Sums of Cubes and Squares of Sums

Today I turned 25321 days old, a prime number of days, and a prime with the property that the sum of the cubes of its digits equals the square of the sum of its digits. In other words: $$2^3+5^3+3^3+2^3+1^3 = (2+5+3+2+1)^2 = 169$$ This prime is a member of OEIS A225567: Primes with nonzero digits such that sum of cubes of digits equal to square of sums. It's initial members are: $$1423, 2143, 2341, 4231, 12253, 21523, 22153, 22531, 23251, 25321, ...$$ Serendipitously I then rediscovered the old and famous connection between the sum of the cubes of the first n natural numbers and the square of the sum of these some numbers, specifically: $$\sum_1^n i^3=\big (\sum_1^n i \big)^2$$ It's easy to see why the first four members of the sequence (1423, 2143, 2341 and 4231 are members) because these are simply instances of: $$1^3+2^3+3^3+4^3=(1+2+3+4)^2$$ I also discovered a similar relationship involving the divisors $d_i$ of any natural number with $n$ divisors, namely that: $$\sum_1^n (\sigma_0(d_i))^3=\big(\sum_1^n \sigma_0(d_i) \big) ^2$$ The previous looks more difficult than it actually is and a simple example will assist. Let's consider the number 10. It has four divisors 1, 2, 5 and 10. Each of these divisors has 1, 2, 2 and 4 divisors respectively. We find that: $$1^3+2^3+2^3+4^3=(1+2+2+4)^2 = 81$$ It's as simple as that and it applies to every natural number. Of course 1224 and permutations of these digits can be found in OEIS A227073: Positive numbers without the digit 0 such that sum of cubes of the digits equals the square of the sum of the digits. The initial members of this sequence are: $$1, 12, 21, 22, 123, 132, 213, 231, 312, 321, 333, 1224, ...$$