Consider the following sequence:

6211, 4214, 6321, 4135, 6359

What is the next number in that sequence? What are the rules of that sequence?

I hope you'll enjoy my first question :)


Sorry guys, with those numbers, it is impossible to find the next one. So I have added the next one so it becomes possible.

The new sequence is:

6211, 4214, 6321, 4135, 6359, 4182


Here is a hint:

Some numbers where chosen to confuse you. You shouldn't focus on them, but focus on the whole.

  • $\begingroup$ Are you sure it's supposed to be 6359 and not 6350? $\endgroup$ – Joe Z. Apr 20 '15 at 22:47
  • $\begingroup$ Yes, I am sure. $\endgroup$ – A.D. Apr 20 '15 at 22:50
  • $\begingroup$ Are you sure it's supposed to be 6359 and not 6459? $\endgroup$ – Joe Z. Apr 20 '15 at 23:06
  • $\begingroup$ Yes, I am sure. $\endgroup$ – A.D. Apr 20 '15 at 23:12

The seventh number is 6136.

The last digit of each number is the $n$th digit after the decimal point of $\pi$.

The first three digit numbers follow this pattern:

  • Place the $n$th Fibonacci number so that the ones' digit is the third digit of the number.

  • Then, insert digits so that the sum of the digits of the sum of the digits of the number is equal to to $n$. If the $n$th Fibonacci number is less than 10, then use a sequence alternating $6$ and $4$ to fill in the first digit, and then fill in the second digit as appropriate.

So the eleven terms of the sequence would be:

6211, 4214, 6321, 4135, 6359, 4182, 6136, 9215, 8343, 4555, 4898

The sequence ends there since the next Fibonacci number is 144 which would no longer ensure the sum-of-digits rule to be intact.

  • $\begingroup$ Argh, beat me to it. $\endgroup$ – Tryth Apr 20 '15 at 22:51
  • $\begingroup$ Your solution pinning $4$ as the second digit seems a bit iffy. $\endgroup$ – Joe Z. Apr 20 '15 at 22:51
  • $\begingroup$ If you look at the update I've made, you'll see you're on the right tracks :) $\endgroup$ – A.D. Apr 20 '15 at 22:54
  • $\begingroup$ You have all the information you need to solve it. There is no unknown sequence of number involved. $\endgroup$ – A.D. Apr 20 '15 at 23:02
  • 1
    $\begingroup$ The fact the all the 2nd numbers are $\le 4$ makes me think the 2nd number might be self-referential in some way. $\endgroup$ – Tryth Apr 20 '15 at 23:02

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