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The sequence starts with the following:

a1 = 1, a2 = 2, a3 = 5, a4 = 4, a5 = 6, a6 = 10, a7 = 9, a8 = 8, a9 = 21, a10 = 12,

a11 = 13, a12 = 20, a13 = 33, a14 = 15, a15 = 42, a16 = 16, a17 = 19, a18 = 63, a19 = 34, a20 = 24, ...

Identify the rule of this sequence and the next five terms.


The same list of numbers, without indices:

1, 2, 5, 4, 6, 10, 9, 8, 21, 12, 13, 20, 33, 15, 42, 16, 19, 63, 34, 24

Hints are spoilered so that people can choose to solve it without seeing them.

Hint 1

Indices are important. There are two important keywords in the title besides "sequence".

Hint 2

a26 through a30 are 27, 78, 29, 84, 31. Still, your task is to find the rule and the values for a21 through a25.

Hint 3-1 (continuation of Hint 1)

One of the two important keywords mentioned in Hint 1 is "root". What is the notation for a function $f$ applied twice to $x$? How does it relate to the word "root"?

Hint 3-2 (continuation of Hint 2)

AnilGoyal's answer is on the right track. Additionally, a21 = 18, a36 = 378, and a63 = 36.

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  • $\begingroup$ I am a first timer here. How should I post answer? directly here or through some other method I mean hidden like your spoiler? $\endgroup$
    – AnilGoyal
    Apr 16 at 9:06
  • $\begingroup$ shouldn't a19 be 51 instead of 34? $\endgroup$
    – AnilGoyal
    Apr 16 at 9:13
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    $\begingroup$ @AnilGoyal No, a19 is 34. $\endgroup$
    – Bubbler
    Apr 16 at 9:31
  • $\begingroup$ Can you please recheck it once again. My logic goes exactly upto a17 and it fails at a17? I may surely be wrong but please recheck it once. $\endgroup$
    – AnilGoyal
    Apr 18 at 3:26
  • $\begingroup$ @AnilGoyal I checked the values again and they are all correct under my rule. Added a few more terms under "Hint 2". I do think you've got pretty good progress though. $\endgroup$
    – Bubbler
    Apr 19 at 0:32
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Before I start, consider function $f$ that satisfies $A_x = f(x)$. That is, to find the $x$-th number in the sequence, we can plug in $x$ to $f$.

So we want to find the values of $f(21), f(22), f(23), f(24), f(25)$.

Inspired from hint 2, instead of mapping x to $f(x)$, what if we try to map the relation of $x$ and $f(f(x))$ instead? We get this following pattern.

$x$ $f(x)$ $f(f(x))$ $x$ in binary
$1$ $f(1)=1$ $f(1)=1=1\times1$ $1$
$2$ $f(2)=2$ $f(2)=2=1\times2$ $10$
$3$ $f(3)=5$ $f(5)=6=2\times3$ $11$
$4$ $f(4)=4$ $f(4)=4=1\times4$ $100$
$5$ $f(5)=6$ $f(6)=10=2\times5$ $101$
$6$ $10$ $12=2\times6$ $110$
$7$ $9$ $21=3\times7$ $111$
$8$ $8$ $8=1\times8$ $1000$
$9$ $21$ $18=2\times9$ $1001$

and so on. Based on the provided sequence, it can be said that

$f(f(x)) = n_x \times x$ where $n_x$ is the number of 1s that appear when $x$ is represented in binary.

That is possibly why $f(f(5))=10=2\times5$ but $f(f(7))=21=3\times7$.

BONUS:

By constructing the table, we have found that $f(21) = 18$ (row 9). It is exactly as stated in hint 3-2.


From rearranging the function we can obtain a new relation

$f(x)=f^{-1}(n_x\times x)$.

For the values of $x=21,22,23,24,25$, it can be calculated that $n_{21}=3, n_{22}=3, n_{23}=4,n_{24}=2,n_{25}=3$.

Therefore,

$f(23) = f^{-1}(78)$. Because we know from hint 3-1 that $f(27)=78$, we know that $f^{-1}(78)$ is $27$. Thus, $f(23)=27.$

Finding $f(24)$ is fun because if we try to construct $x, f(x), f(f(x)), f(f(f(x))), ...$ starting from $x=3$, we get the sequence

$3,5,6,10,12,20,24,40$. So we can know that the value of $f(24)$ is $40$.

So far I can't use $f(22)=f^{-1}(66)$ and $f(25)=f^{-1}(75)$ to my advantage and I'm stuck..

However, I made the connection that the 'root' in title and hint 1 is perhaps meaning 'half-iterate' or 'functional square root', see this Wikipedia page.

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  • $\begingroup$ The rule is mostly correct, and the meaning of "root" is also correct. The reasoning for $f(21)$ and the value of $f(24)$ are correct, but $f(23)$ is not, because $n_{23} × 23 = 92 ≠ 78$. "basic" was referring to "base". You just need a small addition to the rule to define a single sequence. $\endgroup$
    – Bubbler
    Apr 19 at 23:17
  • $\begingroup$ @Bubbler Uh yeah that one was my bad. Thanks for catching the error! $\endgroup$ Apr 20 at 1:20
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It is very clear that -

If $i$ is power of 2, it is $A_i$ indeed. Thus $A_2$ = 2, $A_4$ = 4, and so on..

Now,

$A_1$ takes first of the remaining values which is $1$. The sequence cannot be taken any further and hence stops.

Further,

$A_3$ gets next available value $5$ (some logic seems I am missing here). This value is used to calculate next $A$ with index $5$ which will be double of previous index. Thus, $A_5 = 3*2 = 6$, $A_6 = 5*2 = 10$, $A(10) = 6*2 = 12$, $A(12) = 10*2 = 20$, $A(20) = 12*2 = 24$ which reveals our fourth desired number as $A(24) = 20*2 = 40$

Next,

remaining of our index is 7 thus $A_7$ gets next available value i.e. $9$ here. So $A_7 = 9$. Similarly, $A_9 = 7(previous index) * 3 (again I cannot figured it out why?) = 21$. thus, $A(21) = 9*3 = 27$.

Other sequences

$A(11) = 13$, $A(13) = 33$, $A(33) = 39$.... $A(14) = 15$, $A(15) = 42$, $A(42) = 45$.... $A(17) = 19$ why not 18?...

From here I got confused and couldn't get it

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1
  • $\begingroup$ You're on the right track. To help you identify the rules behind the multiplier, I added three more terms (to specifically help you with the fourth spoiler block) under Hint 3-2. $\endgroup$
    – Bubbler
    Apr 19 at 7:49

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