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A demonstration of the abundance of the number 12 using Cuisenaire rods (Wikipedia).

Euler problem 23 asks us to solve a problem with abundant or excessive numbers.

These are numbers for which the sum of its proper divisors is greater than the number itself.

12 is an abundant number because the sum of its proper divisors (the aliquot sum) is larger than 12: (1 + 2 + 3 + 4 + 6 = 16).

All highly composite numbers or anti-primes greater than six are abundant numbers. These are numbers that have so many divisors that they are considered the opposite of primes, as explained in the Numberphile video below.

## Euler Problem 23 Definition

A perfect number is a number for which the sum of its proper divisors is exactly equal to the number. For example, the sum of the proper divisors of 28 would be 1 + 2 + 4 + 7 + 14 = 28, which means that 28 is a perfect number.

A number n is called deficient if the sum of its proper divisors is less than n and it is called abundant if this sum exceeds n.

As 12 is the smallest abundant number, 1 + 2 + 3 + 4 + 6 = 16, the smallest number that can be written as the sum of two abundant numbers is 24. By mathematical analysis, it can be shown that all integers greater than 28123 can be written as the sum of two abundant numbers. However, this upper limit cannot be reduced any further by analysis, even though it is known that the greatest number that cannot be expressed as the sum of two abundant numbers is less than this limit.

Find the sum of all the positive integers which cannot be written as the sum of two abundant numbers.

## Solution

This solution repurposes the divisors function that determines the proper divisors for a number, introduced for Euler Problem 21. The first code snippet creates the sequence of all abundant numbers up to 28123 (sequence A005101 in the OEIS). An abundant number is one where its aliquot sum is larger than n.

# Generate abundant numbers (OEIS A005101)
A005101 <- function(x){
abundant <- vector()
a <- 1
for (n in 1:x) {
aliquot.sum <- sum(proper.divisors(n)) - n
if (aliquot.sum > n) {
abundant[a] <- n
a <- a + 1
}
}
return(abundant)
}

abundant <- A005101(28123)


The solution to this problem is also a sequence in the Online Encyclopedia of Integer Sequences (OEIS A048242). This page states that the highest number in this sequence is 20161, not 28123 as stated in the problem definition.

The second section of code creates a list of all potential numbers not the sum of two abundant numbers. The next bit of code sieves any sum of two abundant numbers from the list. The answer is determined by adding remaining numbers in the sequence.

# Create a list of potential numbers that are not the sum of two abundant numbers
A048242 <- 1:20161

# Remove any number that is the sum of two abundant numbers
for (i in 1:length(abundant)) {
for (j in i:length(abundant)) {
if (abundant[i] + abundant[j] <= 20161) {
A048242[abundant[i] + abundant[j]] <- NA
}
}
}
A048242 <- A048242[!is.na(A048242)]