Jump and hop randomness tests for binary sequences

Abstract

Linear complexity test included in the NIST test suite only checks whether or not the observed linear complexity is close to the expected linear complexity. To take full advantage of the information of the linear complexity profile, we propose two randomness tests including a jump test based on the jump complexity, i. e., the number of changes in the linear complexity profile of a sequence, and a hop test checking the sum of jump heights at intervals. By using an iterative algorithm we calculate some necessary statistical measures of a random sequence of length M, and combining with hypothesis test we determine whether a given binary sequence is random or not. Moreover, we provide the explicit formulae of the expectation and the variance of the statistical measure of the hop test. The computational complexity of the jump test and the hop test is the same as that of the linear complexity test. Additionally, we provide a type of sample which, having passed all tests in the NIST test suite, is rejected by the jump test and hop test. So the proposed tests deserve to be considered.

Appendix

In order to calculate the expectation and variance of $O_n$, we need to know its counting functions. Denote by $N_n(\sigma )$ the number of n-bit binary sequences with the odd hop sum $\sigma$, and $NL_n(L,\sigma )$ by the number of n-bit binary sequences with the linear complexity L and the odd hop sum $\sigma$.

From Proposition 1 or from the continued fraction method in [16], the number of n-bit binary sequences with a prescribed jump height sequence $jh_1, jh_2, \dots , jh_m$, and the linear complexity L is

$$\begin{aligned} \left\{ \begin{array}{ll} 2^L, &{}2L\le n \\ 2^{n-L} , &{}2L>n\\ \end{array}\right. , \end{aligned}$$

(9)

where $L=jh_1+jh_2+\cdots +jh_m$.

Let the last m-th jump be in position k of such an n-bit sequence, we have $jh_m=2L-k\ge 2L-n$, and obtain

$$\begin{aligned} jh_m\ge \max \{1, 2L-n\}. \end{aligned}$$

(10)

By (9), for calculating $NL_n(L,\sigma )$ we should count the number of sequences $\{jh_1, jh_2, \ldots , jh_m\}$ with $jh_1+jh_2+\cdots +jh_m=L$, $jh_1+jh3+\cdots =\sigma$, and $jh_m \ge \max \{1, 2L-n\}$. Next, we provide related lemmas.

Lemma 1

The number of positive integer sequences $S=\{s_1,s_2,\dots ,s_m\}$ satisfying $\sum _{i=1}^m s_i=h$ is

$$\begin{aligned} \left\{ \begin{array}{lll} \mathrm {C}_{h-1}^{m-1}, &{}0<m\le h\\ 0, &{}otherwise \end{array}\right. . \end{aligned}$$

Lemma 2

The number of positive integer sequences $S=\{s_1,\dots ,s_{last}\}$ satisfying $\sum _{i\ge 1} s_i=h$ and $\sum _{j \ge 0} s_{2j+1}=\bar{h}$ is

$$\begin{aligned} \left\{ \begin{array}{lll} \mathrm {C}_{h-1}^{\bar{h}-1}, &{}0<\bar{h}\le h\\ 0, &{}otherwise \end{array}\right. . \end{aligned}$$

Proof

Let $\mathcal {A}$ be the set of all positive integer sequences $S_a=\{s_1,\dots ,s_{last}\}$ satisfying $\sum _{i\ge 1} s_i=h$ and $\sum _{j \ge 0} s_{2j+1}=\bar{h}$, and $\mathcal {B}$ be the set of all positive integer sequences $S_b=\{s_1,s_2,\dots ,s_{\bar{h}}\}$ satisfying $\sum _{i=1}^{\bar{h}} s_i=h$.

Define $f: \mathcal {A} \rightarrow \mathcal {B}$ as follows. For any sequence $S_a=\{s_1,\dots ,s_{last}\}$ in $\mathcal {A}$,

$$\begin{aligned}f(S_a)=\left\{ \begin{array}{lll} \{\underbrace{1,\dots ,1}_{s_1-1},1+s_2,\underbrace{1,\dots ,1}_{s_3-1},1+s_4,\dots \dots , \underbrace{1,\dots ,1}_{s_{last-1}-1},1+s_{last}\},&{}\quad \mathrm {if}\,\, last\,\,\mathrm {is\, even}\\ \{\underbrace{1,\dots ,1}_{s_1-1},1+s_2,\underbrace{1,\dots ,1}_{s_3-1},1+s_4,\dots \dots , 1+s_{last-1},\underbrace{1,\dots ,1}_{s_{last}}\},&{}\quad \mathrm {if}\,\, last\,\,\mathrm {is\, odd} \end{array}\right. \end{aligned}$$

Once $s_{2i-1}-1=0$, the corresponding terms will disappear. Regarding $f(S_a)$, its sum is still h and its length becomes $s_1-1+1+s_3-1+1+\cdots =\bar{h}$, so f is a mapping from $\mathcal {A}$ to $\mathcal {B}$.

Define $g: \mathcal {B} \rightarrow \mathcal {A}$: for any sequence $S_b\in \mathcal {B}$, if its last term is larger than 1, we may write it in the following unique format

$$\begin{aligned} \{\underbrace{1,\dots ,1}_{s_1-1},1+s_2,\underbrace{1,\dots ,1}_{s_3-1},1+s_4,\dots \dots , \underbrace{1,\dots ,1}_{s_{2k-1}-1},1+s_{2k}\}, \end{aligned}$$

where $s_1,s_2,\dots ,s_{2k}$ are all positive integers. Then, we have $g(S_b)=\{s_1,s_2,\dots ,s_{2k}\}$. If its last term is 1, we write it as

$$\begin{aligned} \{\underbrace{1,\dots ,1}_{s_1-1},1+s_2,\underbrace{1,\dots ,1}_{s_3-1},1+s_4,\dots \dots , \underbrace{1,\dots ,1}_{s_{2k-1}}\}. \end{aligned}$$

Then, we have $g(S_b)=\{s_1,s_2,\dots ,s_{2k-1}\}$. In both cases the length of $S_b$ is $s_1+s_3+\ldots +s_{2k-1}$, which equals $\bar{h}$, the sum of those terms with odd subscripts of $g(S_b)$.

Therefore, we have $|\mathcal {A}|=|\mathcal {B}|$, and derive the conclusion from Lemma 1.

Lemma 3

The number of positive integer sequences $S=\{ s_1, \ldots , s_{last}\}$ satisfying $\sum _{i\ge 1} s_i =h$, $\sum _{j \ge 0} s_{2j+1}=\bar{h}$ and $s_{last}\ge t$ ($t\in Z,0<t<h$) is

$$\begin{aligned} \delta _{\bar{h}:(t,h]}\cdot \mathrm {C}_{h-t-1}^{\bar{h}-t-1}+\delta _{\bar{h}:(0,h-t]}\cdot \mathrm {C}_{h-t-1}^{\bar{h}-1}, \end{aligned}$$

where

$$\begin{aligned} \delta _{x:(a,b]}=\left\{ \begin{array}{ll} 1, &{} a<x\le b\\ 0 , &{} otherwise \end{array}\right. . \end{aligned}$$

(11)

Proof

Let $\mathcal {R}$ be the set of all positive integer sequences $S=\{ s_1, \ldots , s_{last}\}$ satisfying $\sum _{i\ge 1} s_i =h$, $\sum _{j \ge 0} s_{2j+1}=\bar{h}$ and $s_{last}\ge t$. And let

$\mathcal {C}: =\{S_c|S_c\in \mathcal {R}, \mathrm {the length of}\, S\, \mathrm {is odd}\}$, and $\mathcal {D}: =\{S_d|S_d\in \mathcal {R}, \mathrm {the length of}\,\, S\,\ \mathrm {is even}\}$.

Then, $\mathcal {C}\cap \mathcal {D}=\emptyset$ and $\mathcal {C}\cup \mathcal {D}=\mathcal {R}$, so we have

$$\begin{aligned} |\mathcal {R}|=|\mathcal {C}|+|\mathcal {D}|. \end{aligned}$$

(12)

Let $\mathcal {C}^{\prime }$ be the set of all positive integer sequences $S=\{ s_1, \ldots , s_{last}\}$ satisfying $\sum _{i\ge 1} s_i =h-t$, $\sum _{j \ge 0} s_{2j+1}=\bar{h} - t$. Then, we define a mapping $f_o: \mathcal {C}\rightarrow \mathcal {C}^{\prime }$: for any sequence $S_c=\{ s_1, \ldots , s_{last}\}\in \mathcal {C}$,

$$\begin{aligned} f_o(S_c)=\left\{ \begin{array}{lll} \{s_1,s_2,\dots ,s_{last-1},s_{last}-t\}, \quad &{}\mathrm {if}\,\,s_{last}>t\\ \{s_1,s_2,\dots ,s_{last-1}\}.\quad &{}\mathrm {if}\,\, s_{last}=t \end{array}\right. . \end{aligned}$$

In both cases $f_o(S)$ is still a positive integer sequence since $t<h$. And $S_c\in \mathcal {C}$, $s_{last}$ has an odd subscript, so the sum of all terms of $f_o(S_c)$ is $h-t$, and the sum of terms with odd subscripts of $f_o(S_c)$ is $\bar{h}-t$. Thus, $f_o(S_c)\in \mathcal {C}^{\prime }$.

Conversely, we define the mapping $g_o: \mathcal {C}^{\prime } \rightarrow \mathcal {C}$: for any sequence $S_{c^{\prime }}=\{ s_1, \ldots , s_{last}\}$ in $\mathcal {C}^{\prime }$,

$$\begin{aligned} g_o(S_{c^{\prime }})=\left\{ \begin{array}{lll} \{s_1,s_2,\ldots ,s_{last}+t\}, &{}\quad \mathrm {if}\,\, last\,\,\mathrm {is\,odd}\\ \{s_1,s_2,\ldots ,s_{last},t\}. &{}\quad \mathrm {if}\,\,last\,\,\mathrm {is\,even} \end{array}\right. . \end{aligned}$$

In both cases the length of $g_o(S_{c^{\prime }})$ is odd, the sum of all terms of $g_o(S_{c^{\prime }})$ is t larger than that of $S_{c^{\prime }}$, and the sum of terms with odd subscripts of $g_o(S_{c^{\prime }})$ is also t larger than that of $S_{c^{\prime }}$, thus $g_o(S_{c^{\prime }})\in \mathcal {C}$.

Because $f_o=g_o^{-1}$, they both are one-to-one mappings. Thus, we obtain $|\mathcal {C}|=|\mathcal {C}^{\prime }|$. And by Lemma 2 we have

$$\begin{aligned} |\mathcal {C}|=|\mathcal {C}^{\prime }|=\left\{ \begin{array}{ll} \mathrm {C}_{h-t-1}^{\bar{h}-t-1},&{} 0<\bar{h}-t\le h-t\\ 0, &{}otherwise \end{array}\right. . \end{aligned}$$

(13)

Let $\mathcal {D}^{\prime}$ be the set of all positive integer sequences $S=\{ s_1, \ldots , s_{last}\}$ satisfying $\sum _{i\ge 1} s_i =h-t$, $\sum _{j \ge 0} s_{2j+1}=\bar{h}$. Similarly, we define a pair of inverse mappings $f_e:\mathcal {D}\rightarrow \mathcal {D}^{\prime}$ and $g_e:\mathcal {D}^{\prime}\rightarrow \mathcal {D}$. For any sequence $S_d=\{ s_1, s_2, \ldots , s_{last}\}\in \mathcal {D}$,

$$\begin{aligned} f_e(S_d)=\left\{ \begin{array}{ll} \{s_1,s_2,\dots ,s_{last}-t\},&{}\quad \mathrm {if}\,\,s_{last}>t\\ \{s_1,s_2,\dots ,s_{last-1}\},&{}\quad \mathrm {if}\,\,s_{last}=t \end{array}\right. . \end{aligned}$$

And for any sequence $S_{d'}=\{ s_1,s_2, \ldots , s_{last}\} \in \mathcal {D}'$,

$$\begin{aligned} g_e(S_{d'})=\left\{ \begin{array}{ll} \{s_1,s_2,\ldots ,s_{last},t\},&{}\quad \mathrm {if}\,\, last\,\,\mathrm {is\,\,odd}\\ \{s_1,s_2,\ldots ,s_{last}+t\},&{}\quad \mathrm {if}\,\, last\,\,\mathrm {is\,\,even} \end{array}\right. . \end{aligned}$$

We obtain $|\mathcal {D}|=|\mathcal {D}'|$ analogously. And from Lemma 2 we derive

$$\begin{aligned} |\mathcal {D}|=|\mathcal {D}^{\prime }|=\left\{ \begin{array}{ll} \mathrm {C}_{h-t-1}^{\bar{h}-1},&{} 0<\bar{h}\le h-t\\ 0, &{}otherwise \end{array}\right. . \end{aligned}$$

(14)

Combining (12), (13) and (14), we have the conclusion.

For $NL_n(L,\sigma )$, it is straightforward that $L\in [0,n]$ and $o\in [1,L]$. By Lemma 2 and (9), if $0<L\le n/2$ and $1\le \sigma \le L$, we have

$$\begin{aligned} NL_n(L,\sigma )=\mathrm {C}_{L-1}^{\sigma -1}2^{L}. \end{aligned}$$

If $n/2<L<n$ and $1\le \sigma \le L$, considering the restriction on the last jump height (10), and from (9) and Lemma 3 we obtain

$$\begin{aligned} NL_n(L,\sigma )=\left( \delta _{\sigma :(2L-n,L]}\cdot \mathrm {C}_{n-L-1}^{\sigma -2L+n-1} +\delta _{\sigma :(0,n-L]}\cdot \mathrm {C}_{n-L-1}^{\sigma -1}\right) 2^{n-L}. \end{aligned}$$

Special cases include $NL_n(0,0)=1$ and $NL_n(n,n)=1$. Therefore, we conclude in the following lemma.

Lemma 4

The number of n-bit binary sequences with the linear linear complexity L and the odd hop sum o is

$$\begin{aligned} NL_n(L,\sigma )=\left\{ \begin{array}{lll} 1, &{}L=\sigma =0\\ \mathrm {C}_{L-1}^{\sigma -1}2^{L}, &{}0<L\le n/2, 1\le \sigma \le L\\ (\delta _{\sigma :(2L-n,L]}\cdot \mathrm {C}_{n-L-1}^{\sigma -2L+n-1} +\delta _{\sigma :(0,n-L]}\cdot \mathrm {C}_{n-L-1}^{\sigma -1})2^{n-L}, &{}n/2<L<n, 1\le \sigma \le L\\ 1, &{}L=\sigma =n\\ 0, &{}otherwise \end{array}\right. . \end{aligned}$$

(15)

Because $L \ge \sigma$, we get $N_n(\sigma )=\sum _{L=0}^nNL_n(L,\sigma )=\sum _{L=\sigma }^nNL_n(L,\sigma ).$ Before deriving the explicit formulae of $E(O_n)$ and $V(O_n)$, we need the following identities.

$$\begin{aligned} \sum \limits _{i=0}^n \mathrm {C}_n^i= & {} 2^n.\end{aligned}$$

(16)

$$\begin{aligned} \sum \limits _{i=0}^n \mathrm {C}_i^1\mathrm {C}_n^i= & {} n2^{n-1}.\end{aligned}$$

(17)

$$\begin{aligned} \sum \limits _{i=0}^n \mathrm {C}_i^2\mathrm {C}_n^i= & {} n(n-1) 2^{n-3}.\end{aligned}$$

(18)

$$\begin{aligned} \sum \limits _{i=\alpha }^{\beta }q^i= & {} \frac{q^{\beta +1}-q^\alpha }{q-1}.\end{aligned}$$

(19)

$$\begin{aligned} \sum \limits _{i=\alpha }^{\beta }\mathrm {C}_i^1q^i= & {} \frac{(\beta (q-1)-1)q^{\beta +1}-(\frac{\alpha }{q} (q-1)-1)q^{\alpha +1}}{(q-1)^2}.\end{aligned}$$

(20)

$$\begin{aligned} \sum \limits _{i=\alpha }^{\beta }\mathrm {C}_i^2q^i= & {} \frac{(\frac{(q-1)^2}{q}\mathrm {C}_\beta ^2-\frac{q-1}{q}\beta +1)q^{\beta +2}-(\frac{(q-1)^2}{q^2}\mathrm {C}_\alpha ^2-\frac{q-1}{q}\alpha +1)q^{\alpha +2}}{(q-1)^3}, \end{aligned}$$

(21)

where the combinatorial numbers $\mathrm {C}_i^1=i$ and $\mathrm {C}_i^2=\frac{i(i-1)}{2}$. Also, we let $\mathrm {C}_i^j:=0$ if $j>i$. Proof of Theorem 1

Proof

It is straightforward that

$$\begin{aligned} E(O_n) = 2^{-n}\sum \limits _{L=0}^n \sum \limits _{\sigma =0}^L \sigma \cdot NL_n(L,\sigma ). \end{aligned}$$

(22)

Because of (15), we divide $\sum _{L=0}^n \sum _{\sigma =0}^L \sigma \cdot NL_n(L,\sigma )$ into different parts:

$$\begin{aligned} \sum \limits _{L=0}^n \sum \limits _{\sigma =0}^L \sigma \cdot NL_n(L,\sigma )= & {} \sum \limits _{\sigma =0}^0 \sigma \cdot NL_n(0,\sigma )+\sum \limits _{L=1}^{\lfloor n/2\rfloor } \sum \limits _{\sigma =1}^L \sigma \cdot NL_n(L,\sigma )\nonumber \\&+ \sum \limits _{L=\lfloor n/2\rfloor +1}^{n-1} \sum \limits _{\sigma =1}^L \sigma \cdot NL_n(L,\sigma )+\sum \limits _{\sigma =1}^n \sigma \cdot NL_n(n,\sigma )\nonumber \\= & {} n+\sum \limits _{L=1}^{\lfloor n/2\rfloor } \sum \limits _{\sigma =1}^L \sigma \cdot NL_n(L,\sigma )+\sum \limits _{L=\lfloor n/2\rfloor +1}^{n-1} \sum \limits _{\sigma =1}^L \sigma \cdot NL_n(L,\sigma ). \end{aligned}$$

(23)

The last identity is due to $NL_n(n,\sigma ) \ne 0$ only when $\sigma =n$. Then, we compute the two sums in the right side of (23).

Initially, we set $n>2$. Now let n be even. By (17), (18), (19) and (20) we obtain

$$\begin{aligned} \sum \limits _{L=1}^{\lfloor n/2\rfloor } \sum \limits _{\sigma =1}^L \sigma \cdot NL_n(L,\sigma )= & {} \sum \limits _{L=1}^{n/2} \sum \limits _{\sigma =1}^L \sigma \cdot \mathrm {C}_{L-1}^{\sigma -1}2^L\nonumber \\= & {} \sum \limits _{L=1}^{n/2}2^L \sum \limits _{i=0}^{L-1} (i+1)\cdot \mathrm {C}_{L-1}^{i} \nonumber \\= & {} \frac{(3n+4)2^{n-1}-2}{9}. \end{aligned}$$

(24)

And by (15) we have

$$\begin{aligned} \sum \limits _{L=\lfloor n/2\rfloor +1}^{n-1} \sum \limits _{\sigma =1}^L \sigma \cdot NL_n(L,\sigma )= & {} \sum \limits _{L=n/2+1}^{n-1} \sum \limits _{\sigma =1}^L \sigma \cdot (\delta _{\sigma :(2L-n,L]}\cdot \mathrm {C}_{n-L-1}^{\sigma -2L+n-1} +\delta _{\sigma :(0,n-L]}\cdot \mathrm {C}_{n-L-1}^{\sigma -1})2^{n-L}\nonumber \\= & {} \sum \limits _{L=n/2+1}^{n-1}2^{n-L}\left( \sum \limits _{\sigma =1}^L \delta _{\sigma :(2L-n,L]}\cdot \sigma \cdot \mathrm {C}_{n-L-1}^{\sigma -2L+n-1}+\sum \limits _{\sigma =1}^L \delta _{\sigma :(0,n-L]}\cdot \sigma \cdot \mathrm {C}_{n-L-1}^{\sigma -1}\right) . \end{aligned}$$

(25)

Because of (11) we have

$$\begin{aligned} \sum \limits _{\sigma =1}^L \delta _{\sigma :(2L-n,L]}\cdot \sigma \cdot \mathrm {C}_{n-L-1}^{\sigma -2L+n-1}=\sum \limits _{\sigma =2L-n+1}^L \sigma \cdot \mathrm {C}_{n-L-1}^{\sigma -2L+n-1} =\sum \limits _{i=0}^{n-L-1} (i+2L-n+1)\cdot \mathrm {C}_{n-L-1}^{i}, \end{aligned}$$

(26)

and

$$\begin{aligned} \sum \limits _{\sigma =1}^L \delta _{\sigma :(0,n-L]}\cdot \sigma \cdot \mathrm {C}_{n-L-1}^{\sigma -1}=\sum \limits _{\sigma =1}^{n-L} \sigma \cdot \mathrm {C}_{n-L-1}^{\sigma -1} =\sum \limits _{i=0}^{n-L-1} (i+1)\cdot \mathrm {C}_{n-L-1}^{i}. \end{aligned}$$

(27)

Combining (25), (26) and (27), we get

$$\begin{aligned} \sum \limits _{L=n/2+1}^{n-1} \sum \limits _{\sigma =1}^L \sigma \cdot NL_n(L,\sigma )= & {} \sum \limits _{L=n/2+1}^{n-1}2^{n-L}\sum \limits _{i=0}^{n-L-1}(2i+2L-n+2)\mathrm {C}_{n-L-1}^{i}\\= & {} \sum \limits _{L=n/2+1}^{n-1}2^{n-L}\left( 2(n-L-1)2^{n-L-2}+(2L-n+2)2^{n-L-1}\right) . \end{aligned}$$

By (16), (17), (19) and (20), we have

$$\begin{aligned} \sum \limits _{L=n/2+1}^{n-1} \sum \limits _{\sigma =1}^L \sigma \cdot NL_n(L,\sigma )=2^{-1}\sum \limits _{i=1}^{n/2-1}(n-i+1)4^{i}=\frac{(3n+14)2^{n-2}-6n-8}{9}. \end{aligned}$$

(28)

Therefore, combing (22), (23), (24) and (28), if n is even, we have

$$\begin{aligned} E(O_n)= & {} 2^{-n}\left( n+\frac{(3n+4)2^{n-1}-2}{9}+\frac{(3n+14)2^{n-2}-6n-8}{9}\right) \nonumber \\= & {} \frac{n}{4}+\frac{11}{18}+\frac{3n-10}{9\cdot 2^n}. \end{aligned}$$

(29)

If n is odd, since $\lfloor n/2\rfloor =(n-1)/2$, similarly we have $E(O_n) =\frac{n}{4}+\frac{23}{36}+\frac{3n-10}{9\cdot 2^n}$.

Now we compute $V(O_n)$. Because $V(O_n)=E(O_n^2)-E^2(O_n)$, we calculate $E(O_n^2)$ first.

$$\begin{aligned} E(O_n^2)=\sum \limits _{\sigma =0}^n\sigma ^2\cdot Pr\{O_n=\sigma \}=2^{-n}\sum \limits _{L=0}^n\sum \limits _{\sigma =0}^L\sigma ^2\cdot NL_n(L,\sigma ). \end{aligned}$$

(30)

Because of (15) we have

$$\begin{aligned} \sum \limits _{L=0}^n \sum \limits _{\sigma =0}^L \sigma ^2\cdot NL_n(L,\sigma )= & {} \sum \limits _{\sigma =0}^0 \sigma ^2\cdot NL_n(0,\sigma )+\sum \limits _{L=1}^{\lfloor n/2\rfloor } \sum \limits _{\sigma =1}^L \sigma ^2\cdot NL_n(L,\sigma )\nonumber \\&+ \sum \limits _{L=\lfloor n/2\rfloor +1}^{n-1} \sum \limits _{\sigma =1}^L \sigma ^2\cdot NL_n(L,\sigma )+\sum \limits _{\sigma =1}^n \sigma ^2\cdot NL_n(n,\sigma )\nonumber \\= & {} n^2+\sum \limits _{L=1}^{\lfloor n/2\rfloor } \sum \limits _{\sigma =1}^L \sigma ^2\cdot NL_n(L,\sigma )+\sum \limits _{L=\lfloor n/2\rfloor +1}^{n-1} \sum \limits _{\sigma =1}^L \sigma ^2\cdot NL_n(L,\sigma ). \end{aligned}$$

(31)

Next we calculate the two sums in the right side of (31).

First, let n be even. Then we have

$$\begin{aligned} \sum _{L=1}^{\lfloor n/2\rfloor } \sum \limits _{\sigma =1}^L \sigma ^2\cdot NL_n(L,\sigma )= & {} \sum \limits _{L=1}^{n/2} \sum \limits _{\sigma =1}^L \sigma ^2\cdot \mathrm {C}_{L-1}^{\sigma -1}2^L\\= & {} \sum \limits _{L=1}^{n/2}2^L \sum \limits _{i=0}^{L-1} (2\mathrm {C}_i^2+3\mathrm {C}_i^1+1)\cdot \mathrm {C}_{L-1}^{i}\\= & {} \sum \limits _{L=1}^{n/2} 2^L(2\mathrm {C}_{L-1}^22^{L-3}+3\mathrm {C}_{L-1}^12^{L-2}+2^{L-1}), \end{aligned}$$

where the last identity is due to (16), (17) and (18). Combining (20) and (21), we have

$$\begin{aligned} \sum \limits _{L=1}^{\lfloor n/2\rfloor } \sum \limits _{\sigma =1}^L \sigma ^2\cdot NL_n(L,\sigma )=2^{-2}\sum _{L=1}^{n/2} (\mathrm {C}_{L}^2+2\mathrm {C}_{L}^1)4^L =\frac{(9n^2+42n-16)2^{n-3}+2}{27}. \end{aligned}$$

(32)

And from (15) we derive

$$\begin{aligned} \sum \limits _{L=\lfloor n/2\rfloor +1}^{n-1} \sum \limits _{\sigma =1}^L \sigma ^2\cdot NL_n(L,\sigma )= & {} \sum _{L=n/2+1}^{n-1} \sum _{\sigma =1}^L \sigma ^2\cdot \left( \delta _{\sigma :(2L-n,L]}\cdot \mathrm {C}_{n-L-1}^{\sigma -2L+n-1} +\delta _{\sigma :(0,n-L]}\cdot \mathrm {C}_{n-L-1}^{\sigma -1}\right) 2^{n-L}\nonumber \\= & {} \sum \limits _{L=n/2+1}^{n-1}2^{n-L}\left( \sum \limits _{\sigma =1}^L \delta _{\sigma :(2L-n,L]}\cdot \sigma ^2\cdot \mathrm {C}_{n-L-1}^{\sigma -2L+n-1}+\sum _{\sigma =1}^L \delta _{\sigma :(0,n-L]}\cdot \sigma ^2\cdot \mathrm {C}_{n-L-1}^{\sigma -1}\right) . \end{aligned}$$

(33)

We have

$$\begin{aligned} \sum \limits _{\sigma =1}^L \delta _{\sigma :(2L-n,L]}\cdot \sigma ^2\cdot \mathrm {C}_{n-L-1}^{\sigma -2L+n-1}=\sum \limits _{\sigma =2L-n+1}^L \sigma ^2\cdot \mathrm {C}_{n-L-1}^{\sigma -2L+n-1} =\sum \limits _{i=0}^{n-L-1} (i+2L-n+1)^2\cdot \mathrm {C}_{n-L-1}^{i}, \end{aligned}$$

(34)

and if $n\le 2 L$ we have

$$\begin{aligned} \sum \limits _{\sigma =1}^L \delta _{\sigma :(0,n-L]}\cdot \sigma ^2\cdot \mathrm {C}_{n-L-1}^{\sigma -1}=\sum \limits _{\sigma =1}^{n-L} \sigma ^2\cdot \mathrm {C}_{n-L-1}^{\sigma -1} =\sum \limits _{i=0}^{n-L-1} (i+1)^2\cdot \mathrm {C}_{n-L-1}^{i}. \end{aligned}$$

(35)

Combining (33), (34) and (35), we get

$$\begin{aligned} \sum \limits _{L=n/2+1}^{n-1} \sum \limits _{\sigma =1}^L \sigma ^2\cdot NL_n(L,\sigma )= & {} \sum _{L=n/2+1}^{n-1}2^{n-L}\sum _{i=0}^{n-L-1}\left( 4\mathrm {C}_i^2+2(2L-n+3)\mathrm {C}_i^1+(2L-n+1)^2+1\right) \mathrm {C}_{n-L-1}^{i}\nonumber \\= & {} \sum \limits _{L=n/2+1}^{n-1}2^{n-L}\left( 4\mathrm {C}_{n-L-1}^22^{n-L-3}+2(2L-n+3)\mathrm {C}_{n-L-1}^12^{n-L-2}+((2L-n+1)^2+1)2^{n-L-1} \right) \nonumber \\= & {} 2^{-1}\sum \limits _{i=1}^{n/2-1}(5\mathrm {C}_i^2+(2-3n)\mathrm {C}_i^1+n^2+n)4^i =\frac{(9n^2+102n+448)2^{n-4}-18n^2-36n-28}{27}. \end{aligned}$$

(36)

Combing (30), (31), (32) and (36), we have

$$\begin{aligned} E(O_n^2)=\frac{n^2}{16}+\frac{31n}{72}+\frac{26}{27}+\frac{9n^2-36n-26}{27\cdot 2^n}, \end{aligned}$$

then if n is even, from (29) we derive

$$\begin{aligned} V(O_n)=E(O_n^2)-E^2(O_n) =\frac{n}{8}+\frac{191}{324}+\frac{27n^2-192n+64}{81\cdot 2^{n+1}}-\frac{(3n-10)^2}{81\cdot 4^n}. \end{aligned}$$

If n is odd, similarly we obtain $V(O_n) =\frac{n}{8}+\frac{367}{648}+\frac{27n^2-195n+74}{162\cdot 2^n}-\frac{(3n-10)^2}{81\cdot 4^n}$.

Also, we can verify that the conclusion is still correct for $n=1,2$.