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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

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