类Armstrong数及相关立方和拼接数的推广问题问询

Great question! This is a fun twist on Armstrong numbers, and there are some interesting generalizations to explore.

First, let's formalize the problem clearly: For positive integers (a), (b), (c), let (N) be the number formed by concatenating (a), (b), (c) in sequence. If (a) has (p) digits, (b) has (q) digits, and (c) has (r) digits, we can write:
$$
N = a \times 10^{q+r} + b \times 10^r + c
$$
We want to find all triples ((a,b,c)) where:
$$
a^3 + b^3 + c^3 = N
$$

Known Infinite Families: Repeating-Digit Patterns #

You've already highlighted one elegant infinite family: for any positive integer (n), take:

• (a = 1\underbrace{66\cdots66}_{n \text{ times}}) (e.g., 16, 166, 1666, ...)
• (b = 5\underbrace{00\cdots00}_{n \text{ times}}) (e.g., 50, 500, 5000, ...)
• (c = \underbrace{33\cdots33}_{n \text{ times}}) (e.g., 33, 333, 3333, ...)

For example, when (n=1):

16³ + 50³ + 33³ = 4096 + 125000 + 35937 = 165033

Which is exactly the concatenation of 16, 50, 33.

This isn't the only such family. We can construct similar infinite sets by finding fixed digit sequences that maintain the identity when scaled to (n)-digit numbers. The critical reason these work is that the growth rate of the cubic sum matches the concatenated number: both are roughly (10^{3n}) when (a), (b), (c) are (n)-digit numbers, so the equation doesn't break down as (n) increases.

Beyond Repeating Digits: Scattered Solutions & Open Questions #

While repeating-digit families are straightforward to construct, we also need to consider non-patterned "scattered" solutions. For example, small triples like ((1,2,3)) don't work ((1+8+27=36 \neq 123)), but there might be larger non-repeating triples that do.

A special case is when (a=b=c): this reduces to finding Armstrong numbers, which we know are finite (only 153, 370, 371, 407). But for non-equal (a), (b), (c), there's no known upper bound on the number of solutions. The matching growth rates suggest there could be infinitely many scattered solutions, though proving this is still an open problem in number theory.

Key Conclusions #

• Yes, generalizations exist: We can explicitly construct infinite families of triples ((a,b,c)) using repeating-digit patterns (like the one you mentioned) that satisfy the cubic concatenation identity.
• Growth rate alignment is critical: The cubic sum of three (n)-digit numbers grows at the same order of magnitude as the 3(n)-digit concatenated number, which enables infinite solutions to exist.
• Open questions remain: We don't yet know if all solutions belong to repeating-digit families, or if there are infinitely many non-patterned scattered solutions waiting to be found.

内容的提问来源于stack exchange，提问作者Mathejunior