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Here's a sequence I know of that offhand isn't obvious.

$0.14,0.14,3.28,3.28,9.57,3.28,9.57,3.28,9.57,15.85,3.28,15.85,...$

Keep in mind that this sequence originally looked like something else, but I did something to it.

I want to know what the sequence is (it's fairly well known), what the next $5$ numbers in the modified sequence are and finally, what I did to it to modify it.

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  • $\begingroup$ Are those the exact numbers or have you rounded them off? $\endgroup$
    – Gareth McCaughan ♦
    Jun 10, 2016 at 21:03
  • $\begingroup$ (feel free not to answer if working that out is part of the challenge) $\endgroup$
    – Gareth McCaughan ♦
    Jun 10, 2016 at 21:03
  • $\begingroup$ (I am fairly sure whether you have or not, but will not say in case you prefer it to be part of the challenge) $\endgroup$
    – Gareth McCaughan ♦
    Jun 10, 2016 at 21:05
  • $\begingroup$ Interesting how the apparent common denominator of 1/7 is a clue $\endgroup$
    – humn
    Jun 10, 2016 at 22:26

2 Answers 2

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The sequence is:

the difference between consecutive prime numbers ($1, 2, 2, 4, 2, 4, 2, 4, 6, 2, 6, \ldots$), which apparently is sequence A001223.

The next five numbers in the modified sequence are:

$9.57, 3.28, 9.57, 15.85, 15.85$

And the modified sequence is:

$a_n\pi-3$ rounded to two decimal places, where $a_n$ is the corresponding element in the original sequence. (This only works if the original sequence starts with a $1$, though, so maybe you had thought that $1$ was a prime number? In which case, the first element would be $2-1=1$ followed by another $1$.)

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  • $\begingroup$ Correct. I always forget that 1 doesn't count, so I tend to put it in with the others. $\endgroup$
    – dcfyj
    Jun 10, 2016 at 21:47
  • $\begingroup$ No worries! Good puzzle :) $\endgroup$
    – DooplissForce
    Jun 10, 2016 at 21:48
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    $\begingroup$ If you look at sequence A075526 instead, it looks like you can fix up the 1-not-a-prime issue as follows: write $\pi(n)$ = number of primes <=n for n=1,2,3,... and then your $a_n$ is the number of $m$ with $\pi(m)=n$. So now you have two sorts of pi in the definition of your sequence, which presumably makes it even tastier. $\endgroup$
    – Gareth McCaughan ♦
    Jun 10, 2016 at 21:50
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Well, this looks like

pi,pi,2pi,2pi,
4pi,2pi,4pi,2pi,4pi,
6pi,2pi,6pi,?

I couldn't find it's origin still (sorry for looking dumb?), but it would look like the next ones are 2pi,6pi,8pi,2pi,8pi..., and in yours they would be the same numbers minus three.

What did you do.

Subtracted ~3 from every number...

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  • $\begingroup$ You're on the right track, But I did more than what you said I did. $\endgroup$
    – dcfyj
    Jun 10, 2016 at 21:30

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