Squares with Ascending Digits

Yesterday's date, November 22, 2025, prompted me to think about square numbers with ascending digits (in base 10). In the month-day-year abbreviation, the digits of yesterday's date form 112225, which is a square number! In particular, 112225 = 335². Interestingly, both 335 and 112225 have an ascending digit sequence. It turns out that such squares, while rare, occur infinitely often. 112225 is a member of one of several similar infinite sequences. This one begins:

25, 1225, 112225, 11122225,...

Here the nth entry of the sequence has n-1 1's, followed by n 2's, and then a single 5 at the end. These are the squares of the sequence:

5,35,335,3335,....

Let's prove this fact by induction. Clearly, it is true that 5² = 25. Assume the nth term of the bottom sequence, a_n squares to the nth of the top sequence, b_n. Then we notice:

a_n+1 = 3*10ⁿ + a_n

so that

a_n+1² = (3*10ⁿ + a_n)² = 9*10²ⁿ + 6*10ⁿ*a_n + a_n². Now the key point is that 3*a_n is just 6 more than 333...33*3 = 999...99, so that 3*a_n = 10...05 = 10ⁿ + 5, and hence 6*a_n = 20...010 = 2*10ⁿ + 10. Simplifying the above equation then gives

a_n+1² = 9*10²ⁿ + 2*10²ⁿ + 10ⁿ⁺¹ + a_n² = 1*10²ⁿ⁺¹ + 1*10²ⁿ + 1*10ⁿ⁺¹ + a_n².

Finally, we see that to get from a_n² = b_n to a_n+1², we add two 1's to the very beginning of the number (1*10²ⁿ⁺¹ + 1*10²ⁿ) and then change the final 1 in the sequence of 1's to a 2, by adding 1*10ⁿ⁺¹. This results in precisely b_n+1, as claimed.

It'd be interesting to see whether there are higher perfect powers with ascending digit sequences; perhaps that will be the subject of another post.