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What's the next two numbers and how did you get to the answer?

The series:

2, 4, 4, 8, 12, 16, 48, 64, 288, 448, 1728

Hint 1

The first two digits are the beginning of the sequence and were not created by the logic of the sequence.

Hint 2

He was a no one
A zero, zero
Now he's a hot shot
He's a hero

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  • 9
    $\begingroup$ pokes the second hint box, a bit, but nothing rattles out $\endgroup$ May 18, 2016 at 21:48
  • $\begingroup$ Hmm, is "Coming Soon..." some kind of cryptic hint to confuse us, or is the second hint being added soon. ;) $\endgroup$ May 19, 2016 at 9:54
  • $\begingroup$ A few remarks just to save other solvers doing redundant work. First: the sequence is not in OEIS. Second: every term so far is a multiple of the term two before it. Third: the ratios are 2 2 3 2 4 4 6 7 6. Fourth: the sequence 2,4,12,48,288,1728 (i.e., alternate terms from this sequence) is also not in OEIS. $\endgroup$
    – Gareth McCaughan
    May 19, 2016 at 13:32
  • $\begingroup$ This puzzle feels Herculean. $\endgroup$ May 19, 2016 at 13:55

2 Answers 2

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The next two numbers are

$3136$ and $15552$.

Reasoning:

Denote by $f(n)$ the number of $0$s in the binary representation of $n$. For example:
$n=4$: binary representation = $100$; hence $f(4)=2$
$n=8$: binary representation = $1000$; hence $f(8)=3$
$n=12$: binary representation = $1100$; hence $f(12)=2$
$n=16$: binary representation = $1000$; hence $f(16)=4$
$n=48$: binary representation = $110000$; hence $f(48)=4$
$n=64$: binary representation = $1000000$; hence $f(64)=6$
$n=288$: binary representation = $100100000$; hence $f(288)=7$
$n=448$: binary representation = $111000000$; hence $f(448)=6$
$n=1728$: binary representation = $11011000000$; hence $f(1728)=7$
$n=3136$: binary representation = $110001000000$; hence $f(3136)=9$

The recursion is:

$a_n~=~ a_{n-2}*f(a_{n-1})$

Example:

$a_3~=~ a_1*f(a_2) ~=~ 2*f(4) ~=~ 2*2 ~=~ 4$
$a_4~=~ a_2*f(a_3) ~=~ 4*f(4) ~=~ 4*2 ~=~ 8$
$a_5~=~ a_3*f(a_4) ~=~ 4*f(8) ~=~ 4*3 ~=~12$
$a_6~=~ a_4*f(a_5) ~=~ 8*f(12) ~=~ 8*2 ~=~16$
$a_7~=~ a_5*f(a_6) ~=~ 12*f(16) ~= 12*4 ~=~48$
$a_8~=~ a_6*f(a_7) ~=~ 16*f(48) ~=~16*4 ~=~64$
$a_9~=~ a_7*f(a_8) ~=~ 48*f(64) ~=~48*6 ~=~288$
$a_{10}~=~ a_8*f(a_9) ~=~ 64*f(288) ~=~ 64*7 ~=~ 448$
$a_{11}~=~ a_9*f(a_{10}) ~=~ 288*f(448) ~=~ 288*6 ~=~ 1728$
$a_{12}~=~ a_{10}*f(a_{11}) ~=~ 448*f(1728) ~=~ 448*7 ~=~ 3136$
$a_{13}~=~ a_{11}*f(a_{12}) ~=~ 1728*f(3136) ~=~ 1728*9 ~=~ 15552$

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  • $\begingroup$ Ah, I just had it too. Well done! $\endgroup$
    – hexomino
    May 19, 2016 at 13:53
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Edit: This is an old answer based on the first $10$ entries in the sequence.

I think the next two numbers in the sequence are

$2016$ and $4032$

Reasoning

For a number $x$ define $i(x) = x - \lfloor x \rfloor$
and define $f(x) = \lfloor x \rfloor - \lceil i(x) \rceil$

As per the hint, set $a_1 = 2$ and $a_2 = 4$.
Then, for $n>2$, the $n$th term in the sequence is given by

$$ a_n = a_{n-2} * f(\log_2 (a_{n-1})) $$

Example

$a_9 = 288$ and $a_{10} = 448$.
$\log_2(a_{10}) = \log_2(448) = 8.8073549\ldots$
$ \Rightarrow f(\log_2(448)) = 8 - 1 = 7$

Hence, $a_{11} = 288 * 7 = 2016$

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  • $\begingroup$ Another way to define your function f: f(x) = x if x is an integer, otherwise f(x) = floor(x)-1. $\endgroup$
    – Gareth McCaughan
    May 19, 2016 at 11:02
  • $\begingroup$ You're math and logic are sound, it matches the given numbers but it is not the next two in my sequence. Maybe I haven't given enough numbers to differentiate the two sequences. I will add another number. I'm sorry about that. $\endgroup$ May 19, 2016 at 13:13
  • $\begingroup$ @TheReffered given any finite sequence, an infinite amount of sequences can be made to fit it. $\endgroup$
    – KoA
    May 19, 2016 at 13:19
  • $\begingroup$ @KoA What's the protocol for sequences then? Should I skip two numbers in my sequence and give one on the other side? Or just mark this answer as correct? $\endgroup$ May 19, 2016 at 13:22
  • $\begingroup$ @TheReffered as long as the 'rule' you've come up with is clearly the most correct, then you don't have to do anything - generally with sequence questions it's best to leave another hint as to what the sequence is, as you have :) $\endgroup$
    – KoA
    May 19, 2016 at 13:24

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