More on the Fascinating Characterizations of Mulatu’s Numbers

Lemma 1.

Any two consecutive Mulatu numbers are relatively prime.

Proof.

Mathematically, the lemma is $gcd(M_n,M_{n+1})=1$ for any non-negative integer $n,$ where gcd denotes greatest common divisor.

We use induction on $n.$

For $n=0,gcd(M_0,M_1)=gcd\left(41\right)=1.$ Assume the lemma holds for $n=k>0.$ That is

Claim: $gcd(M_{k+1},M_{k+2})=1,$ for all $k\ge 0.$

From the Euclidean algorithm, we have:

$gcd(M_n,M_{n+1})=gcd(M_n,M_{n+1}\mathit{mod}{M}_n),$ for all $n\ge 0.$

Now, $M_{k+2}\mathit{mod}{M}_{k+1}=\left(M_k+M_{k+1}\right)\mathit{mod}{M}_{k+1}=M_k.$ This implies that

$gcd(M_{k+1},M_{k+2})=gcd(M_k,M_{k+1})=1,$ for all $k\ge 0.$ Therefore, the lemma holds by the principle of mathematical induction.

Theorem 1.

Adding any ten consecutive Mulatu numbers together will always result in a number that is divisible by 11.

Proof.

Let $q={\sum }_{n=i}^{i+9}{M}_n$ for any number $i$ . We then show that $q$ is divisible by $11$ .

Using the definition of Mulatu numbers, we have $M_{i+j}=F_{j-1}{M}_i+F_j{M}_{i+1}$ for $j\ge 2$ in the set of natural numbers, where $F_n$ denotes the $n^{\mathit{th}}$ Fibonacci number.

This completes the proof.

Theorem 2.

Multiplying any Mulatu’s number by two and subtracting the next Mulatu’s number in the sequence for a number greater than or equal to two will result in the answer being Mulatu’s number, that is, $2M_n-M_{n+1}=M_{n-2}$ , $n\ge 2.$

Proof.

Claim: $2M_n-M_{n+1}=M_{n-2}$ , $n\ge 2.$

Now, $M_{n+1}+M_{n-2}=M_n+M_{n-1}+M_{n-2}=2M_n.$

Theorem 3.

When adding consecutive, even-positioned Mulatu numbers beginning with $M_{2i}$ , for $i\ge 1$ , the result is a number that is one less than the Mulatu number succeeding the last Mulatu number in the sum. This provides a general formula for a simple way to find the sum of any finite even-positioned Mulatu number.

Proof.

By definition, $M_{2i}=M_{2i+1}-M_{2i-1}$ , for any natural number i.

This implies that

So, we are done.

Theorem 4.

When adding consecutive, odd-positioned Mulatu’s numbers beginning with $M_1$ , only this time, the result is a number that is four less than the Mulatu number following the last even number in the sum. This provides a general formula for a simple way to find the sum of any finite odd-positioned Mulatu number.

Proof.

We use induction on $n$ .

Theorem 5.

When any four consecutive numbers in the Mulatu sequence are considered, the difference between the squares of the two numbers in the middle is equal to the product of the two outer numbers. Mathematically,

Corollary 1.

For $n\ge 1$ , $2M_n+M_{n-1}=M_{n+2}.$

Proof.

By using Theorem 5 above and the relation $M_{n+1}=M_n+M_{n-1}$ , the result follows.

Theorem 6.

Adding any number of consecutive Mulatu numbers will result in a number that is one less than the Mulatu number of two places beyond the last summand. This provides a general formula for a simple way to find the sum of any number of Mulatu numbers.

Theorem 7.

Let $M_n$ , $F_n$ and $L_n$ be the Mulatu’s, Fibonacci’s, and Lucas’s numbers for each $n\ge 0$ . The number $(M_n+M_{n+3})(F_n+F_{n+3})(L_n+L_{n+3})$ is divisible by 8.

Proof.

We use induction on $n$ .

For $n=0,M_0+M_3=10,F_0+F_3=2$ and $L_0+L_3=6$ . This implies that

$(M_0+M_3)(F_0+F_3)(L_0+L_3)=120$ is divisible by 8.

Assume that the statement holds for $n=k$ . As $M_k+M_{k+3}=2(M_k+M_{k+1})$ , we have

Thus, $2|(M_{k+1}+M_{k+4})$ . Similarly, $2|(F_{k+1}+F_{k+4})$ and $2|(L_{k+1}+L_{k+4})$ .

Hence, the theorem holds by the principle of mathematical induction.

Theorem 8.

Half of the sum of any two even consecutive Mulatu numbers yields Mulatu’s number preceding the larger one. i.e.,

Proof.

The above equation represents the statement of the theorem as it is easy to show that every (3n) ^th Mulatu number is even for any non-negative integer n.

We use induction on $n$ .

For $n=0,\frac{1}{2}(M_0+M_3)=5=M_2.$ Let it be true for $n=k$ . That is

This assumption with the definition of Mulatu’s sequence leads:

$\frac{1}{2}(M_{3n}+M_{3(n+1)})=M_{3(n+1)-1},$ for all $n.$

The following definitions are given to study Mulatu numbers in relation with natural numbers, and to define a characteristic number to it.

Definition 1.

a) Two natural numbers are said to be Mulatu summable if their sum is a Mulatu number.

b) A natural number $k$ is said to be $\gamma$ -Mulatu summable if a natural number $\gamma \ge 2$ exists such that $\mathit{\gamma k}$ is a Mulatu number.

Definition 2.

Let $\gamma$ be the smallest natural number such that a natural number $k$ is $\gamma$ -Mulatu summable. Then, $\gamma$ is said to be the Mulatu characteristic of $k$ , denoted by $m^{\ast }\left(k\right).$

Lemma 2.

The Mulatu characteristics neither preserve nor reverse any inequality.

Proof.

Let $k$ and $n$ be natural numbers with $k<n$ . Suppose that the Mulatu characteristic either preserve or reverse an inequality. That is, the Mulatu characteristics preserve or reverse an inequality. Thus, it must preserve or reverse the inequality $k<n$ . Thus, $m^{\ast }\left(k\right)<m^{\ast }\left(n\right)$ or $m^{\ast }\left(k\right)>m^{\ast }\left(n\right)$ for each $k<n$ .

Neither is true because, for instance, $2<3$ but $m^{\ast }$ (2) $=$ 2 = $m^{\ast }$ (3). For the remaining inequalities, we can see counter examples $1<2$ but $m^{\ast }\left(1\right)=4>1=m^{\ast }\left(2\right)$ ; and $3<4$ , with $m^{\ast }\left(3\right)=2<7=m^{\ast }\left(4\right)$ . Thus, the negation of the given statement is not true, and this completes the proof.

Theorem 9.

Let $k$ and $n$ be natural numbers such that $k<n$ , then

Proof.

Suppose not. That is $k{m}^{\ast }\left(k\right)\le {\mathit{nm}}^{\ast }\left(n\right)$ for all $k<n.$

Dividing both sides by $k{m}^{\ast }\left(n\right)$ gives

This implies

$m^{\ast }\left(k\right)>m^{\ast }\left(n\right)$ > m ∗ (n), $\forall k<n$ , contradicting Lemma 2.

Theorem 10.

Let ${\gamma }_i\ge 2$ and ${\gamma }_{i+1}>2$ for $i\ge 0$ be consecutive natural numbers such that 2 is ${\gamma }_i$ -Mulatu summable and is ${\gamma }_{i+1}$ -Mulatu summable, respectively, and let ${\gamma }_i$ and ${\gamma }_{i+1}$ be Mulatu summable.

Proof.

We use induction on $i$ .

For $i=0,{\gamma }_0=2,{\gamma }_1=3$ and ${\gamma }_0+{\gamma }_1=5=M_2$ . Assuming it holds for $i=k$ and Theorem 8, we get

Definition 3.

The series ${\sum }_{n=0}^{\infty }{M}_n$ is called Mulatu’s series.

Theorem 11.

Mulatu’s series is divergent.

Proof.

Using the ratio test, $\underset{n\to \infty }{lim}\frac{M_{n+1}}{M_n}=1.618>1.$ Thus, it diverges.

Theorem 12.

$f\left(x\right)={\sum }_{n=0}^{\infty }{M}_n{x}^n=\frac{4-3x}{1-x-x^2}$ is a generating function for the Mulatu’s sequence ${\left\{M_n\right\}}_{n=0}^{\infty }.$

Proof.

Let $f\left(x\right)={\sum }_{n=0}^{\infty }{M}_n{x}^n$ .

Claim: $f\left(x\right)=\frac{4-3x}{1-x-x^2}$

Using the properties of Mulatu’s numbers, specifically, $M_{n+2}=M_n+M_{n+1},$ we have

This implies,

That is

Substituting $M_0$ and $M_1$ completes the proof.