This is my first post so I'll keep it simple and easy.
The following numbers form a series but a number is skipped in it.
Find that missing number and its position (The number can also be at the beginning or at the end).

2, 2, 3, 8, 56, 63

  • $\begingroup$ Can the sequence go on forever? Or is it just these six terms and the missing one? $\endgroup$ Commented Mar 13, 2020 at 18:20
  • $\begingroup$ @Randal'Thor Yes, the sequence can go on forever. $\endgroup$ Commented Mar 13, 2020 at 18:24
  • $\begingroup$ Forever in both directions? To infinity and negative infinity? $\endgroup$ Commented Mar 13, 2020 at 22:17
  • $\begingroup$ @ChaseRoberts I can tell you but as this question has been answered, you can check for yourself. $\endgroup$ Commented Mar 14, 2020 at 4:24

1 Answer 1


If we insert

a 6 between the 3 and the 8, yielding 2 2 3 6 8 56 63

then the resulting sequence

obtains each number from its predecessor by the operations x1, +1, x2, +2, x7, +7.

Where do these numbers come from? Well,

they are 1 less than the numbers they are first applied to. That is, the sequence can be described as follows. Start with 2. Get the next number by multiplying by (current number minus 1), and the next by adding the number we just multiplied by. Then repeat those two steps for ever.

We would then continue with

63x62 = 3906 and then 3906+62 = 3968; then
3968x3967 = 15741056 and then 15741056+3967 = 15745023;
and so on.


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