I've been playing around with sequences lately and found a pretty evenly doubled sequence of single digit numbers:

1, 12, 30, 64, 65, 156, 175, 368, 369, 371, 752, 753, 1524, 1525, 3060, 3073, 6168, 6219, 6221, 12444, 12453, ...


The given sequence is generated using a repeating sequence. If we let $G$ represent this generating sequence, then the numbers $G_0$ and $G_9$ = $1$.

Can you tell me what number comes next, and why?

  • 1
    $\begingroup$ single digit numbers? $\endgroup$ Oct 17 at 1:42
  • $\begingroup$ @DanielMathias yes, the answer is a sequence of single digit numbers, exposed by understanding the presented sequence. $\endgroup$ Oct 17 at 11:45
  • $\begingroup$ I assume it's not just the last digit, but it would technically work. $\endgroup$
    – PiGuy314
    Oct 17 at 17:03
  • $\begingroup$ @PiGuy314 you’ll have to elaborate on what you mean, but you are correct that it’s not just the last digit. $\endgroup$ Oct 17 at 17:05
  • $\begingroup$ I merely meant that the final digits of G0 and G9 were both 1. $\endgroup$
    – PiGuy314
    Oct 17 at 17:10


It must be around ....

24904 = = Double 1245 (= 2490) with 4 inserted somewhere

.... give or take a small increment.


In the Sequence, the Digits 1 to 9 appear in cyclic order, but at various Positions in each number.
The remaining Digits in each number form a Sequence that is either doubling or incrementing or both.
The Next number must be something like 2490 with 5 inserted somewhere and one Digit incremented like 24915 or 25905
Then the Next number must be something like Double with 6 inserted somewhere


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