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Another way to approach this from the generating function perspective. Notice that $$\sum_{k=l}^u (-1)^k [x^k]\, f(x) = [x^u]\,\frac{f(-x)}{1-x} - [x^{l-1}]\,\frac{f(-x)}{1-x},$$ and so the question r …

Here is an approach that leads to an integral equation for $$A(\nu,t) := \sum_{n=0}^{\infty}(-1)^{n}\frac{a_{n}(\nu)}{\nu} e^{Int}.$$ First we notice that $$(-1)^{n}\frac{a_{n}(\nu)}{\nu} = \frac{1}{ …

Rewriting the l.h.s. of the conjectured identity and using the properties of beta function, we have: \begin{split} \text{l.h.s.} &= \sum_{B, b} (-1)^{a+b}{b\choose a} {B-1\choose A-1}\frac{1}{(B+b)\bi …

.$$ As Brendan McKay pointed out, the asymptotics for $S(n)$ is the same as for $S'(n)$. …

From the generating function: $$\frac1{1-(x^a+x^{a+1}+\dots+x^b)}=\frac{1-x}{1-x-x^a+x^{b+1}}$$ it follows that the number of compositions of $n$ with parts in $[a,b]$ (given by the coefficient of $x^ …

The equation can be solved explicitly in terms of Lambert W function, which then enables accurate approximation of $n^{n/x}$. Let's introduce $y:=x^2$ to rewrite the equation as $$\Big(\frac{y}{c^2 n^ …

Just a rough idea. Let $\alpha, \beta$ be the zeros of $1+Ax+Bx^2$, then for $j\geq 1$ $$c_j = \left.\left(\frac{\partial}{\partial s}\right)^j \log( B(\alpha - e^s)(\beta-e^s) )\right|_{s=0} = \left. …

Here is a proof that if an odd integer $x>1$ satisfies (1), then $x$ is a perfect number. First, by using the property that $\varphi(nm)=n\varphi(m)$ whenever $\mathrm{rad}(n)\mid\mathrm{rad}(m)$, we …

I assume that the initial conditions $a_0,a_1,b_0,b_1$ and that $n\to +\infty$. Let $A(x):=\sum_{n\geq 0} a_n x^n$ and $B(x):=\sum_{n\geq 0} a_n x^n$. Then the recurrence relations become: $$\begin{ca …

Let $f(n,k):=\sum_{i=0}^k\binom{n}{i}$. It can be seen that $$\sum_{i=0}^k i\binom{n}{i} = \sum_{i=1}^k n\binom{n-1}{i-1} = n\cdot f(n-1,k-1).$$ So, it remains to evaluate $$\frac{n\cdot f(n-1,k-1)}{ …

Notice that the product of binomial coefficients can be expressed as a multinomial coefficient: $$\prod_{i=1}^m \binom{n_i}{n_{i+1}} = \binom{n_1}{n_1-n_2,n_2-n_3,\dots,n_{m-1}-n_m,n_m}.$$ Denoting $d …