6
$\begingroup$

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, __

$\endgroup$
  • 1
    $\begingroup$ Hmm. Is it purely mathematical or not? $\endgroup$ – Don Thousand Feb 25 at 2:39
  • $\begingroup$ No non-mathematical shenanigans occur in the above sequence. @DonThousand $\endgroup$ – Zaenille Feb 25 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 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 at 3:07
  • $\begingroup$ Then yes, it is purely mathematical. Good luck! $\endgroup$ – Zaenille Feb 25 at 3:14
9
$\begingroup$

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:
$a_7 = 252 = 7 \times 36$

| improve this answer | |
$\endgroup$
  • $\begingroup$ Teamwork! I like it! $\endgroup$ – Ébe Isaac Feb 25 at 16:08
  • $\begingroup$ Awesome job! You got it. $\endgroup$ – Zaenille Feb 25 at 22:19
  • $\begingroup$ good job. i didn't think about digits :)) $\endgroup$ – Mahdi Mahmoodian Feb 27 at 18:58
5
$\begingroup$

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.

| improve this answer | |
$\endgroup$
  • $\begingroup$ 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 at 22:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.