In this section, an existence of breaking point for termination of *Q *_{d}_{,ℓ}(*n*) is investigated. It is a relatively easy fact that *Q*_{2,3}(*n*) is a finite sequence due to *Q*_{2,3}(66) = 73 [21]. On the other hand, our following analysis confirms that the termination of *Q*_{2,3}(*n*) is an exceptional behaviour for *Q*_{d}_{,ℓ}(*n*), at least, in the range of experiments that the next section focuses on.

#### Proposition 2.1

*Q*_{d}_{,ℓ}(*n*) *dies for*$d<\frac{\ell \cdot (\ell -1)-1+(-1{)}^{\ell}}{2}$*where ℓ* ≥ 3.

*Proof*. It is clear that if *Q*_{d}_{,ℓ}(*d* · *ℓ* + 1) > *d* · *ℓ* + 1, then *Q*_{d}_{,ℓ}(*n*) dies immediately thereafter. Since${Q}_{d,\ell}(n)=\u2308\frac{n\cdot (\ell -1)}{\ell}\u2309$for *n* ≤ *d* · *ℓ*,
$$\begin{array}{rl}{Q}_{d,\ell}(d\cdot \ell +1-i)& =\u2308\frac{(d\cdot \ell +1-i)\cdot (\ell -1)}{\ell}\u2309\\ & =\u2308d\cdot \ell +1-(d+i)+\frac{i-1}{\ell}\u2309\\ & =d\cdot \ell +1-(d+i)+\u2308\frac{i-1}{\ell}\u2309\end{array}$$

for 1 ≤ *i* ≤ *d* · *ℓ*.

So *Q*_{d}_{,ℓ}(*d* · *ℓ* + 1) can be expressed as follows:
$$\begin{array}{rl}{Q}_{d,\ell}(d\cdot \ell +1)& =\sum _{i=1}^{\ell}{Q}_{d,\ell}(d\cdot \ell +1-{Q}_{d,\ell}(d\cdot \ell +1-i))\\ & =\sum _{i=1}^{\ell}{Q}_{d,\ell}\phantom{\rule{negativethinmathspace}{0ex}}\left(d\cdot \ell +1-\left(d\cdot \ell +1-(d+i)+\u2308\frac{i-1}{\ell}\u2309\right)\right)\\ & =\sum _{i=1}^{\ell}{Q}_{d,\ell}\phantom{\rule{negativethinmathspace}{0ex}}\left(d+i-\u2308\frac{i-1}{\ell}\u2309\right)\\ & ={Q}_{d,\ell}(d+1)+\sum _{i=1}^{\ell -1}{Q}_{d,\ell}(d+i)\\ & =\u2308\frac{(d+1)\cdot (\ell -1)}{\ell}\u2309+\sum _{i=1}^{\ell -1}\u2308\frac{(d+i)\cdot (\ell -1)}{\ell}\u2309\\ & =d+1-\u230a\frac{d+1}{\ell}\u230b+\sum _{i=1}^{\ell -1}(d+i)-\sum _{i=1}^{\ell -1}\u230a\frac{d+i}{\ell}\u230b\\ & =d\cdot \ell +1+\frac{\ell \cdot (\ell -1)}{2}-\u230a\frac{d+1}{\ell}\u230b-\sum _{i=1}^{\ell -1}\u230a\frac{d+i}{\ell}\u230b.\end{array}$$

So, the inequality *Q*_{d}_{,ℓ}(*d* · *ℓ* + 1) > *d* · *ℓ* + 1 is equivalent to the inequality
$$\u230a\frac{d+1}{\ell}\u230b+\sum _{i=1}^{\ell -1}\u230a\frac{d+i}{\ell}\u230b<\frac{\ell \cdot (\ell -1)}{2}.$$(1)

If$d=\frac{\ell \cdot (\ell -1)}{2}$in Inequality 1, total value of *ℓ* terms of the left side is$\frac{\ell \cdot (\ell -1)}{2}$for *l* ≥ 3, and the inequality is not satisfied. Note that$\frac{\ell \cdot (\ell -1)}{2}\equiv 0$(mod *ℓ*) if *ℓ* is odd and$\frac{\ell \cdot (\ell -1)}{2}\equiv \frac{\ell}{2}$(mod *ℓ*) if *ℓ* is even. So, if$d=\frac{\ell \cdot (\ell -1)}{2}-1,$the left side decreases from$\frac{\ell \cdot (\ell -1)}{2}$if *ℓ* is even (thereby satisfying the inequality), but the left side does not change from before if *ℓ* is odd. In the odd case, if$d=\frac{\ell \cdot (\ell -1)}{2}-2,$the left side would decrease, though. So, if *ℓ* is odd, the largest value of *d* that satisfies Inequality 1 is$\frac{\ell \cdot (\ell -1)}{2}-2,$and if *ℓ* is even, the largest value of *d* that satisfies Inequality 1 is$\frac{\ell \cdot (\ell -1)}{2}-1.$These two cases can be combined into one inequality:$d<\frac{\ell \cdot (\ell -1)-1+(-1{)}^{\ell}}{2}.$

#### Proposition 2.2

${Q}_{d,\ell}(d\cdot \ell +2)=d\cdot (\ell -1)+\lfloor \frac{(\ell +3)}{2}\rfloor $*for*$d=\frac{\ell \cdot (\ell -1)-1+(-1{)}^{\ell}}{2}$*where ℓ* ≥ 5.

*Proof*. We follow similar steps with previous proof thanks to its result. So *Q*_{d}_{,ℓ}(*d* · *ℓ* + 2) can be expressed as follows:
$$\begin{array}{rl}{Q}_{d,\ell}(d\cdot \ell +2)& =\sum _{i=1}^{\ell}{Q}_{d,\ell}(d\cdot \ell +2-{Q}_{d,\ell}(d\cdot \ell +2-i))\\ & ={Q}_{d,\ell}(d\cdot \ell +2-{Q}_{d,\ell}(d\cdot \ell +1))+\sum _{i=2}^{\ell}{Q}_{d,\ell}(d\cdot \ell +2-{Q}_{d,\ell}(d\cdot \ell +2-i))\\ & ={Q}_{d,\ell}(1)+\sum _{i=2}^{\ell}{Q}_{d,\ell}\phantom{\rule{negativethinmathspace}{0ex}}\left(d\cdot \ell +2-\left(d\cdot \ell +2-(d+i)+\u2308\frac{i-2}{\ell}\u2309\right)\right)\\ & =1+\sum _{i=2}^{\ell}{Q}_{d,\ell}\phantom{\rule{negativethinmathspace}{0ex}}\left(d+i-\u2308\frac{i-2}{\ell}\u2309\right)\\ & =1+{Q}_{d,\ell}(d+2)+\sum _{i=2}^{\ell -1}{Q}_{d,\ell}(d+i)\\ & =1+\u2308\frac{(d+2)\cdot (\ell -1)}{\ell}\u2309+\sum _{i=2}^{\ell -1}\u2308\frac{(d+i)\cdot (\ell -1)}{\ell}\u2309\\ & =d+3-\u230a\frac{d+2}{\ell}\u230b+\sum _{i=2}^{\ell -1}(d+i)-\sum _{i=2}^{\ell -1}\u230a\frac{d+i}{\ell}\u230b\\ & =d\cdot (\ell -1)+2+\frac{\ell \cdot (\ell -1)}{2}-\u230a\frac{d+2}{\ell}\u230b-\sum _{i=2}^{\ell -1}\u230a\frac{d+i}{\ell}\u230b.\end{array}$$

Since$\u230a\frac{d+1}{\ell}\u230b=\u230a\frac{d+2}{\ell}\u230b$for *l* ≥ 5 and$d=\frac{\ell \cdot (\ell -1)-1+(-1{)}^{\ell}}{2},$from Proposition 2.1,
$$\frac{\ell \cdot (\ell -1)}{2}-\u230a\frac{d+2}{\ell}\u230b-\sum _{i=2}^{\ell -1}\u230a\frac{d+i}{\ell}\u230b=\u230a\frac{d+1}{\ell}\u230b.$$(2)

Thanks to Equation 2, *Q*_{d}_{,ℓ}(*d* · *ℓ* + 2) can be expressed as follows for *l* ≥ 5,
$$\begin{array}{rl}{Q}_{d,\ell}(d\cdot \ell +2)& =d\cdot (\ell -1)+2+\u230a\frac{d+1}{\ell}\u230b\\ & =d\cdot (\ell -1)+2+\u230a\frac{\frac{\ell \cdot (\ell -1)-1+(-1{)}^{\ell}}{2}+1}{\ell}\u230b\\ & =d\cdot (\ell -1)+\lfloor \frac{(\ell +3)}{2}\rfloor .\end{array}$$

#### Proposition 2.3

${Q}_{d,\ell}(d\cdot \ell +3)=d\cdot (\ell -1)+\frac{7+(-1{)}^{\ell +1}}{2}$*for*$d=\frac{\ell \cdot (\ell -1)-1+(-1{)}^{\ell}}{2}$*where ℓ* ≥ 7.

*Proof*. We follow similar steps with the previous proof thanks to result of it. So *Q*_{d}_{,ℓ}(*d*·*ℓ*+3) can be expressed as follows:
$$\begin{array}{rl}{Q}_{d,\ell}(d\cdot \ell +3)& =\sum _{i=1}^{\ell}{Q}_{d,\ell}(d\cdot \ell +3-{Q}_{d,\ell}(d\cdot \ell +3-i))\\ & ={Q}_{d,\ell}(d\cdot \ell +3-{Q}_{d,\ell}(d\cdot \ell +2))+{Q}_{d,\ell}(d\cdot \ell +3-{Q}_{d,\ell}(d\cdot \ell +1))\\ & \phantom{\rule{0.2in}{0ex}}+\sum _{i=3}^{\ell}{Q}_{d,\ell}(d\cdot \ell +3-{Q}_{d,\ell}(d\cdot \ell +3-i))\\ & ={Q}_{d,\ell}(d\cdot \ell +3-d\cdot (l-1)-\lfloor \frac{(l+3)}{2}\rfloor )+{Q}_{d,\ell}(2)\\ & \phantom{\rule{0.2in}{0ex}}+\sum _{i=3}^{\ell}{Q}_{d,\ell}\phantom{\rule{negativethinmathspace}{0ex}}\left(d\cdot \ell +3-\left(d\cdot \ell +3-(d+i)+\u2308\frac{i-3}{\ell}\u2309\right)\right)\\ & =2+{Q}_{d,\ell}(d+3-\lfloor \frac{(\ell +3)}{2}\rfloor )+\sum _{i=3}^{\ell}{Q}_{d,\ell}\phantom{\rule{negativethinmathspace}{0ex}}\left(d+i-\u2308\frac{i-3}{\ell}\u2309\right)\\ & =2+{Q}_{d,\ell}(d+3-\lfloor \frac{(\ell +3)}{2}\rfloor )+{Q}_{d,\ell}(d+3)+\sum _{i=3}^{\ell -1}{Q}_{d,\ell}(d+i)\\ & =2+{Q}_{d,\ell}(d+3-\lfloor \frac{(\ell +3)}{2}\rfloor )+\u2308\frac{(d+3)\cdot (\ell -1)}{\ell}\u2309\\ & \phantom{\rule{0.2in}{0ex}}+\sum _{i=3}^{\ell -1}\u2308\frac{(d+i)\cdot (\ell -1)}{\ell}\u2309\\ & =d+5+{Q}_{d,\ell}(d+3-\lfloor \frac{(l+3)}{2}\rfloor )-\u230a\frac{d+3}{\ell}\u230b+\sum _{i=3}^{\ell -1}(d+i)-\sum _{i=3}^{\ell -1}\u230a\frac{d+i}{\ell}\u230b\\ & =d\cdot (\ell -2)+2+{Q}_{d,\ell}(d+3-\lfloor \frac{(l+3)}{2}\rfloor )+\frac{\ell \cdot (\ell -1)}{2}-\u230a\frac{d+3}{\ell}\u230b\\ & \phantom{\rule{0.2in}{0ex}}-\sum _{i=3}^{\ell -1}\u230a\frac{d+i}{\ell}\u230b.\end{array}$$

Since$\u230a\frac{d+2}{\ell}\u230b=\u230a\frac{d+3}{\ell}\u230b$for *l* ≥ 7 and$d=\frac{\ell \cdot (\ell -1)-1+(-1{)}^{\ell}}{2},$from Proposition 2.2,
$$\frac{\ell \cdot (\ell -1)}{2}-\u230a\frac{d+3}{\ell}\u230b-\sum _{i=3}^{\ell -1}\u230a\frac{d+i}{\ell}\u230b=\u230a\frac{d+1}{\ell}\u230b+\u230a\frac{d+2}{\ell}\u230b.$$(3)

Thanks to Equation 3, *Q*_{d}_{,ℓ}(*d* · *ℓ* + 3) can be expressed as follows for *l* ≥ 7,
$$\begin{array}{rl}{Q}_{d,\ell}(d\cdot \ell +3)& =d\cdot (\ell -2)+{Q}_{d,\ell}(d+3-\lfloor \frac{(l+3)}{2}\rfloor )+2+\u230a\frac{d+1}{\ell}\u230b+\u230a\frac{d+2}{\ell}\u230b\\ & =d\cdot (\ell -2)+{Q}_{d,\ell}(d+3-\lfloor \frac{(l+3)}{2}\rfloor )+2+\u230a\frac{\frac{\ell \cdot (\ell -1)-1+(-1{)}^{\ell}}{2}+1}{\ell}\u230b\\ & \phantom{\rule{0.2in}{0ex}}+\u230a\frac{\frac{\ell \cdot (\ell -1)-1+(-1{)}^{\ell}}{2}+2}{\ell}\u230b\\ & =d\cdot (\ell -2)+{Q}_{d,\ell}(d+3-\lfloor \frac{(\ell +3)}{2}\rfloor )+2\cdot \lfloor \frac{(\ell +1)}{2}\rfloor .\end{array}$$

In here if *ℓ* is odd, then$d=\frac{\ell \cdot (\ell -1)}{2}-1$and *Q*_{d}_{,ℓ}(*d* · *ℓ* + 3) can be expressed as follows,
$$\begin{array}{rl}{Q}_{d,\ell}(d\cdot \ell +3)& =d\cdot (\ell -2)+{Q}_{d,\ell}(\frac{(\ell -1{)}^{2}}{2})+\ell +1\\ & =d\cdot (\ell -2)+\u2308\frac{(\ell -1{)}^{3}}{2\cdot \ell}\u2309+\ell +1\\ & =d\cdot (\ell -2)+\frac{\ell \cdot (\ell -3)}{2}+\ell +3\\ & =d\cdot (\ell -2)+\frac{\ell \cdot (\ell -1)}{2}+3\\ & =d\cdot (\ell -1)+4.\end{array}$$

And if *ℓ* is even, then$d=\frac{\ell \cdot (\ell -1)}{2}$and similar way can be used,
$$\begin{array}{rl}{Q}_{d,\ell}(d\cdot \ell +3)& =d\cdot (\ell -2)+{Q}_{d,\ell}(\frac{(\ell -1{)}^{2}+3}{2})+\ell \\ & =d\cdot (\ell -2)+\frac{\ell \cdot (\ell -3)}{2}+\ell +3\\ & =d\cdot (\ell -2)+\frac{\ell \cdot (\ell -1)}{2}+3\\ & =d\cdot (\ell -1)+3.\end{array}$$

These two cases can be combined into one equation that is${Q}_{d,l}(d\cdot \ell +3)=d\cdot (\ell -1)+\frac{7+(-1{)}^{\ell +1}}{2}$

Since each computed term *Q*_{d}_{,ℓ}(*d* · *ℓ* + *k*) is also determinative for *Q*_{d}_{,ℓ}(*d* · *ℓ* + *k* + 1) for all *k* ≥ 1 by definition of our nested recurrences, the properties of further terms can also be shown with similar propositions. contains some related patterns which guarantee the successive computations of recursions without termination in these short intervals as below. See also decreasing order of corresponding bursts in Figure 2 for an example *Q*_{209,21}(*n*).

Fig. 2 Graph of *Q*_{209,21}$(n)-\frac{20}{21}\cdot n$for 4000 ≤ *n* ≤ 10000.

Table 1 Behaviour of *Q*_{d},_{ℓ}(*d* · *ℓ* + *k*) for *ℓ* ≥ 2 · *k* + 1 and *k* ≤ 10 where$d=\frac{\ell \cdot (\ell -1)-1+(-1{)}^{\ell}}{2}.$

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