New math problem

[loveschak]

Let C(n) be the number of ways to write n as the sum of consecutive non-negative integers. The terms from C(0) to C(10) are 1, 2, 1, 3, 1, 2, 3, 2, 1, 3, 3. Find an ordinary generating function for C(n) in closed form, in terms of elementary functions if possible. Prove that the series 1/C(n) diverges. When about the series 1/(n C(n)). What is the

Comments

[tellumo]

Wow. I'm going to have to think about that one a little bit more, but for now, allow me to add that this sequence (starting at C(1)) is Sloane's A054843.

Actually, I can do a little more, but only because a Sloane's submitter did the hard bit, which is proving that the terms of this sequence are the sums of those of A001227 (the number of odd divisors of n) and A010054 (1 if n is triangular, 0 if not).

For all n, there exists an m such that m > n and A001227m > A001227n. To see this, consider that the opposite implies that there are a finite quantity of prime numbers (proof left as an exercise for someone less tired than I). As A001227n > A001227n for all n, as all terms of A010054 are nonnegative, C(n) grows without bound, and 1/C(n) diverges.

I think that's right, if rather cavalier with the whole "proof" business. I'm curious to see the submitter's result on which I'm basing all this.

[madcaptenor]

Here's a sketch of the proof of the OEIS result. Consider, say, 90. The divisors of 90 are

1, 2, 3, 5, 6, 9, 10, 15, 18, 30, 45, 90

and there are six odd divisors: 1, 3, 5, 9, 15, 45. Write [m, n] for m + ... + n. There are six sums: [90], [29, 31], [16, 20], [6, 14], [-1, 13] = [2, 13], [-20, 24] = [21, 24]
In general, the odd divisor k corresponds to a sum with k terms with middle term n/k. That is, the divisor k corresponds to the sum [n/k - (k-1)/2, n/k + (k-1)/2]; in the case where this sum includes negative terms we cancel them in the appropriate way. (I omit the proof that these are actually distinct, which isn't quite obvious but which I'm too lazy to do.) If n is triangular, with n = 1 + 2 + ... + r, then this process generates one of [0, r] and [1, r] and we want to count both.

[arturis]

What do you mean by the sum of consecutive non-negative integers? I can conceive of no interpretation that allows C(0)=1, because there are no ways to write 0 as the sum of any two non-negative numbers, unless you can use numbers again to get 0 = 0 + 0 but then why not keep reusing zeroes (0 = 0 + 0 + 0) to have C(0)= infinity?

I guess I'm not clear on what you mean.

[tborchert]

You are summing sequences of non-negative numbers, not pairs of them. You are allowed to sum a sequence of a single number, so C(n) is always at least one.

[arturis]

Sequences! I see.

[loveschak]

C(0) should really be 2, since 0 is the sum of the following two trivial sets of consecutive numbers:
{}
{0}

modular forms

[anonymous]

This is a good problem and deserves a better response than what I'm going to give. Sorry, but I've got other math I really need to do.

My answer is: this is exactly the sort of problem that the theory of modular forms solves so spectacularly. If I was going to search the literature to see if this has already been done, my keywords would be "partition" and "modular form." If it hasn't already been done, it would be the perfect sort of problem for a modular form expert to give at an REU.

Dr. Andrew Shallue
not a modular forms expert

[madcaptenor]

The generating function is