# Number sequence: 202500, 5184, 32400, 12960,?

Can you solve my first selfmade number sequence ?

202500, 5184, 32400, 12960, ?

1. Hint:

The first three are all square numbers.

1. Hint:

The tail of the solution might be confusing.

1. Hint

How to get $$n$$ with $$(n-1)$$ and $$(n-2)$$ ?

• Just to double check, the order matters? – RShields Jun 20 '19 at 22:59
• Note: these are all rot13(erthyne ahzoref nxn unzzvat ahzoref) – RShields Jun 20 '19 at 23:01
• @RShields rot13(nyy jvgu gur fnzr ahzore bs cevzr snpgbef) – Arnaud Mortier Jun 20 '19 at 23:06
• rot13(guerr gb gur cbjre bs sbhe) @ArnaudMortier – RShields Jun 21 '19 at 1:31
• Yes, the order is important. – Matti Jun 21 '19 at 7:59

Building on the observation by rhsquared, the next number is

$$\sqrt{2^9 3^8 5^3} = 6480 \sqrt{10} \approx 20491.559$$

Reasoning

Each element of the sequence is the geometric mean of the previous two.
That is for $$n>2$$, $$a_n = \sqrt{a_{n-1}a_{n-2}}$$

(Partial) It looks like all the numbers are

Regular numbers, i.e. they can be represented as 2^i·3^j·5^k or
202500 = 2^2 * 3^4 * 5^4
5184 = 2^6 * 3^4 * 5^0
32400 = 2^4 * 3^4 * 5^2
12960 = 2^5 * 3^4 * 5^1
We can see that all the numbers contain 3^4 and also for all of them the sum of the powers equals 10.
So the answer will be in the form 2^i * 3^4 * 5^j, where i + j = 6, with one of the following combinations: 0,6 where the result is 1265625
1,5 where the result is 506250
3,3 where the result is 81000

• Good observation ! But the solution has decimal places. – Matti Jun 21 '19 at 8:04