# Finding the rank of a particular number in a sequence of the sum of numbers and their highest prime factor

This question comes from a maths contest (infer no calculators or other electronic calculating aids) for 14-16 year olds (infer no use of complicated theorems, but those accessible to high-school students).

Consider the sequence $\;a_1,a_2,a_3,a_4,... \;$ such that $\;a_1=2 \;$ and for every positive integer $n$,

$a_{n+1}=a_n+p_n \;,\;$ where $\;p_n\;$ is the largest prime factor of $\;a_n\;$ The first few terms of the sequence are $\;2,4,6,9,12,15,20.\;$ What is the largest value of $\;n\;$ such that $\;a_n\;$ is a four-digit number ?

Continuing the sequence:

$\underbrace{2,4,\color{red}{6}}_{\text{add}\; 2},\underbrace{9,12,\color{red}{15}}_{\text{add}\; 3},\underbrace{20,25,30,\color{red}{35}}_{\text{add}\; 5},\underbrace{42,49,56,63,70,\color{red}{77}}_{\text{add}\; 7},\underbrace{88,99,110,121,132,\color{red}{143}}_{\text{add}\; 11},\underbrace{156,\ldots}_{\text{add}\; 13}$

Inferring from the above:

1. The numbers to be added to get the next number in the sequence are the prime numbers: $\qquad \qquad \qquad \qquad \qquad2,3,5,7,11,13,17,\ldots$
2. The numbers in red are the product of two successive prime numbers:\begin{align} 2 &\times 3=6\\ 3 &\times 5=15\\ 5 &\times 7=35\\ &\vdots\\ \text{thence}&\\ &\vdots\\ 91 &\times 97=8827\\ 97 &\times 101=\color{blue}{9797} \longleftarrow \\ 101 &\times 103=10403\\ \end{align} and $9797$ is the largest $4$-digit number in the sequence.
3. $$\begin{array}{c|ccccc} n & 3 & 6 & 10 &16& 22 \\ \hline \hline \text{current}\; n-\text{previous}\; n & - &\color{DarkOrange}{ 3 }& \color{DarkOrange}{4} &\color{DarkOrange}{6}& \color{DarkOrange}{6}\\ \hline a_n & 6&15&35&77&143 \\ \hline \text{highest prime factor of}\; a_n & 3&5&7&11&13\\ \text{lowest prime factor of}\; a_{n-1} & -&2&3&5&7&\\ \text{difference} & -&\color{DarkOrange}{ 3 }& \color{DarkOrange}{4} &\color{DarkOrange}{6}& \color{DarkOrange}{6}\\ \end{array}$$

My problem is how do I now correlate the numbers in $\;\color{DarkOrange}{\text{orange}}\;$ to find the $\;n\;$ which results in $\;a_n=9797\;$.

Just observe that to get a product of the form $p_np_{n+1}$ from $p_np_{n-1}$ you have to wait for $p_{n+1}-p_{n-1}$ terms. So the required $n$ is, to get to a prime product $p_kp_{k+1}$ is $\sum_{j=1}^k(p_{j+1}-p_{j-1})+1=p_{k+1}+p_k-2$ where I have defined $p_0=1$. So for your case the answer is $103+101-2=202$.