An Early Christmas Algebraic Sequence

The key fact is: if $p$ and $q$ are consecutive primes, then $a_{p+q-2} = pq$ .

You can see this by induction: it's true for $p = 2, q = 3$ , and then if it's true for $p$ and $q$ , suppose $r$ is the next prime after $q$ . Then the terms of the sequence between $a_{p+q-2}$ and $a_{q+r-2}$ are pretty clearly an arithmetic sequence with common difference $q$ , as $q$ is the largest prime factor of all of them. So $a_{q+r-2} = qr$ and the inductive hypothesis holds.

So $a_{196} = 97 \cdot 101$ , and then $a_{198} = 99 \cdot 101 = 9999$ , so $\fbox{198}$ is the answer.

observed that the sequence is

2×1 , 2×2 , 2×3 , 3×3 , 3×4 , 3×5 , 4×5 , 5×5

There are some perfect square of prime , Consider from a part of sequence (p,q are primes)

p×(p+1) , p×(p+2) , ... , p×q , p+1×q , p+2×q , ... , q×q

There are 2(q-p) terms , So the Sequence

2×1 , 2×2 , 2×3 , 3×3 , ... p×p , ... , q×q , ... , r×r

there are 2(r-q) + 2(q-p) + 2(p-3) + 2(3-2) + 2 terms , which simplify to 2r-2

now find a prime number that squared is less than 9999

that prime is 97 (97×97 = 9409)

so the sequence

2 , 4 ,6 , 9 , .... , 9409

has 192 terms , now keep finding few next terms

a192 = 97×97

a193 = 98×97

a194 = 99×97

a195 = 100×97

a196 = 101×97

a197 = 101×98

a198 = 101×99

so the answer is 198

We notice that the sequence of $a_n$ is as follows: $S_n = \{2,4,6,9,12,15,20,25,30,35...\}$ $\quad \space = \{1\dot{}2,2\dot{}2,2\dot{}3,3\dot{}3,4\dot{}3,3\dot{}5,4\dot{}5,5\dot{}5,6\dot{}5,5\dot{}7...\}$

If $q_n$ denotes the $n^{th}$ prime, then:

$S_n = \{q_1,2q_1,q_1q_2,3q_2,4q_2,q_2q_3,4q_3,5q_3,6q_3,q_3q_4...q_kq_{k+1}...\}$

When $a_n=q_kq_{k+1}$ then

$n = (q_2-0)+(q_3-q_1)+(q_4-q_2)+ ...+(q_{k+1}-q_k) = q_{k+1}+q_k-q_1$

For example:

When $k = 1$ or $a_n = 2\dot{}3 = 6 \quad \Rightarrow n = q_2 + q_1 - q_1 = 0 + 3 + 2 -2 = 3$ .

When $k = 2$ or $a_n = 15 \quad \Rightarrow n = q_3 + q_2 - q_1 = 5 + 3 - 2 = 6$

When $k = 3$ or $a_n = 35$ $\quad \Rightarrow n = q_4 + q_3 - q_1 = 7 + 5 - 2 = 10$

The largest $k$ for $a_n < 10000$ is $25$ , when $a_{n_1} = q_{25}q_{26} = 97\dot{}101 = 9797$ , when $n = 101 + 97 -2 = 196$ .

Then $a_{197} = 9797 + 101 = 9898$ and $a_{198} = 9898 + 101 = 9999$ .

Therefore the largest $n = \boxed {198}$ .

Lemma : For every prime $p$ , the perfect square $p^2$ occupies position $2p-2$ in the sequence.

Thus, the last perfect square less than $100^2$ is $p = 97$ at position 192.

The sequence then continues: $a_{192} = 97^2, a_{193} = 97\cdot 98, a_{194} = 97\cdot 99, a_{195} = 97\cdot 100, \\ a_{196} = 97\cdot 101, a_{197} = 98\cdot 101, a_{198} = 99\cdot 101 = 9999.$ Thus the answer is $\boxed{198}$ .

(Or, once you detect the pattern, count backward from $a_{200} = 101^2$ .)

Proof of the lemma, by induction. Suppose the statement is true for some prime $p$ , and let $p'$ be the next prime number. Then $\begin{array}{ll} a_{2p-2+1} = p\cdot (p+1), \\ a_{2p-2+n} = p\cdot (p+n) & \text{for}\ p+ n \leq p', \\ a_{p+p'-2} = p\cdot p', \\ a_{p+p'-1} = (p+1)\cdot p', \\ a_{p+p'+n-2} = (p+n)\cdot p' & \text{for}\ p+n \leq p', \\ a_{2p'-2} = p'\cdot p'.\end{array}$ Since the statement is true for $p = 2$ ( $a_2 = 2^2$ ), it is true for all prime numbers.

2×1,2×2,2×3|3×3,3×4,3×5|...97×101|101×98,101×99 so we see that answer is 3+(5-2)+(7-3)+(11-5)+...+(101-89)+2= 101+97=198 a198=9999

I observed that every other nimber in the pattern is either a perfect square of a prime, or one below the perfect square of a composite, for each even term of the sequence $a_{n}$ it is the either the square or $1$ less the square of $\frac{n}{2}+1$ , the closest square to $9999$ is $10000=100^{2}$ so $100=\frac{n}{2}+1$ and $n=198$ , because $10000$ is the square of a composite and not a prime, $a_{198}$ is $1$ less than $10000$ , or $9999$ and the answer is $\boxed{198}$

Disclaimer: I'm aware this isn't a mathematically stable argument as I gave no proof for the pattern I found however as this problem was from a mathematic competition I had speed in mind and went with the pattern I first saw