How is this set of 8 numbers - $6, 8, 15, 20, 36, 48, 90$, and $120$ - derived?
I could provide hints but it would make it too easy.
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2 Answers
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I don't know if it is the right solution, but if we consider:
a die
Then I got this explanation:
Roll a die in front of you. You should see three sides, like this:
Now, multiply the numbers appearing on the three faces. You should get a number between $6$ and $120$.
In fact, there are only 8 possibilities: the ones in the sequence.
Why? Because, if you see a number $x$, $1 \le x\le 6$, then you can't see the number $7-x$, because it is on the opposite side.
So if we list all the possible products obtained from multiplying three numbers on a die, and if we cross out the non-obtainable ones, we get the following list :
$1*2*3 = 6$
$1*2*4 = 8$
1*2*5 = 10
1*2*6 = 12
1*3*4 = 12
$1*3*5 = 15$
1*3*6 = 18
$1*4*5 = 20$
1*4*6 = 24
1*5*6 = 30
2*3*4 = 24
2*3*5 = 30
$2*3*6 = 36$
2*4*5 = 40
$2*4*6 = 48$
2*5*6 = 60
3*4*5 = 60
3*4*6 = 72
$3*5*6 = 90$
$4*5*6 = 120$
And of course, you will recognize the sequence in the question. Plus, this sequence is finite.
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$\begingroup$ Nice! It explains, why something like x/(7-x) appeared in the formula I've found, and also why there is a binary-like symmetry. $\endgroup$– eliasAug 30, 2016 at 9:46
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$\begingroup$ @elias Exactly! Thanks for correcting the typo by the way :) $\endgroup$– IAmInPLSAug 30, 2016 at 9:49
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1$\begingroup$ @trentcl Yes, thank you. I was not sure : oxforddictionaries.com/definition/english/dice. It states that in modern standard English dice is both the singular and the plural: throw the dice could mean a reference to either one or more than one dice $\endgroup$– IAmInPLSAug 30, 2016 at 18:17
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3$\begingroup$ @IAmInPLS Huh. For something that is "modern standard English", I've never seen "a dice" in print (board game manuals, for example, pretty much unanimously use "a die"). Maybe it's a British/American split. I would still recommend using "a die". A question about this on english.se $\endgroup$– trentAug 30, 2016 at 19:28
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If we note them as
$a_0=6$, $a_1=8$, and so on,
they match the formula:
$a_n=a_{n-b_n}\times\frac{4+\log_2b_n}{3-\log_2b_n}$, where $b_n$ is the largest possible power of two, which divides $n$
For example:
$n=4$: $b_n=4$
$a_4=a_{4-4}\times\frac{4+2}{3-2}=a_0\times\frac61=36$
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$\begingroup$ it does, as calculating the next term would include a division by 0 $\endgroup$– eliasAug 30, 2016 at 5:09
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$\begingroup$ I understand now. The series is derived in a different way. $\endgroup$– MotiAug 30, 2016 at 5:14
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$\begingroup$ could you elaborate on why this does not work? maybe by adding some of the next terms in your sequence? $\endgroup$– eliasAug 30, 2016 at 5:17
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4$\begingroup$ Ok. I'm interested in the solution you have in mind and how it is related to the formula I gave: if it is a purely accidental match, or one being a direct conclusion of the other. I will wait for the hints patiently. $\endgroup$– eliasAug 30, 2016 at 6:11