Proving the Sequence of Numbers is Always Composite

The sequence of numbers consists of terms such as 1, 1001, 1001001, 1001001001, and so on. Each term in this sequence follows a specific pattern. Let's denote the nth term in this sequence as Nn. Specifically, Nn is a number where each nth term contains n zeros and n ones. Mathematically, this can be expressed as:

Expression of Nn

Nn 10(2n) - 1

Step 1: Factorization of Nn

Using the difference of squares identity a2 - b2 (a b)(a - b), we can factorize Nn as follows:

Nn 10(2n) - 1

Step 2: Using the Factorization of 10(2n) - 1

For n ≥ 1, we can factorize it as:

Nn (10n 1)(10n - 1)

This factorization proves that Nn can be expressed as a product of two terms that are both greater than 1 for all n ≥ 1.

Step 3: Proving Nn is Composite

Let's verify this for specific values of n.

For n 1

N1 1001 7 × 143 7 × 11 × 13

For n 2

N2 1001001 1001 × 1001 10012

For n ≥ 3

The expression Nn 10(2n) - 1 can be factored as:

Nn (10n 1)(10n - 1)

Since both 10n - 1 and 10n 1 are greater than 1 for all n ≥ 1, it follows that Nn is a composite number.

Conclusion

Therefore, for all n ≥ 1, Nn can be expressed as a product of two integers greater than 1. As a result, the numbers 1, 1001, 1001001, 1001001001, and so on, are never prime numbers.

Proof Using Another Formulation

Let the sequence be an. Clearly, a1 1, a2 7 × 11 × 13, and a3 3 × 333667 are not prime numbers. Now, let's consider an for n ≥ 4: a_n 110^{3}10^{6} ... 10^{3n-1} frac{(10^3)^n - 1}{10^3 - 1} frac{(10^n)^3 - 1^3}{999} frac{(10^n - 1)(10^{2n} 10^n 1)}{999}

For any n ≥ 4, both 10^n - 1 and 10^{2n} 10^n 1 are greater than 999. Hence, an can be factorized as a product of two integers greater than 1, and therefore, an is never a prime number.