1
$\begingroup$

It's a simple pattern I thought of. Here it is:

1,1,1,3,3,6,6,10,10,22,22,44,44,84,?,?,?,?,?

Whoever gives the correct answer and a good explanation gets best answer.

$\endgroup$
3
  • $\begingroup$ If somebody gives a better solution I may change the best answer status. $\endgroup$
    – warspyking
    Sep 28, 2014 at 22:34
  • 1
    $\begingroup$ and what is criterion for "better" solution? $\endgroup$
    – klm123
    Sep 29, 2014 at 8:01
  • $\begingroup$ A better explaination $\endgroup$
    – warspyking
    Sep 29, 2014 at 18:00

2 Answers 2

5
$\begingroup$

Start with S(1) = 1.

To construct the next 6 numbers (S(2) - S(7)), use n = 1 and apply the following formulas:

S(2) = S(1) * 2 + (-1)*n
S(3) = S(2)
S(4) = S(3) * 2 + (+1)*n
S(5) = S(4)
S(6) = S(5) * 2 + ( 0)*n
S(7) = S(6)

To construct the next 6 numbers, double n to 2 and apply the above pattern again.

To construct the next 6 numbers (84 and your 5 ?s), double n to 4 and apply the pattern again. This gives:

S(14) = S(13) * 2 + (-1)*n = 44 * 2 - 4 = 84
S(15) = S(14) = 84
S(16) = S(15) * 2 + (+1)*n = 84 * 2 + 4 = 172
S(17) = S(16) = 172
S(18) = S(17) * 2 + ( 0)*n = 172 * 2 + 0 = 343
S(19) = S(18) = 343

So the sequence is:

1,1,1,3,3,6,6,10,10,22,22,44,44,84,84,172,172,343,343

$\endgroup$
4
  • $\begingroup$ Nice, I was going to close this question as I made the wrong sequence, but that was great! $\endgroup$
    – warspyking
    Sep 29, 2014 at 18:21
  • 3
    $\begingroup$ I guess it just goes to show that you can trick yourself into seeing a pattern in almost any data. Out of curiosity, what was the actual sequence? $\endgroup$ Sep 29, 2014 at 22:53
  • $\begingroup$ The actual sequence was 1,1,1,3,3,5,5,11,11,21,21,43,43, etc. $\endgroup$
    – warspyking
    Sep 30, 2014 at 22:00
  • $\begingroup$ That would be S(n) = S(n-2) + 2 S(n-4). $\endgroup$
    – Florian F
    Oct 3, 2014 at 8:59
3
$\begingroup$

You can simplify it into a single formula:

$s_{n} = 2 s_{n-2} + 2 s_{n-6} - 4 s_{n-8}$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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