**4. Numerical Examples**

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**Example 1.** *Consider a tensor* A = (*ai*1*i*2*i*3 )*, with order three and dimension six, defined as follows:*

*a*111 = 3.9, *a*112 = *a*121 = *a*122 = 1, *a*113 = *a*131 = *a*133 = *a*144 = *a*155 = 0.2, *a*222 = 5, *a*211 = *a*212 = *a*221 = *a*235 = *a*236 = 1, *a*214 = *a*215 = *a*232 = *a*256 = *a*266 = 0.2, *a*333 = 4, *a*334 = *a*343 = 0.5, *a*344 = 1, *a*311 = *a*322 = *a*331 = *a*326 = *a*352 = 0.2, *a*444 = 6.5, *a*433 = *a*434 = *a*443 = 1, *a*412 = *a*414 = *a*421 = *a*441 = 0.5, *a*555 = 6.3, *a*556 = 1, *a*512 = *a*515 = *a*521 = *a*523 = *a*551 = *a*565 = *a*566 = 0.5, *a*666 = 8, *a*655 = *a*656 = *a*665 = 1, *a*625 = *a*626 = *a*652 = *a*662 = 0.5

*and other ai*1*i*2*i*3 = 0. *By calculations, we have:*

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$$\begin{aligned} \mathbb{R}\_1(\mathcal{A}) &= 4, \; \mathbb{R}\_2(\mathcal{A}) = 6, \; \mathbb{R}\_3(\mathcal{A}) = 3, \; \mathbb{R}\_4(\mathcal{A}) = 5, \; \mathbb{R}\_5(\mathcal{A}) = 4.5, \; \mathbb{R}\_6(\mathcal{A}) = 5, \\\\ \mathbb{N}\_1 &= \{1, 2\}, \; \mathbb{N}\_2 = \{3, 4\}, \; \mathbb{N}\_3 = \{5, 6\}, \; \mathbb{N}\mathbb{N}\_1 = \{3, 4, 5, 6\} = \mathbb{N}\_2 \bigcup \mathbb{N}\_3. \end{aligned}$$

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N3−<sup>1</sup> 1 = N21 = {11, 12, 21, <sup>22</sup>}, N3−<sup>1</sup> 2 = N22 = {33, 34, 43, <sup>44</sup>}, N3−<sup>1</sup> 3 = N23 = {55, 56, 65, <sup>66</sup>}, (N\N1)<sup>3</sup>−<sup>1</sup> = (N\N1)<sup>2</sup> = {33, 34, 35, 36, 43, 44, 45, 46, 53, 54, 55, 56, 63, 64, 65, <sup>66</sup>}, <sup>N</sup>3−<sup>1</sup>\(N\N1)<sup>3</sup>−<sup>1</sup> = <sup>N</sup><sup>2</sup>\(N\N1)<sup>2</sup> = {11, 12, 13, 14, 15, 16, 21, 22, 23, 24, 25, 26, 31, 32, 41, 42, 51, 52, 61, <sup>62</sup>}.

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$$a\_{\mathbf{i}\_1}^{(1)} = \sum\_{\substack{i\_{23} \in \mathcal{C}\_1^2 \\ i\_{23} = 0}} |a\_{12i\_3}| = |a\_{121}| + |a\_{121}| + |a\_{122}| = 1 + 1 + 1 = 3.$$

$$a\_{\mathbf{i}\_{12} = 0}^{(2)} = \sum\_{\substack{i\_{23} \in \mathcal{C}\_1^2 \\ i\_{23} = 0}} |a\_{2 i\_{23}}| = |a\_{211}| + |a\_{212}| + |a\_{221}| = 1 + 1 + 1 = 3.$$

$$a\_{\mathbf{i}\_{12}}^{(3)} = \sum\_{\substack{i\_{23} \in \mathcal{C}\_2^2 \\ i\_{23} = 0}} |a\_{3 i\_{23}}| = |a\_{334}| + |a\_{343}| + |a\_{344}| = 0.5 + 0.5 + 1 = 2.$$

$$\begin{array}{l} a\_{\mathbf{i}\_{12}}^{(4)} = \sum\_{\substack{i\_{23} \in \mathcal{C}\_2^2 \\ i\_{23} = 0}} |a\_{4 i\_{23}}| = |a\_{433}| + |a\_{434}| + |a\_{443}| = 1 + 1 + 1 = 3.$$

$$\delta\_{\mathbf{i}\_{23}}^{(5)} = \sum\_{\substack{i\_{23} \in \mathcal{C}\_2^2 \\ i\_{23} = 0}} |a\_{5 i\_{23}}| = |a\_{5 5 6}| + |a\_{5 6}| + |a\_{5 6}| = 1 + 0.5 + 0.5 = 2.$$

$$\delta\_{\mathbf{i}\_{23}}^{(5)} = \sum\_{\substack{i\_{23} \in \mathcal{C}\_2^2 \\ i\_{23} = 0}} |a\_{6 i\_{2$$

∑ *<sup>i</sup>*2*i*3∈N<sup>2</sup>\(N\N1)<sup>2</sup> *<sup>δ</sup>*1*i*2*i*3<sup>=</sup><sup>0</sup>|*<sup>a</sup>*1*i*2*i*3 | = |*<sup>a</sup>*112| + |*<sup>a</sup>*121| + |*<sup>a</sup>*122| + |*<sup>a</sup>*113| + |*<sup>a</sup>*131| + 0 = 1 + 1 + 1 + 0.2 + 0.2 = 3.4.

∑ *<sup>i</sup>*2*i*3∈N<sup>2</sup>\(N\N1)<sup>2</sup> *<sup>δ</sup>*2*i*2*i*3<sup>=</sup><sup>0</sup> |*<sup>a</sup>*2*i*2*i*3 | = |*<sup>a</sup>*211| + |*<sup>a</sup>*212| + |*<sup>a</sup>*221| + |*<sup>a</sup>*214| + |*<sup>a</sup>*215| + |*<sup>a</sup>*232| + 0 = 1 + 1 + 1 + 0.2 + 0.2 + 0.2 = 3.6.

$$\sum\_{i\_2 i\_3 \in \backslash(\mathbb{N} \backslash \mathbb{N}\_1)^2} |a\_{1i\_2 i\_3}| = |a\_{133}| + |a\_{144}| + |a\_{155}| + 0 = 0.2 + 0.2 + 0.2 = 0.6.$$

∑ *<sup>i</sup>*2*i*3<sup>∈</sup>(N\N1)<sup>2</sup> |*<sup>a</sup>*2*i*2*i*3 | = |*<sup>a</sup>*235| + |*<sup>a</sup>*236| + |*<sup>a</sup>*256| + |*<sup>a</sup>*266| + 0 = 1 + 1 + 0.2 + 0.2 = 2.4. *α*(1) N1 = 1, *α*(2) N1 = 3, *α*(3) N2 = 1, *α*(4) N2 = 2, *α*(5) N3 = 2.5, *α*(6) N3 = 2. (|*<sup>a</sup>*111| − *α*(1) N1 ) · (|*<sup>a</sup>*333| − *α*(3) N2 )=(3.9 − 3) × (4 − 2) = 1.8 > *α*(1) N1 · *α*(3) N2 = 1 × 1 = 1. (|*<sup>a</sup>*111| − *α*(1) N1 ) · (|*<sup>a</sup>*444| − *α*(4) N2 )=(3.9 − 3) × (6.5 − 3) = 3.15 > *α*(1) N1 · *α*(4) N2 = 1 × 2 = 2. (|*<sup>a</sup>*111| − *α*(1) N1 ) · (|*<sup>a</sup>*555| − *α*(5) N3 )=(3.9 − 3) × (6.3 − 2) = 3.87 > *α*(1) N1 · *α*(5) N3 = 1 × 2.5 = 2.5. (|*<sup>a</sup>*111| − *α*(1) N1 ) · (|*<sup>a</sup>*666| − *α*(6) N3 )=(3.9 − 3) × (8 − 3) = 4.5 > *α*(1) N1 · *α*(6) N3 = 1 × 2 = 2. (|*<sup>a</sup>*222| − *α*(2) N1 ) · (|*<sup>a</sup>*333| − *α*(3) N2 )=(5 − 3) × (4 − 2) = 4 > *α*(2) N1 · *α*(3) N2 = 3 × 1 = 3. (|*<sup>a</sup>*222| − *α*(2) N1 ) · (|*<sup>a</sup>*444| − *α*(4) N2 )=(5 − 3) × (6.5 − 3) = 7 > *α*(2) N1 · *α*(4) N2 = 3 × 2 = 6. (|*<sup>a</sup>*222| − *α*(2) N1 ) · (|*<sup>a</sup>*555| − *α*(5) N3 )=(5 − 3) × (6.3 − 2) = 8.6 > *α*(2) N1 · *α*(5) N3 = 3 × 2.5 = 7.5.

(|*<sup>a</sup>*222| − *α*(2) N1 ) · (|*<sup>a</sup>*666| − *α*(6) N3 )=(5 − 3) × (8 − 3) = 10 > *α*(2) N1 · *α*(6) N3 = 3 × 2 = 6. *From Definition 4, we have* A ∈ *LDDT. Furthermore,* (|*<sup>a</sup>*333| − *α*(3) N2 ) · (|*<sup>a</sup>*111| − ∑ *<sup>i</sup>*2*i*3∈N<sup>2</sup>\(N\N1)<sup>2</sup> *<sup>δ</sup>*1*i*2*i*3<sup>=</sup><sup>0</sup> |*<sup>a</sup>*1*i*2*i*3 |)=(4 − 2) × (3.9 − 3.4) = 1 > *α*(3) N2 · ( ∑ *<sup>i</sup>*2*i*3<sup>∈</sup>(N\N1)<sup>2</sup> |*<sup>a</sup>*1*i*2*i*3 |) = 1 × 0.6 = 0.6, (|*<sup>a</sup>*333| − *α*(3) N2 ) · (|*<sup>a</sup>*222| − ∑ *<sup>i</sup>*2*i*3∈N<sup>2</sup>\(N\N1)<sup>2</sup> *<sup>δ</sup>*2*i*2*i*3<sup>=</sup><sup>0</sup> |*<sup>a</sup>*2*i*2*i*3 |)=(4 − 2) × (5 − 3.6) = 2.8 > *α*(3) N2 · ( ∑ *<sup>i</sup>*2*i*3<sup>∈</sup>(N\N1)<sup>2</sup> |*<sup>a</sup>*2*i*2*i*3 |) = 1 × 2.4 = 2.4, (|*<sup>a</sup>*444| − *α*(4) N2 ) · (|*<sup>a</sup>*111| − ∑ *<sup>i</sup>*2*i*3∈N<sup>2</sup>\(N\N1)<sup>2</sup> *<sup>δ</sup>*1*i*2*i*3<sup>=</sup><sup>0</sup> |*<sup>a</sup>*1*i*2*i*3 |)=(6.5 − 3) × (3.9 − 3.4) = 1.75

$$\gg \quad \overline{\alpha}\_{\mathbb{M}\_2}^{(4)} \cdot (\sum\_{i\_2 i\_3 \in (\mathbb{M}/\mathbb{N}\_1)^2} |a\_{1i\_2 i\_3}|) = 2 \times 0.6 = 1.2\sqrt{}$$

(|*<sup>a</sup>*444| − *α*(4) N2 ) · (|*<sup>a</sup>*222| − ∑ *<sup>i</sup>*2*i*3∈N<sup>2</sup>\(N\N1)<sup>2</sup> *<sup>δ</sup>*2*i*2*i*3<sup>=</sup><sup>0</sup> |*<sup>a</sup>*2*i*2*i*3 |)=(6.5 − 3) × (5 − 3.6) = 4.9 > *α*(4) N2 · ( ∑ *<sup>i</sup>*2*i*3<sup>∈</sup>(N\N1)<sup>2</sup> |*<sup>a</sup>*2*i*2*i*3 |) = 2 × 2.4 = 4.8, (|*<sup>a</sup>*555| − *α*(5) N3 ) · (|*<sup>a</sup>*111| − ∑ *<sup>i</sup>*2*i*3∈N<sup>2</sup>\(N\N1)<sup>2</sup> *<sup>δ</sup>*1*i*2*i*3<sup>=</sup><sup>0</sup> |*<sup>a</sup>*1*i*2*i*3 |)=(6.3 − 2) × (3.9 − 3.4) = 2.15 > *α*(5) N3 · ( ∑ *<sup>i</sup>*2*i*3<sup>∈</sup>(N\N1)<sup>2</sup> |*<sup>a</sup>*1*i*2*i*3 |) = 2.5 × 0.6 = 1.5, (|*<sup>a</sup>*555| − *α*(5) N3 ) · (|*<sup>a</sup>*222| − ∑ *<sup>i</sup>*2*i*3∈N<sup>2</sup>\(N\N1)<sup>2</sup> *<sup>δ</sup>*2*i*2*i*3<sup>=</sup><sup>0</sup> |*<sup>a</sup>*2*i*2*i*3 |)=(6.3 − 2) × (5 − 3.6) = 6.02 > *α*(5) N3 · ( ∑ *<sup>i</sup>*2*i*3<sup>∈</sup>(N\N1)<sup>2</sup> |*<sup>a</sup>*2*i*2*i*3 |) = 2.5 × 2.4 = 6, (|*<sup>a</sup>*666| − *α*(6) N3 ) · (|*<sup>a</sup>*111| − ∑ *<sup>i</sup>*2*i*3∈N<sup>2</sup>\(N\N1)<sup>2</sup> *<sup>δ</sup>*1*i*2*i*3<sup>=</sup><sup>0</sup> |*<sup>a</sup>*1*i*2*i*3 |)=(8 − 3) × (3.9 − 3.4) = 2.5 > *α*(6) N3 · ( ∑ *<sup>i</sup>*2*i*3<sup>∈</sup>(N\N1)<sup>2</sup> |*<sup>a</sup>*1*i*2*i*3 |) = 2 × 0.6 = 1.2, (|*<sup>a</sup>*666| − *α*(6) N3 ) · (|*<sup>a</sup>*222| − ∑ *<sup>i</sup>*2*i*3∈N<sup>2</sup>\(N\N1)<sup>2</sup> *<sup>δ</sup>*2*i*2*i*3<sup>=</sup><sup>0</sup> |*<sup>a</sup>*2*i*2*i*3 |)=(8 − 3) × (5 − 3.6) = 7 > *α*(6) N3 · ( ∑ *<sup>i</sup>*2*i*3<sup>∈</sup>(N\N1)<sup>2</sup> |*<sup>a</sup>*2*i*2*i*3 |) = 2 × 2.4 = 4.8.

*These imply that* A *satisfies all the conditions of Theorem 1. Therefore,* A *is an* H*-tensor.*
