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:$$\sum _1^n{i}^3=(\sum _1^{n}i{)}^2$$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:$$\sum _1^n({\sigma }_0(d_i){)}^3=(\sum _1^n{\sigma }_0(d_i){)}^2$$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:$$1,12,21,22,123,132,213,231,312,321,333,1224,...$$