2025 Fermat Contest About Sequences

2025 Fermat Contest: Exploring Sequences and Prime Factorization

The Centre of Mathematics and Computing Contest recently concluded the 2025 Fermat contest, which brought together math enthusiasts from around the world to showcase their problem-solving skills. Among the many challenging questions, Question No. 25 stood out for its unique blend of sequence analysis and prime factorization. In this article, we'll delve into the details of this intriguing problem and explore its solutions.

Problem Statement

$25.$ A sequence $a_1, a_2, a_3, \ldots$ is defined as follows:

$$a_1 = 2, \quad a_2 = 3, \quad a_3 = 5, \quad a_4 = 7, \quad a_5 = 11, \quad a_6 = 13, \quad a_7 = 17, \quad a_8 = 19, \quad a_9 = 23, \quad a_{10} = 29, \quad \ldots$$

Each term $a_n$ is a prime number, and the sequence is constructed by taking the next prime number after $a_{n-1}$. The problem asks us to find the value of the sum of the first $n$ terms of this sequence, denoted by $S_n$.

Understanding the Sequence

At first glance, the sequence appears to be a simple list of prime numbers. However, upon closer inspection, we notice that each term is not only a prime number but also a prime number that is one more than a multiple of 6. This observation leads us to the following:

$$a_n = 6k + 1$$

where $k$ is a positive integer. This relationship between the terms of the sequence and the multiples of 6 is crucial in understanding the behavior of the sequence.

Prime Factorization

The problem requires us to find the sum of the first $n$ terms of the sequence. To do this, we need to find a way to express the sum in terms of the prime factorization of the terms. We can start by expressing each term as a product of its prime factors:

$$a_n = p_1^{e_1} \cdot p_2^{e_2} \cdot \ldots \cdot p_m^{e_m}$$

where $p_i$ are distinct prime numbers and $e_i$ are positive integers.

Finding the Sum

Using the prime factorization of each term, we can express the sum of the first $n$ terms as:

$$S_n = \sum_{i=1}^n a_i = \sum_{i=1}^n p_1^{e_1} \cdot p_2^{e_2} \cdot \ldots \cdot p_m^{e_m}$$

To simplify this expression, we can use the fact that the sum of the first $n$ prime numbers is given by:

$$\sum_{i=1}^n p_i = \frac{n^2 \ln n}{2} + O(n \ln n)$$

Using this result, we can rewrite the sum as:

$$S_n = \sum_{i=1}^n p_i^{e_i} = \sum_{i=1}^n (p_i)^{e_i}$$

Solving for $S_n$

To find the value of $S_n$, we need to evaluate the sum:

$$S_n = \sum_{i=1}^n (p_i)^{e_i}$$

Using the fact that the sum of the first $n$ prime numbers is given by:

$$\sum_{i=1}^n p_i = \frac{n^2 \ln n}{2} + O(n \ln n)$$

we can rewrite the sum as:

$$S_n = \sum_{i=1}^n (p_i)^{e_i} = \sum_{i=1}^n (p_i)^{e_i} \cdot \frac{n^2 \ln n}{2} + O(n \ln n)$$

Simplifying this expression, we get:

$$S_n = \frac{n^3 \ln n}{2} \sum_{i=1}^n (p_i)^{e_i} + O(n^2 \ln n)$$

Final Answer

Using the result from the previous step, we can find the value of $S_n$:

$$S_n = \frac{n^3 \ln n}{2} \sum_{i=1}^n (p_i)^{e_i} + O(n^2 \ln n)$$

Substituting the values of $n$ and $\ln n$, we get:

$$S_n = \frac{10^3 \ln 10}{2} \sum_{i=1}^{10} (p_i)^{e_i} + O(10^2 \ln 10)$$

Evaluating the sum, we get:

$$S_n = 2500 \ln 10 + O(100 \ln 10)$$

Therefore, the final answer is:

$S_n = 2500 \ln 10 + O(100 \ln 10)$

Conclusion

In this article, we explored the 2025 Fermat contest problem No. 25, which involved finding the sum of the first $n$ terms of a sequence defined by prime numbers. We used the prime factorization of each term to express the sum in terms of the prime numbers and then used the result to find the value of the sum. The final answer was $S_n = 2500 \ln 10 + O(100 \ln 10)$. This problem showcases the importance of understanding the properties of prime numbers and their relationship to the sequence.

Additional Resources

For those interested in learning more about prime numbers and their properties, we recommend the following resources:

• "Prime Numbers" by G.H. Hardy and E.M. Wright: This classic book provides an in-depth introduction to prime numbers and their properties.
• "The Prime Number Theorem" by G.H. Hardy and J.E. Littlewood: This paper provides a detailed proof of the Prime Number Theorem, which states that the number of prime numbers less than or equal to $x$ is approximately $\frac{x}{\ln x}$.
• "Prime Numbers and the Riemann Hypothesis" by Michael Atiyah: This book provides an introduction to the Riemann Hypothesis, which is a famous unsolved problem in number theory that deals with the distribution of prime numbers.

We hope this article has provided a useful introduction to the 2025 Fermat contest problem No. 25 and the properties of prime numbers.
2025 Fermat Contest: Q&A on Sequences and Prime Factorization

In our previous article, we explored the 2025 Fermat contest problem No. 25, which involved finding the sum of the first $n$ terms of a sequence defined by prime numbers. We used the prime factorization of each term to express the sum in terms of the prime numbers and then used the result to find the value of the sum. In this article, we'll answer some of the most frequently asked questions about this problem.

Q: What is the sequence $a_n$ defined by?

A: The sequence $a_n$ is defined as follows:

$$a_1 = 2, \quad a_2 = 3, \quad a_3 = 5, \quad a_4 = 7, \quad a_5 = 11, \quad a_6 = 13, \quad a_7 = 17, \quad a_8 = 19, \quad a_9 = 23, \quad a_{10} = 29, \quad \ldots$$

Each term $a_n$ is a prime number, and the sequence is constructed by taking the next prime number after $a_{n-1}$.

Q: How is the sum of the first $n$ terms of the sequence defined?

A: The sum of the first $n$ terms of the sequence is denoted by $S_n$ and is defined as:

$$S_n = \sum_{i=1}^n a_i$$

Q: What is the relationship between the terms of the sequence and the multiples of 6?

A: Each term $a_n$ is not only a prime number but also a prime number that is one more than a multiple of 6. This can be expressed as:

$$a_n = 6k + 1$$

where $k$ is a positive integer.

Q: How can we use the prime factorization of each term to express the sum in terms of the prime numbers?

A: We can express each term as a product of its prime factors:

$$a_n = p_1^{e_1} \cdot p_2^{e_2} \cdot \ldots \cdot p_m^{e_m}$$

where $p_i$ are distinct prime numbers and $e_i$ are positive integers.

Q: How can we simplify the expression for the sum using the result from the Prime Number Theorem?

A: We can use the fact that the sum of the first $n$ prime numbers is given by:

$$\sum_{i=1}^n p_i = \frac{n^2 \ln n}{2} + O(n \ln n)$$

Using this result, we can rewrite the sum as:

$$S_n = \sum_{i=1}^n (p_i)^{e_i} = \sum_{i=1}^n (p_i)^{e_i} \cdot \frac{n^2 \ln n}{2} + O(n \ln n)$$

Simplifying this expression, we get:

$$S_n = \frac{n^3 \ln n}{2} \sum_{i=1}^n (p_i)^{e_i} + O(n^2 \ln n)$$

Q: What is the final answer to the problem?

A: Using the result from the previous step, we can find the value of $S_n$:

$$S_n = \frac{n^3 \ln n}{2} \sum_{i=1}^n (p_i)^{e_i} + O(n^2 \ln n)$$

Substituting the values of $n$ and $\ln n$, we get:

$$S_n = \frac{10^3 \ln 10}{2} \sum_{i=1}^{10} (p_i)^{e_i} + O(10^2 \ln 10)$$

Evaluating the sum, we get:

$$S_n = 2500 \ln 10 + O(100 \ln 10)$$

Therefore, the final answer is:

$S_n = 2500 \ln 10 + O(100 \ln 10)$

Conclusion

In this article, we answered some of the most frequently asked questions about the 2025 Fermat contest problem No. 25. We hope this Q&A article has provided a useful introduction to the problem and its solution.

Additional Resources

For those interested in learning more about prime numbers and their properties, we recommend the following resources:

• "Prime Numbers" by G.H. Hardy and E.M. Wright: This classic book provides an in-depth introduction to prime numbers and their properties.
• "The Prime Number Theorem" by G.H. Hardy and J.E. Littlewood: This paper provides a detailed proof of the Prime Number Theorem, which states that the number of prime numbers less than or equal to $x$ is approximately $\frac{x}{\ln x}$.
• "Prime Numbers and the Riemann Hypothesis" by Michael Atiyah: This book provides an introduction to the Riemann Hypothesis, which is a famous unsolved problem in number theory that deals with the distribution of prime numbers.

We hope this article has provided a useful introduction to the 2025 Fermat contest problem No. 25 and the properties of prime numbers.