6
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First time entering a puzzle so I have no idea if what I've made is too easy. But it's not on OEIS at least.

Find the next term!

1, 2, 9, 48, 120, 162, __

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5
  • 1
    $\begingroup$ Hmm. Is it purely mathematical or not? $\endgroup$ – Don Thousand Feb 25 '20 at 2:39
  • $\begingroup$ No non-mathematical shenanigans occur in the above sequence. @DonThousand $\endgroup$ – Zaenille Feb 25 '20 at 2:54
  • $\begingroup$ Actually, I guess it depends what you mean by purely mathematical. I'll probably just drop hints over time if it doesn't get answered. $\endgroup$ – Zaenille Feb 25 '20 at 3:06
  • 1
    $\begingroup$ I mean, can I predict the next term if all I know is math. Honestly, the term that's throwing me off is 120, since it has a divisor of 5, unlike all the other terms. $\endgroup$ – Don Thousand Feb 25 '20 at 3:07
  • $\begingroup$ Then yes, it is purely mathematical. Good luck! $\endgroup$ – Zaenille Feb 25 '20 at 3:14
9
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Strongly inspired by Mahdi Mahmoodian's answer:

$a_n = n \times s$ where $s$ is the sum of digits in all previous numbers in the sequence

$a_1 = 1 \rightarrow$ sum of digits $= 1$

$a_2 = 2 = 2 \times 1 \rightarrow$ sum of digits $= 1 + 2 = 3$

$a_3 = 9 = 3 \times 3 \rightarrow$ sum of digits $= 1 + 2 + 9 = 12$

$a_4 = 48 = 4 \times 12 \rightarrow$ sum of digits $= 1 + 2 + 9 + 4 + 8 = 24$

$a_5 = 120 = 5 \times 24 \rightarrow$ sum of digits $= 1 + 2 + 9 + 4 + 8 + 1 + 2 + 0 = 27$

$a_6 = 162 = 6 \times 27 \rightarrow$ sum of digits $= 1 + 2 + 9 + 4 + 8 + 1 + 2 + 0 + 1 + 6 + 2 = 36$

The answer is:
$a_7 = 252 = 7 \times 36$

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3
  • $\begingroup$ Teamwork! I like it! $\endgroup$ – Ébe Isaac Feb 25 '20 at 16:08
  • $\begingroup$ Awesome job! You got it. $\endgroup$ – Zaenille Feb 25 '20 at 22:19
  • $\begingroup$ good job. i didn't think about digits :)) $\endgroup$ – Mahdi Mahmoodian Feb 27 '20 at 18:58
5
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I got an idea but it's not complete yet. I will update it as soon as I find something else:

The $i$th number has $i$ as the divisor.

$f(1) = 1 * 1 = 1$
$f(2) = 2 * 1 = 2$
$f(3) = 3 * 3 = 9$
$f(4) = 4* 12 = 48$
$f(5) = 5*24 = 120$
$f(6) = 6 * 27 = 162$

Also, I find something that applies to the first 4 number:

The second divisor of $i$th number is $\sum_{n=1}^{i-1}f(n)$.

Example: $f(4) = 4 * \sum_{n=1}^{3}f(n) =4 * (1 + 2 + 9) = 48$

But after $48$ it doesn't work.

So after all I know one thing about the answer:

It's divisible by 7.

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1
  • $\begingroup$ Good job for the first to get so close to the mark. Upvote from me. Just missing that last bit of the equation which Michal has finished. $\endgroup$ – Zaenille Feb 25 '20 at 22:20

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