Approximate recursion defined by alternating summation?

Question

Problem & Question

Given a constant $k\ge 2$, we define the sequence $a_k(0)=0,a_k(1)=1$ and for $n\ge2$:

$$ a_k(n)=\sum_{i=1}^{\left[\frac{n}{k}\right]} (-1)^{n-i+1} a_k(i) $$

Where $\left[\frac{n}{k}\right]$ gives integer part of $\frac{n}{k}$. (The $\left[\space\right]$ represents truncation.)

For example, for $k=2$ we have the following scatterplot: [Click to see animated gif].

It appear as two interchanging curves repeat [following shape] in larger forms as we increase $n$.

Given $k$, can we find a real equation that would approximate "the curves" given by $a_k(n)$?

That is, the plot given by $a_k(n)$ can be defined as $\Gamma_1\cup\Gamma_2$ where $\Gamma_1,\Gamma_2$ are those "the curves".

Specifically, $\Gamma_1,\Gamma_2$ are graphs of two real functions $\gamma_1(x),\gamma_2(x)$ where $\gamma_1(x)=-\gamma_2(x)$ as they are symmetric over $y$-axis. The goal is to define them for $x\in\mathbb R$. Currently, you can say that $a_k(n)$ approximates $\gamma_1,\gamma_2$ at points $x=n\in\mathbb N$.

If it is not clear what I mean, clicking on the previous link "[following shape]", you can see the curve above the $y$-axis as $\Gamma_1$ given by $\gamma_1(n)$, and the curve below $y$-axis as $\Gamma_2$ given by $\gamma_2(n)$.

For simplicity, we can restrict ourselves to $k=2$ if needed.

My idea

The curves $\gamma_1(n)\approx-\gamma_2(n)$ generated by $a_k(n)$ remind me of a sine function whose period is being stretched by some function $g(n)$, and whose amplitude is being increased by some function $h(n)\ne 0$.

Is it possible to find such $g,h$ such that the following "sine form":

$$f_k(x):=h_k(x)\sin\left(\pi g_k(x)\right)\approx \gamma_1(x)=-\gamma_2(x)\space ?$$

To answer my question? This needs to be defined for all real $x\gt 0$.

Notice the above $f_k(x)$ form would have roots (be zero) at $x=g_k^{-1}(m),m\in\mathbb N$.

That is, lets observe "near-zeros" of the sequence $a_k(n)$ - The points where the curves pass over the $y$-axis and and are closest to it. Here are first couple "near-zeros" for $k=2$:

$$a_2(n)\approx 0 \text{ at } n\approx 2,4.5,12.5,33.5,84.5,204.5,480.5,1102.5,2494.5,5568.5,\dots$$

Meaning that the $f_k$ (the $\gamma_1,\gamma_2$) should be zero (have a root) somewhere near these "near-zeros". In the context of my "sine form" $f$ mentioned above, and for $k=2$, this means that we want to find $g_2$ such that for $m=1,2,3,\dots$ we have:

$$ g_2^{-1}(m)=2,4.5,12.5,33.5,84.5,204.5,480.5,1102.5,2494.5,5568.5,\dots $$

For example, here is the plot of $a(n)$ for $n\le215$ showing a near-zero around $n\approx204.5$, where we have $a(204)=-a(205)=-40\approx0$. (Hence I've taken $n\approx \frac{204+205}{2} = 204.5$)

But, the problem here is:

• How to determine at which $n$ will $a_k(n)$ reach a "near-zero"?

The other problem that needs to be solved, is to find $h_k(n)$. For starters we need to know the growth of the absolute value of the sequence, $|a_k(n)|$? Then, the question remains:

• Can we interpolate (approximate) a "closed form" of $h_k(n)$, to get the "amplitude" of $f_k$?

Alternatively, is there a better $f_k$ form to search for, than my "sine form" $f_k$?

Answer 1

Not a complete answer, but we're getting somewhere.

First note that for $l < k$ and $n>0$, we have \begin{align*} a_k(kn+l) = \sum_{i=1}^{n} (-1)^{kn+l-i+1}a_k(i) = (-1)^l a_k(kn) \end{align*} Therefore, for any $j \geq 1$, we have \begin{align*} \sum_{i=jk}^{jk+k-1} (-1)^{n-i+1}a_k(i) = k(-1)^{n-jk+1}a_k(jk) \end{align*} We use this to find the following sum. \begin{align*} a_k(k^2n+k^2-k) =& \sum_{i=1}^{kn+k-1} (-1)^{k^2n+k^2-k-i+1}a_k(i) \\ =& \sum_{i=1}^{kn+k-1} (-1)^{k^2n-i+1}a_k(i) \\ =& (-1)^{k^2n} + \sum_{j=1}^{n} \sum_{i=jk}^{jk+k-1} (-1)^{k^2n-i+1}a_k(i) \\ =& (-1)^{k^2n} + k\sum_{j=1}^{n}(-1)^{k^2n-jk+1}a_k(jk) \end{align*} We also calculate the following sum. \begin{align*} a_k(k^2n+k^2) =& \sum_{i=1}^{kn+k}(-1)^{k^2n+k^2-i+1}a_k(i) \\ =& (-1)^ka_k(k^2n+k^2-k) + (-1)^{k+1}a_k(kn+k) \end{align*}

Now, for odd $k$, this is where my work stops, but if $k$ is even, we can do the following. \begin{align*} a_k(k^2n+k^2)-a_k(k^2n+k^2-k) =& -a_k(kn+k) \\ \frac{a_k(k^2n+k^2)-a_k(k^2n+k^2-k)}{k} =& \frac{-1}{k}a_k(kn+k) \end{align*} Supposing there is some differentiable $\gamma_k$ that approximates $a_k$, we use Lagrange's intermediate value theorem to state that \begin{align*} \frac{\gamma_k(k^2n+k^2)-\gamma_k(k^2n+k^2-k)}{k} =& \frac{-1}{k}\gamma_k(kn+k) \\ \gamma_k'(\xi) =& \frac{-1}{k}\gamma_k(kn+k) \end{align*} for some $\xi \in [k^2n+k^2-k,k^2n+k^2]$. Which means that as we take $n$ very large, we approximately get \begin{align*} \gamma_k'(\xi) = \frac{-1}{k}\gamma_k \left( \frac{\xi}{k} \right). \end{align*} We use the method of power series to solve this DE, assuming $\gamma_k(x) = \sum_{i=0}^{\infty}\alpha^{(k)}_i x^i$, and we find $(i+1)\alpha^{(k)}_{i+1} = \frac{-\alpha^{(k)}_i}{k^{i+1}}$. By induction, we find $\alpha^{(k)}_i = \frac{(-1)^i \alpha^{(k)}_0}{i!k^{\frac{i^2+i}{2}}}$.

Therefore, we must have \begin{align*} \gamma_k(x) =& \alpha_0^{(k)} \sum_{i=0}^{\infty} \frac{(-1)^i}{i!k^{\frac{i^2+i}{2}}} x^i \end{align*} Now, if you could check if for some $\alpha_0^{(k)}$, this comes close to the curve your sequence described, I would be happy.