The Chebyshev polynomials $(T\_k)\_{k \in \mathbb{N}\_0}$ are defined recursively by
$$
T\_0(x)=1 , \ \ T\_1(x)=x, \ \ T\_{k+1}(x)=2x\,T\_k(x)-T\_{k-1}(x) \ .
$$
With this one can find the explicit formulas
\begin{align}
& T\_{2l}(x) = l \sum\limits\_{j=0}^{l} (-1)^{l-j} \frac{(l+j-1)!}{(l-j)!(2j)!} \, (2x)^{2j}\\
& T\_{2l+1}(x) = \frac{2l+1}{2} \sum\limits\_{j=0}^l (-1)^{l-j} \frac{(l+j)!}{(l-j)!(2j+1)!} \, (2x)^{2j+1}
\end{align}
for all $l \in \mathbb{N}$. However I needed to normalize these polynomials into
\begin{align\*}
& \widetilde{T}\_{2l}(x) = l \sum\limits\_{j=0}^{l} (-1)^{l-j} \frac{(l+j-1)!}{(l-j)!(2j)!} \left( x^{2j} - \frac{1}{j+1} {2j \choose j} \right)\\
& \widetilde{T}\_{2l+1}(x) = \frac{2l+1}{2} \sum\limits\_{j=0}^l (-1)^{l-j} \frac{(l+j)!}{(l-j)!(2j+1)!} \, x^{2j+1}
\end{align\*}
for $l \in \mathbb{N}$ and $\widetilde{T}\_0(x)=0, \, \widetilde{T}\_1(x) = \frac{x}{2}$. I'm pretty sure any link to the recursion formular of the (non-normalized) Chebyshev polynomials is gone. Can anyone help me find out if there is a new recursion formula for $\widetilde{T}\_{k+1}$, depending on $\widetilde{T}\_0,...,\widetilde{T}\_{k}$? I think it might have some interesting applications for a certain graph theoretical setting.


Any help is much appreciated.



Recursive formula from given explicit formula for normalized Chebyshev polynomials
If I read your formulas correctly you define $\tilde T\_{2l}(x)=T\_{2l}(x/2)-C\_l$ and $\tilde T\_{2l+1}(x)=T\_{2l+1}(x/2)$, where
$$C\_l=l\sum\_{j=0}^l (-1)^{l-j}\frac{(l+j-1)!}{(l-j)!(2j)!}\frac 1{j+1}\binom{2j}j.$$
This can be rewritten
$$C\_l=(-1)^l l!\sum\_{j=0}^l\frac{(-l)\_j(l)\_j}{j!(2)\_j},$$
where $(a)\_j=a(a+1)\dotsm(a+j-1)$. By the Chu-Vandermonde summation formula,
$$C\_l=(-1)^ll!\frac{(2-l)\_l}{(2)\_l}=\begin{cases}1, & l=0,\\
-1/2, & l=1,\\ 0, & l\geq 2.\end{cases}$$
So your polynomials are given by
$$\tilde T\_k(x)=T\_k(x/2)-\delta\_{0,k}-\frac 12\delta\_{2,k}.$$
They satisfy the recursion
$$\tilde T\_{k+1}(x)=x\tilde T\_k(x)-\tilde T\_{k-1}(x),\qquad k\geq 4.$$


