Can you solve my first selfmade number sequence ?
202500, 5184, 32400, 12960, ?
- Hint:
The first three are all square numbers.
- Hint:
The tail of the solution might be confusing.
- Hint
How to get $n$ with $(n-1)$ and $(n-2)$ ?
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$\begingroup$ Just to double check, the order matters? $\endgroup$– RShieldsCommented Jun 20, 2019 at 22:59
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1$\begingroup$ Note: these are all rot13(erthyne ahzoref nxn unzzvat ahzoref) $\endgroup$– RShieldsCommented Jun 20, 2019 at 23:01
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1$\begingroup$ @RShields rot13(nyy jvgu gur fnzr ahzore bs cevzr snpgbef) $\endgroup$– Arnaud MortierCommented Jun 20, 2019 at 23:06
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$\begingroup$ rot13(guerr gb gur cbjre bs sbhe) @ArnaudMortier $\endgroup$– RShieldsCommented Jun 21, 2019 at 1:31
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$\begingroup$ Yes, the order is important. $\endgroup$– TraubenzuckerCommented Jun 21, 2019 at 7:59
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2 Answers
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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 $n>2$, $$ a_n = \sqrt{a_{n-1}a_{n-2}}$$
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(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
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2$\begingroup$ Good observation ! But the solution has decimal places. $\endgroup$ Commented Jun 21, 2019 at 8:04