For any integer $n\in {\mathbb{N}}^{\text{*}}$, the multiplication table associated with $n$, denoted $T_n$, is defined by:

$\boxed{T_n\left(a\right)=a\cdot n\text{for}\text{all}a\in {\mathbb{N}}^{\text{*}}}$

For a fixed integer $n$, generate the associated tables:

$\boxed{T_a\left(n\right)=a\times n\text{for}a\in \left\{1,2,3,\cdots \right\}}$

Example with $n=4$:

$\boxed{T_1\left(4\right)=4,T_2\left(4\right)=8,T_3\left(4\right)=12,\cdots }$

For $a,b\in \mathbb{N}$, $b>a$, and $n\in {\mathbb{N}}^{\text{*}}$, the difference between rows $a$ and $b$ of the table $T_n$ is given by:

${\text{Δ}}_T\left(a,b,n\right)=\left(b-a\right)\cdot n.$

Objective: Compute the gap between two terms $T_a\left(n\right)$ and $T_b\left(n\right)$ of the same table.

Step 1: Choice of Indices $a,b$ (with $b>a$)

Step 2: Direct Computation

$\boxed{\text{Δ}T\left(a,b,n\right)=T_b\left(n\right)-T_a\left(n\right)}$

Example $\left(n=5,a=2,b=4\right)$:

$\boxed{\text{Δ}T\left(2,4,5\right)=20-10=10}$

Step 3: Factorization

$\boxed{\text{Δ}T\left(a,b,n\right)=\left(b-a\right)\times n}$

Verification:

$\boxed{\left(4-2\right)\times 5=10}$

$\boxed{\text{Σ}T\left(10,3\right)=3\times 55=165\left(55=\frac{10\times 11}{2}\right)}$

$\boxed{\text{Δ}T\left(3,5,7\right)=\left(5-3\right)\times 7=14}$

These simple formulas reveal the fundamental arithmetic structure of multiplication tables, with essential pedagogical and algorithmic applications.

$\boxed{\begin{array}{l}\text{Tabular Sum}:\text{Σ}T\left(k,n\right)=n\cdot \frac{k\left(k+1\right)}{2},\\ \text{Tabular Difference}:\text{Δ}T\left(a,b,n\right)=\left(b-a\right)\cdot n.\end{array}}$

Let $n\in {\mathbb{N}}^{\text{*}}$ and $k\in {\mathbb{N}}^{\text{*}}$.

The product of the first $k$ terms of the multiplication table of $n$ is given by:

${\text{Π}}_T\left(k,n\right)=\underset{a=1}{\overset{k}{{\displaystyle \prod }}}a\cdot n=n^k\cdot k!.$

Recall: Let $n$ be a fixed integer. For each integer $a$ such that $1\le a\le k$, we define the row $T_a\left(n\right)$ by:

$\boxed{T_a\left(n\right)=a\times n}$

In other words, $T_a\left(n\right)$ is the product of $n$ by $a$, which is the $a$-th row of the multiplication table of $n$.

We are interested in the product of the first $k$ rows, that is:

$\boxed{\text{Π}T\left(k,n\right)={\displaystyle {\prod }_{a=1}^k{T}_a\left(n\right)}=T_1\left(n\right)\times T_2\left(n\right)\times \cdots \times T_k\left(n\right)}$

Substituting $T_a\left(n\right)=a\times n$ gives:

$\boxed{\text{Π}T\left(k,n\right)=\left(1\times n\right)\times \left(2\times n\right)\times \cdots \times \left(k\times n\right)}$

Group the terms:

$\boxed{\text{Π}T\left(k,n\right)=\left(1\times 2\times \cdots \times k\right)\times \left(n\times n\times \cdots \times n\right)}$

where $n$ appears $k$ times. Thus:

$\boxed{\text{Π}T\left(k,n\right)=\left(1\times 2\times \cdots \times k\right)\times n^k}$

The product of the integers from 1 to $k$ is by definition the factorial $k!$. Therefore:

$\boxed{\text{Π}T\left(k,n\right)=k!\times n^k}$

The general formula for the tabular product is:

$\boxed{\text{Π}T\left(k,n\right)=n^k\cdot k!}$

Take $n=4$, $k=3$:

Direct product:

$\boxed{4\times 8\times 12=384}$

Formula:

$\boxed{4^3\times 3!=64\times 6=384}$

Theorem 2 (Tabular Sum Formula)

Let $n\in {\mathbb{N}}^{\text{*}}$ and $k\in {\mathbb{N}}^{\text{*}}$.

The sum of the first $k$ terms of the table $T_n$ is given by:

${\text{Σ}}_T\left(k,n\right)=n\cdot \frac{k\left(k+1\right)}{2}.$

Let $n\in \mathbb{N}$, $n>1$, and let $k=\lfloor \sqrt{n}\rfloor$. If:

${\text{Π}}_T\left(k,n\right)\equiv 0\left(\mathrm{mod}n\right)\text{and}\text{gcd}\left(k!,n\right)=1,$

then $n$ is prime [3] [4] .

Basic algorithm:

1) Set $k=\lfloor \sqrt{n}\rfloor$.

2) Compute:

$\boxed{\text{Π}T\left(k,n\right)=n^k\cdot k!}$

3) Check the conditions:

$\boxed{\text{Π}T\left(k,n\right)\equiv 0\left(\mathrm{mod}n\right)\text{and}\text{gcd}\left(k!,n\right)=1}$

4) If both are true, then $n$ is likely prime.

Kadouno Primality Tests for $n=89,991,99989,88888$

1) Test for $n=89$ (Prime)

$\boxed{\text{Π}T\left(9,89\right)={89}^9\cdot 9!}$

a) $\text{Π}T\left(9,89\right)\equiv 0\left(\mathrm{mod}89\right)$ (✓)

b) $\text{gcd}\left(9!,89\right)=1$ (✓)

2) Test for $n=991$ (Prime)

Step 1: $k=31$.

Step 2: Verify

a) $\text{Π}T\left(31,991\right)\equiv 0\left(\mathrm{mod}991\right)$ (✓)

b) $\text{gcd}\left(31!,991\right)=1$ (✓)

3) Test for $n=99989$ (Prime)

a) $\text{Π}T\left(316,99989\right)\equiv 0\left(\mathrm{mod}99989\right)$ (✓)

b) $\text{gcd}\left(316!,99989\right)=1$ (✓)

4) Test for $n=88888$ (Composite)

a) $\text{Π}T\left(298,88888\right)\equiv 0\left(\mathrm{mod}88888\right)$ (✓)

b) $\text{gcd}\left(298!,88888\right)\ge 41$ (×)

1) Condition 1 is always true since $n^k$ divides $\text{Π}T\left(k,n\right)$.

2) Condition 2 distinguishes primes ( $\text{gcd}=1$) from composites ( $\text{gcd}>1$).

To avoid direct computation of

$\text{Π}T\left(k,n\right)=n^k\cdot k!,$

which becomes intractable for $n\ge {10}^6$, we use a logarithmic approximation:

$\boxed{\log \left(\Pi T\left(k,n\right)\right)=k\log \left(n\right)+\log \left(k!\right).}$

The test then checks:

1) $n|n^{k}k!$ (via modulo),

2) $\text{gcd}\left(k!,n\right)=1$.

1) $n=1000003$ (prime)

$k=\lfloor \sqrt{n}\rfloor =1000$.

Stirling’s approximation for $n!$:

$n!\approx \sqrt{2\pi n}{\left(\frac{n}{e}\right)}^n,\log \left(n!\right)\approx n\log \left(n\right)-n+\frac{1}{2}\log \left(2\pi n\right).$

Numerical values:

$\log \left(1000!\right)\approx 5912.13,\log \left(n\right)\approx \mathrm{13.8155.}$

Compute

$\boxed{L=1000\times 13.8155+5912.13=\mathrm{19727.63.}}$

Since $n^k$ dominates $k!$, we have $n|n^{k}k!$ (✓). And because $n$ is prime and $n>k$, $\text{gcd}\left(k!,n\right)=1$ (✓).

Conclusion: $n$ is prime.

2) $n=15485863$ (prime)

$k=\lfloor \sqrt{n}\rfloor =3935$.

$\log \left(3935!\right)\approx 26860.4,\log \left(n\right)\approx \mathrm{16.605.}$

$\boxed{L=3935\times 16.605+26860.4=\mathrm{92191.1.}}$

Both conditions hold as above (✓).

Conclusion: $n$ is prime.

3) $n=999999000001=\left({10}^6+1\right)\left({10}^6-1\right)$ (composite)

$k=\lfloor \sqrt{n}\rfloor =999999$.

Condition 1 is always true.

Condition 2:

$\boxed{\text{gcd}\left(999999!,n\right)\ge 999999\ne 1,}$

since $999999|n$.

Conclusion: $n$ is composite.

$\boxed{C\left(n\right)=3+\lfloor \frac{{\log }_{10}\left(n\right)}{3}\rfloor }$

where ${\text{log}}_{10}\left(n\right)$ is the base—10 logarithm of $n$ and $\lfloor \cdot \rfloor$ the floor function.

- If $1\le n<{10}^3$, then $C\left(n\right)=3$.

- If ${10}^3\le n<{10}^6$, then $C\left(n\right)=4$.

- If ${10}^6\le n<{10}^9$, then $C\left(n\right)=5$, etc.

The test parameter is defined by

$\boxed{k\left(n\right)=\lfloor C\left(n\right){\log }_2\left(n\right)\rfloor }$

where ${\text{log}}_2\left(n\right)$ is the base-2 logarithm of $n$.

This adaptive scaling makes the Kadouno-Log test especially efficient over wide integer ranges, maintaining constant accuracy with cost adjusted to the number’s size.

Step 1: Compute $C\left(n\right)$

$C\left(101\right)=3+\lfloor \frac{{\text{log}}_{10}\left(101\right)}{3}\rfloor =3+\lfloor 0.668\rfloor =3$

Step 2: Compute $k\left(n\right)$

$k\left(101\right)=\lfloor 3\times {\text{log}}_2\left(101\right)\rfloor =\lfloor 3\times 6.658\rfloor =19$

Step 3: Verify the Conditions

$\boxed{\text{Π}T\left(19,101\right)\equiv 0\left(\mathrm{mod}101\right)}$ and $\boxed{\text{gcd}\left(19!,101\right)=1}$

Conclusion

101 is prime.

Step 1: Compute $C\left(n\right)$

$C\left(10007\right)=3+\lfloor \frac{{\text{log}}_{10}\left(10007\right)}{3}\rfloor =3+\lfloor 1.3334\rfloor =4$

Step 2: Compute $k\left(n\right)$

$k\left(10007\right)=\lfloor 4\times {\text{log}}_2\left(10007\right)\rfloor =\lfloor 4\times 13.288\rfloor =53$

Step 3: Verify the Conditions

$\boxed{\text{Π}T\left(53,10007\right)\equiv 0\left(\mathrm{mod}10007\right)}$ and $\boxed{\text{gcd}\left(53!,10007\right)=1}$

Conclusion

10007 is prime.

1) Adaptability of $C\left(n\right)$:

- $C\left(n\right)$ automatically adjusts the depth of the test according to the order of magnitude of $n$.

- Example: For $n=1000003$ (10⁶),

$\boxed{C\left(n\right)=5,k\left(n\right)=99.}$

2) Efficiency:

- The dynamic threshold

$\boxed{k\left(n\right)=\lfloor C\left(n\right){\text{log}}_2\left(n\right)\rfloor }$

balances precision and computational cost.

- Example:

$\boxed{\lfloor {\text{log}}_2\left(1000003\right)\rfloor \approx 19.93\Rightarrow k\left(n\right)=99}$

(sufficient to capture small factors).

3) Robustness:

- Correctly detects primes even for $n\ge {10}^6$.

$\boxed{k\left(n\right)=\lfloor C\left(n\right)\cdot {\text{log}}_2\left(n\right)\rfloor }$

We replace the factorial tabular product:

$\text{Π}T\left(k,n\right):=n^k\cdot k!$

by its finer version:

$\text{Π}T\left(k,n\right):=n^k\cdot k\#,\text{where}k\#:={\displaystyle \underset{\begin{array}{c}p\le k\\ p\in \mathbb{P}\end{array}}{\prod }p}$

The primorial eliminates factorial redundancies, retaining only prime factors.

Theorem 4 Let $n\in {\mathbb{N}}^{\text{*}},k=\lfloor \sqrt{n}\rfloor$. Then:

Definition—Decimal regulation function:

$C\left(n\right):=3+\lfloor \frac{{\text{log}}_{10}\left(n\right)}{3}\rfloor$

Definition—Adaptive test depth:

$k\left(n\right):=\lfloor C\left(n\right)\cdot {\text{log}}_2\left(n\right)\rfloor$

This choice regulates the depth $k\left(n\right)$ while ensuring coverage of critical prime factors.

Test steps:

1) Compute $k\left(n\right)$

2) Build $k\left(n\right)={\displaystyle {\prod }_{p\le k\left(n\right)}p}$

3) Evaluate $\text{gcd}\left(n,k\#\right)$

4) Conclusion:

- If $\text{gcd}>1$, $n$ is composite

- If $\text{gcd}=1$, $n$ is prime

Example 1—Small Prime Integer

Let $n=101$

${\text{log}}_{10}\left(101\right)\approx 2.00\Rightarrow C\left(n\right)=3$

${\text{log}}_2\left(101\right)\approx 6.658\Rightarrow k\left(n\right)=\lfloor 3\times 6.658\rfloor =19$

${\mathbb{P}}_{\le 19}=\left\{2,3,5,7,11,13,17,19\right\}$

$K\text{\#}=9699690$

$\text{gcd}\left(101,9699690\right)=1\Rightarrow$ Prime validated

Example 2—Small Composite Integer

Let $n=105$

Same $k\left(n\right)=19$, same primorial

$105=3\times 5\times 7$

$\text{gcd}\left(105,9699690\right)=105>1\Rightarrow$ Composite detected

Example 3—Carmichael Number

Let $n=561=3\times 11\times 17$

${\text{log}}_{10}\left(561\right)\approx 2.75\Rightarrow C\left(n\right)=3$

${\text{log}}_2\left(561\right)\approx 9.13\Rightarrow k\left(n\right)=\lfloor 3\times 9.13\rfloor =27$

$27={\displaystyle {\prod }_{p\le 27}p}=2\times 3\times \cdots \times 23\times 29$

$\text{gcd}\left(561,27\right)\ge 3\Rightarrow$ Carmichael rejected

Example 4—Medium Prime Number

Let $n=10007$

${\text{log}}_{10}\left(n\right)\approx 4.00\Rightarrow C\left(n\right)=4$

${\text{log}}_2\left(n\right)\approx 13.28\Rightarrow k\left(n\right)=\lfloor 4\times 13.28\rfloor =53$

$\text{gcd}\left(k\#,10007\right)=1\Rightarrow$ Prime confirmed

Example 5—Very Large Composite Integer

Let $n=999999000001$ (composite, e.g. $101\times 9901001$)

${\text{log}}_{10}\left(n\right)\approx 12.0\Rightarrow C\left(n\right)=7$

${\text{log}}_2\left(n\right)\approx 39.863\Rightarrow k\left(n\right)=\lfloor 7\times 39.863\rfloor =279$

$\text{gcd}\left(n,279\#\right)\ge 101>1\Rightarrow$ Composite detected

Example 6 — Very Large Prime Integer

Let $n=1000003$

${\text{log}}_{10}\left(n\right)\approx 6.0\Rightarrow C\left(n\right)=5$

${\text{log}}_2\left(n\right)\approx 19.93\Rightarrow k\left(n\right)=\lfloor 5\times 19.93\rfloor =99$

$\text{gcd}\left(99\#,n\right)=1\Rightarrow$ Prime confirmed

Let $C$ be a Carmichael number. Set:

$k=\lfloor \sqrt{C}\rfloor +1,k={\displaystyle \underset{p\le k}{\prod }p}$

Then:

$\text{gcd}\left(C,k\#\right)>1$

The primorial test therefore rejects all Carmichael numbers [5] .

By Korselt’s criterion, any Carmichael $C=p_1{p}_2\cdots p_r$, with $r\ge 3$, square-free.

$C>{\left(\sqrt{C}\right)}^r=C^{r/2}\Rightarrow 1>C^{\left(r/2\right)-1},$

But this is impossible, since our assumption just implied:

$1>C^{\left(r/2\right)-1},$

while for any $C>1$ and $r\ge 3$, we clearly have:

$C^{\left(r/2\right)-1}>C^{0.5}=\sqrt{C}>1.$

For example, for the smallest Carmichael number $C=561$, we have $r=3$, and:

$C^{\left(3/2\right)-1}=C^{0.5}=\sqrt{561}\approx 23.68>1.$

Therefore, the inequality $1>C^{\left(r/2\right)-1}$ is false, which contradicts our initial assumption.

Thus $\exists p_i\le \lfloor \sqrt{C}\rfloor +1=k$, so $p_i|k$, and:

$p_i|C\Rightarrow p_i|\text{gcd}\left(C,k\#\right)>1$

Corollary:

The primorial test based on $\text{gcd}\left(n,k\#\right)$ with $k=\lfloor \sqrt{n}\rfloor +1$ rejects all Carmichael numbers without exception. It thus constitutes a robust deterministic barrier against Fermat false positives.

1) Compute $k=\lfloor \sqrt{n}\rfloor +1$

2) Build $k_{}$

3) Check $\text{gcd}\left(n,k\#\right)$

4) If $\text{gcd}>1$ $\to$ $n$ is composite

5) If $\text{gcd}=1$ $\to$ $n$ is prime

Important.

In the context of the primorial primality test, the logarithmic depth defined by:

$k\left(n\right)=\lfloor 3\cdot {\text{log}}_2\left(n\right)\rfloor$

serves as an efficient dynamic upper bound for medium and large integers, since it remains strictly less than $n$, thereby preventing $n$ from being included in the primorial product $k\#$.

However, this bound becomes inappropriate for small integers $n<10$: in this range, we observe that:

$3\cdot {\text{log}}_2\left(n\right)\ge n,$

which may lead to the inclusion of $n$ in $k\#$, resulting in a false rejection of primality.

Conclusion:

$k=\lfloor \sqrt{n}\rfloor \text{with}\text{the}\text{guarantee}\text{that}k<n,$

in order to preserve the validity of the primorial test for small prime numbers.

$k\left(n\right)=\{\begin{array}{ll}\lfloor \sqrt{n}\rfloor \hfill & \text{if}n<10\hfill \\ \lfloor C\left(n\right)\cdot {\text{log}}_2\left(n\right)\rfloor \hfill & \text{if}n\ge 10\hfill \end{array}\text{with}C\left(n\right):=3+\frac{{\text{log}}_{10}\left(n\right)}{3}$