-l} = \\ &&-4 \sum\limits_{l_1=l-1}^{k-1} \binom{l_1}{l-1} \theta^{(k-1)}_{l_1} (-2)^{l_1-l+1} + 4 \sum\limits_{l_1=l}^{k-1} \binom{l_1}{l} \theta^{(k-1)}_{l_1} (-2)^{l_1-l} \end{eqnarray} and they read: \begin{eqnarray} \theta^{(k)}_k &=& \frac{(-1)^{k-1}}{1} \cdot (\frac{4}{5})^{k-1} \cdot \frac{k}{(k-0)!}\\ \theta^{(k)}_{k-1} &=& \frac{(-1)^k}{8} \cdot (\frac{4}{5})^{k-1} \cdot \frac{(-10-k+3 k^2)}{(k-1)!}\\ \theta^{(k)}_{k-2} &=& \frac{(-1)^{k-1}}{4224} \cdot (\frac{4}{5})^{k-1} \cdot \frac{(8040-3158 k+31 k^2+297 k^3)}{(k-2)!}\\ \theta^{(k)}_{k-3} &=& \frac{(-1)^k}{33792} \cdot (\frac{4}{5})^{k-1} \cdot \frac{(-75960+26930 k-9073 k^2+390 k^3+297 k^4)}{(k-3)!}\\ \theta^{(k)}_{k-4} &=& \frac{(-1)^{k-1}}{3033169920} \cdot (\frac{4}{5})^{k-1} \cdot \frac{(8778619200 - 1962039736 k + 651072294 k^2 - 174815675 k^3 + 8229870 k^4 + 2499255 k^5)}{(k-4)!}\\ \vdots \end{eqnarray}

Answer 4

HINT

Let $$f_n(x) = P_{n+1}(x),$$ then $$f_{n+1}(x) = x^2f'_n(x) + (n-1)xf_n(x) - (n-1)f'_{n-1}(x)$$ (the index corresponds to the highest power of a polynomial).

One can prove that $$f_{3k}(x)= u_k(x^3),\quad f_{3k+1}(x) = xv_k(x^3), \quad f_{3k+2}(x) = x^2w_k(x^3).$$ Really, $$f_1(x) = xv_0(x^3),\quad f_2(x)= x^2w_0(x^3),$$ where $$v_0(y) = w_0(y)= 1.$$ Then, $$f_{3k+3}(x) = x^2(x^2w_k(x^3))'+(3k+1)x(x^2w_k(x^3)) - (3k+1)(xv_k(x^3))',$$ $$f_{3k+3}(x) = 3x^3(x^3w'_k(x^3) + (k+1)w_k(x^3)) - (3k+1)(3x^3v'_k+v_k(x^3))') = u_{k+1}(x^3),$$ $$\boxed{u_{k+1}(y) = 3y(yw'_k(y)-(3k+1)v'_k(y))+3(k+1)yw_k(y)-(3k+1)v_k(y)},\tag1$$ $$\boxed{u_{k+1} = 3y(yw_k-(3k+1)v_k)'+3k(yw_k-(3k+1)v_k)+(9k^2-1)v_k},\tag{1'}$$ $$f_{3k+1}(x) = x^2(u_k(x^3))'+(3k-1)x(u_k(x^3)) - (3k-1)(x^2w_{k-1}(x^3))',$$ $$f_{3k+1}(x) = 3(x^3u'_k(x^3) + (3k-1)u_k(x^3)) - (3k-1)x(3x^3w'_{k-1}(x^3)+2w_{k-1}(x^3))=xv_k(x^3),$$ $$\boxed{v_k(y) = 3(yu'_k(y)-(3k-1)w'_{k-1}(y))+(3k-1)(u_k(y)-2w_{k-1}(y))},\tag2$$ $$\boxed{v_{k+1} = 3(yu_{k+1}-(3k+2)w_k)'+(3k-1)u_{k+1}-2(3k+2)w_k)},\tag{2'}$$ $$f_{3k+2}(x) = x^2(xv_k(x^3))'+3kx(xv_k(x^3)) - 3k(u_k(x^3))',$$ $$f_{3k+2}(x) = x^2(3x^3v'_k(x^3) + (3k+1)v_k(x^3) - 3kx(3x^2u'_k(x^3)=x^2w_k(x^3),$$ $$\boxed{w_k(y) = 3(yv'_k(y)-3ku'_k(y))+(3k+1)v_k(y)},\tag3$$ $$\boxed{w_k = 3(yv_k-3ku_k)'+(3k-2)v_k},\tag{3'}$$ and one can get $$u_0(y) = 1,\quad u_1(y) = 3y-1,\quad u_2(y) = 945y^2 - 546y +24, \quad u_3(y) = 2027025 y^3 - 1756755 y^2 + 323946 y - 4410\dots,\tag4$$ $$v_0(y)= 1,\quad v_1(y) = 15y-6,\quad v_2(y) = 10935y^2 -6993y +630\dots,\tag5$$ $$w_0(y) = 1,\quad w_1(y) = 105y-51,\quad w_2(y) = 135135y^2 - 103950 y + 14238\dots\tag6$$ Conditions $(1)-(6)$ has the standard form and looks useful.

For example, $$945 = 3\cdot5\cdot7\cdot9,\quad 546 = 5\cdot 7 \cdot 9 + 3\cdot7\cdot11,$$ and there is not a problem to find the coefficients $g_2(y),\ g_1(y),\ a_n$ in the ODE $$g_2(x)u''(x)+g_1(x)u'(x)+a_nu(x) = 0,\tag7$$ where $g_2(x)$ and $g_1(x)$ is the same for all of $u_n(y),$ and $a_n$ is the constant.

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HOW TO FIND $\mathbf{g_2, g_1, a_n}.$

Substitutions of $(4)$ to $(7)$ give $$g_2(x)(3x-1)''+g_1(x)(3x-1)'+a_1(3x-1) = 0,$$ $$g_2(x)(945y^2 - 546y +24)''(x)+g_1(x)(945y^2 - 546y +24)'(x)+a_2(945y^2 - 546y +24) = 0,$$ $$g_2(x)(2027025 y^3 - 1756755 y^2 + 323946 y - 4410)''+g_1(x)(2027025 y^3 - 1756755 y^2 + 323946 y - 4410)'+a_3(2027025 y^3 - 1756755 y^2 + 323946 y - 4410) = 0.$$ These leads to the system of identities $$\begin{cases} 3g_1(x) + a_1 (3x-1) = 0\\ 1890g_2(x) + (1890x-546)g_1(x) + a_2(945y^2 - 546y +24) =0\\ (6\cdot2027025 y - 2\cdot1756755)g_2(x) + (3\cdot2027025 y^2 - 2\cdot1756755 y + 323946 )g_2(x) + a_3(2027025 y^3 - 1756755 y^2 + 323946 y - 4410) = 0\dots \end{cases}\tag8$$ Easy so see that $g_1$ and $g_2$ are polynomials.

The first coefficient $a_1$ is the scale factor and can be chosen arbitrarily, $a_2$ and $a_3$ can be calculated from $(8)$ using the highest coefficients comparison. The others have to be obtained semiempirically, probably using some next polynomials $u_n$ for the control.