An autonumerogram, or self-enumerating number, is a number that can be read as describing itself by providing a pairwise inventory of its own digits. For example, 22 (two 2's) and 14233221 (one 4, two 3's, three 2's, and two 1's) are both autonumerograms.

A pan-digit autonumerogram of order n is an autonumerogram containing all digits "0" through "n". What is the smallest pan-digit autonumerogram?

  • $\begingroup$ Note: "n", the largest digit occurring, need not be 9. $\endgroup$
    – Johannes
    Oct 24, 2015 at 16:41

2 Answers 2


0 can only appear once, since no digit that is included in the autonumerogram will appear 0 times. So our number must include "10".

Then 1 appears at least twice in the autonumerogram, so it must include a digit that is at least 2. If there are exactly 2 1's and exactly 2 2's, then 2 appears at least 3 times, a contradiction. So the largest digit in the autonumerogram must be at least 3.

If we want the largest digit to be 3, then the number is of the form 10A1B2C3 (possibly with some sections rearranged), where A, B, and C are all 1, 2, or 3. We must have A+B+C=7 to make the total correct, so either they are 2, 2, 3 or they are 1, 3, 3.

If A, B, and C are 2, 2, and 3, then there are a total of 2 1's, 3 2's, and 2 3's. The smallest number for these is 10212332.

If A, B, and C are 1, 3, and 3, then there are a total of 3 1's, 1 2, and 3 3's. The smallest number for these is 10123133.

So the smallest pan-digit autonumerogram is 10123133.


Isn't it the same thing as a self-reflective numerical statement?

In which case the answer is 10141516181923273271.

If we're talking pan digital autonumerograms of "arbitrary order $n$", a list of answers follows from that question as well:

  1. 10123133
  2. 1014223133
  3. [no solution]
  4. 10151623243241
  5. 1014161723253251
  6. 101415171823263261
  7. 10141516181923273271
  • $\begingroup$ Ah yes, you're right. This duplicates an earlier problem statement. $\endgroup$
    – Johannes
    Oct 25, 2015 at 12:46

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