Last Sunday, I was browsing Project Euler where I found an interesting line in the statement of Problem 23:

“…it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers.”

An *abundant number* is a positive integer such that the sum of the proper divisors of is greater than . Equivalently, the sum of the divisors of is greater than .

Problem 23 isn’t about the above stated result, but that is what interested me. So, I want to spend the next two or three posts proving this result and then, if we feel like it, providing a solution to problem 23. In this first post, I want to give an exposition of Douglass E. Iannucci’s 2005 article *On the Smallest Abundant Number Not Divisible by the First Primes. *We don’t need the full strength of the algorithm to find said abundant number, but it will be useful for finding odd abundant numbers.

Let denote the arithmetic function

where ranges over the divisors of . We define the *index* of a positive integer by

,

where is the prime factorization of . The right-most expression shows that is a multiplicative function.

Suppose and are primes with . Then

,

where the inequality is strict if . For integers , we obtain the inequality

Consider the product

where . Since the infinite series diverges, where ranges through all the primes, we see that increases without bound as . For , define the quantity by

Let denote the minimal abundant number not divisible by the first primes. Tautologically,

For and , define the quantity by

and by

Since , we see that and therefore .

Lemma 1.is divisible by .

*Proof. *Suppose that . has the unique prime factorization

By definition of , we have that . Since is an enumeration of the primes in ascending order, we see that for all .

I claim that . Indeed, otherwise

But this is a contradiction, as is by definition an abundant number and therefore .

By hypothesis, there exists an index such that . Since , , and , we see that there is at least one index such that . Relabeling if we necessary, we may assume without loss of generality that . Our work above shows that

Since is multiplicative, we conclude that

,

which contradicts the minimality of .

We now describe an algorithm for finding the smallest abundant number not divisible by the first primes (e.g. ). First, compute . Set . As argued above, the divergence of the infinite series , where ranges over all primes, implies that the product , as . This is still true if we require that be coprime to . Hence, there exists a unique minimal integer such that

I claim that . Indeed, by the division algorithm and Lemma 1, we can write , where is a positive integer. Since , we must have that . Since is by definition the minimal abundant number not divisible by , we conclude that .