Generating function for A300483 (related to Chebyshev polynomial of first kind)

Question

• Let $a(n)$ be A300483. Here $$ a(n) = 2\int\limits_{t \geqslant 0}T_n\left(\frac{t+1}{2}\right)\exp(-t)\,dt. $$ where $T_n(x)$ is $n$-th Chebyshev polynomial of first kind.
• Let $b(n)$ be an integer sequence with generating function $B(x)$ such that $$ B(x)=1+\frac{1-x^2}{1-x+x^2}\sum\limits_{i=0}^{\infty}i!\left(\frac{x}{1-x+x^2}\right)^i. $$

I conjecture that $$ b(n)=a(n). $$

Here is the PARI/GP program to compute $b(n)$:

upto1(n) = Vec(1+(1-x^2)/(1-x+x^2)*sum(i=0, n, i!*(x/(1-x+x^2))^i) + x*O(x^n))

Is there a way to prove it?

Answer 1

The integral in $a(n)$ can be recognized as Laplace transform of $T_n(\frac{1+x}{2})$ evaluated at $x=1$. That is, in the expansion of $T_n(\frac{1+x}{2})$ each $x^k$ is replaced with $k!$.

From the generating function for Chebyshev polynomials, we have $$\sum_{n\geq 0} T_n(\frac{x+1}2) t^n = \frac12 \frac{2-t(1+x)}{1-t(1+x)+t^2}=\frac{2-t-tx}{2(1-t+t^2)} \sum_{k\geq 0} \big(\frac{t}{1-t+t^2}\big)^k x^k,$$ which under $2$ times Laplace transform evaluated at $x=1$ gives $$\sum_{n\geq0} a(n)t^n=\frac{1}{1-t+t^2} \sum_{k\geq0} \big(\frac{t}{1-t+t^2}\big)^k ((2-t)k! -t(k+1)!)$$ $$=\frac{2-t}{1-t+t^2} + \frac{1}{1-t+t^2} \sum_{k\geq1} k!\bigg(\big(\frac{t}{1-t+t^2}\big)^k (2-t) - \big(\frac{t}{1-t+t^2}\big)^{k-1}t\bigg)$$ $$=\frac{2-t}{1-t+t^2} + \frac{1-t^2}{1-t+t^2} \sum_{k\geq1} k!\big(\frac{t}{1-t+t^2}\big)^k,$$ which matches the definition of $B(x)$.