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The answer is yes. Indeed, as noted at A007063, $$k(\theta_j)=3\theta_j-(j+1), $$ where $$\theta_j:=\sum_{i=0}^{j-1}2^{\lfloor i/3\rfloor}\ge2^{\lfloor(j-1)/3\rfloor}. $$ So, $$k(\theta_j)-\theta_j …

According to Mathworld (see also Wikipedia), $$P_{k,n}=1-\frac{F_{k,n}}{2^n},\tag{1}$$ where, for each natural $k$, $(F_{k,n})_{n=1}^\infty$ is the sequence of $k$-step Fibonacci numbers, defined rec …

Taking into account that $\binom pq=0$ for nonnegative integers $p$ and $q$ such that $q>p$, write $$S(n,m)=\frac{(n+m+1)!}{n!m!}T(n,m),$$ where \begin{align*} T(n,m)&:=\sum_{l\ge0}\frac{1}{n+m-l+1} …

Let us show that \begin{equation*} a(n)\le\exp(n^{1-A+o(1)}) \tag{1}\label{1} \end{equation*} (as $n\to\infty$). Indeed, for each $q\in(1-A,1)$, \begin{equation*} (k-1)^q-k^q\sim-qk^{q-1},\qua …

The integral in question can be rewritten as $$ \begin{aligned} I&:=\frac1{\sqrt n}\,\int_{-a\sqrt n}^{(1-a)\sqrt n} e^{-u^2}\Big(1-\exp\Big\{-\frac{u^4/n}{1-u^2/n}\Big\}\Big)\,du \\ &\le\frac1{\sqrt …

We have $$F_{-n}=\frac{(-a_-)^n-(-a_+)^n}{\sqrt5} \tag{1}\label{1}$$ with $$a_\pm:=\frac{1\pm\sqrt5}2.$$ We also have the easy formula for $\sum_{j=j_1}^{j_2}(p+qj)x^j$, which yields $$\sum_{i=1}^{n-1 …

We will obtain simple explicit Fibonacci-like expressions for the feasible numbers of the red and white balls in the jar, which we will denote by $r$ and $w$ respectively, with $n:=r+w$. We have $\bin …