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answer
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How to find the coefficient of $x^k$ in the expression $\prod_{p=2}^n (1+xp) $
I got this general formula for $ n\in N$ (I showed it here)
$$\int_0^1 \left(\frac{x}{1-x} \ln x \right)^n dx=n \sum_{p=0}^{n-1}a(n,p+1) (-1)^{n-p} \zeta(p+2)+n! $$
where $a(n,k)$ is the coefficient ...
8
votes
2
answers
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integral transform of Fibonacci polynomials is integral
The Fibonacci polynomials are defined recursively by $F_0(x)=0, F_1(x)=1$ and $F_n(x)=xF_{n-1}(x)+F_{n-2}(x)$, for $n\geq2$.
While computing certain integrals, I observe the following (numerically) ...