September 3, 2017, @jamestanton raised the following question on Twitter:

So, without exercising much, if any, thought, I just wrote a simple bit of *Mathematica*® code to explore if I could find any of these numbers Jim sought:

**n = 2;**

** L = {};**

** While[n <= 1000000,**

** If[Total[Cases[Divisors[n], d_ /; OddQ[d]]] == n, L = {L, n}];**

** n++]**

** L = Flatten[L]**

The result?

Nada!

Not a one.

The tiniest bit of thought tells us that for *odd* numbers n the sum of the odd divisors of n is at least n+1.

So, what about *even* numbers? What are the even numbers n whose sum of odd divisors is greater than n?

The following bit of *Mathematica*® code looked for these n between 2 and 100,000:

**n = 2;**

**L = {};**

**While[n <= 100000,**

** If[EvenQ[n] && Total[Cases[Divisors[n], d_ /; OddQ[d]]] > n, **

** L = {L, n}];**

** n++]**

**L = Flatten[L]**

And the result?

{1890, 3150, 4410, 5670, 6930, 8190, 9450, 10710, 11550, 11970, 12870, 13230, 13650, 14490, 14850, 15750, 16170, 16830, 17010, 17850, 18270, 19110, 19530, 20790, 22050, 23310, 24570, 25410, 25830, 27090, 28350, 29610, 30030, 30870, 32130, 33390, 34650, 35910, 37170, 38430, 38610, 39270, 39690, 40950, 42210, 43470, 43890, 44550, 44730, 45990, 46410, 47250, 48510, 49770, 50490, 51030, 51870, 52290, 53130, 53550, 54810, 56070, 56430, 57330, 57750, 58590, 59670, 59850, 61110, 62370, 62790, 63630, 64350, 64890, 66150, 66690, 66990, 67830, 68250, 68310, 69930, 71610, 72450, 73710, 74250, 74970, 76230, 77490, 78750, 79170, 80850, 81270, 82110, 83790, 84150, 84630, 85050, 85470, 86130, 88830, 89250, 90090, 91350, 91770, 92070, 92610, 94050, 94710, 95550, 96390, 97650, 99330, 99450, 99750}

Aha, quite a few of them. How many? Up to 100,000 there are 114 even numbers n whose sum of odd divisors is greater than n.

Some questions that come to mind are:

- What about a counting function that seeks an answer to: “given K>2 , how many even integers n < K are there whose sum of odd divisors is greater than n?”
- If Even(n) := n
*th*even number whose sum of odd divisors is > n, how does the ratio Even(n+1)/Even(n) behave as n increases?

To tackle the second question computationally, and to keep things simple (for now, at least) let’s just extend the list above to those even numbers n between 2 and 10,000,00 whose sum of odd divisors is greater than n.

Then we make a table of values Even(n+1)/Even(n) and plot the results as a function of n:

So it sort of seems the Even(n+1)/Even(n) might, somewhat lazily, be converging to 1.

Could we- numerically at least – bound the quantity Even(n+1)/Even(n) – 1 *above* by some quantity that we know heads to 0 with increasing n?

After a bit of playing around with the values of Even(n+1)/Even(n) – 1 it seems that 5/n might do the trick as an upper bound (at least so far as we’ve got):

Well, this is pretty much just computation for exploration and doesn’t involve much thinking, and it certainly doesn’t prove anything (because, among other things, proof comes from insights).

But, as the inimitable Steve Irwin was wont to say: