Let , where and . Then the periodic part of the nimsequence is as follows:

, if

– is odd

– is odd

, if

– is odd

– is even

Notice: thus far we’ve handled all cases with odd. This was known before, check for example http://arxiv.org/pdf/1202.2986.pdf.

, if

–

–

, if

– is even

–

, if

– is even

– is odd

–

Note: now we handled all cases with even and odd.

, if

– is even

–

Note: and now we’re also done with .

, if

– is even

–

–

, if

–

–

–

And now we’re also completely done with . All that’s left now is the case even and even, where if , . We will split this case in three: either exactly equals , or , or , and in this last case we have to distinguish between or . Here goes.

, if

– is even

– is even

–

, if

– is even

– is even

–

, if

– is even

–

–

–

– Using the fact that , these last two assumptions can be written as:

, if

– is even

– is even

–

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1) .

2) For all we have: .

Let be a non-negative integer and let be a set of non-negative integers containing . Define . That is, since contains , the set is the set of all integers that can be written as a sum of at most two non-zero elements of . We now have the following theorem:

Theorem. If , then , where is either or , according to whether the smallest element of larger than (which we shall call ) is even or odd.

In particular, if , then some power of two can be written as sum of at most two elements of . In Additive Number Theory: Inverse Problems and the Geometry of Sumsets (which is an awesome book, by the way!) this special case is proved and then used to prove the neat fact that if and , then some power of two can be written as sum of at most four elements of . We will now prove the general version.

Proof. When , then, for , at most one of lies in (why does not lie in ?), so we then immediately have . So we may assume that . Furthermore, for , we obviously have . Now assume that, for all we have that the largest set for which no is the sum of at most two elements of , has cardinality at most . We will use this induction hypothesis to prove our Theorem.

Case I. is even.

First some definitions:

Let be a set such that no element of can be written as a sum of at most two elements of .

Let be such that .

Let be such that (note that ).

Let (on which we will use the induction hypothesis, which is allowed, since ).

Let , for .

Now note that, for all relevant , we get , while for we even know (again, why?). What’s left is a direct computation.

Case II. is odd.

We carry over all definitions of Case I, with the exception of the fact that the sets are now defined for , and we may no longer deduce the existence of an index for which . The obvious computation now yields:

And we’re done.

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If Schanuel’s Conjecture holds, then for every , there exists an integer , such that . Furthermore, if , then the smallest such grows at most linearly in .

I now believe I can prove the above, without assuming Schanuel’s Conjecture! Although I haven’t fully convinced myself that my ideas work, I’m getting more confident by the minute. I will now present an (admittedly very brief) overview of my ideas. I will expand on this as time goes on. Also, for definitions and lemma’s I refer to, check my preprint.

I use Schanuel’s Conjecture for two things:

1) To get an arbitrarily large prime dividing . (Theorem 2 in my preprint)

2) To make sure .

It is, if I’m not mistaken, not hard to do the second part without Schanuel. We can prove that (as I do on page 14, beginning with ‘Now that we have shown that has arbitrarily large prime divisors’), and since is immediate when , we may assume . Since is arbitrarily large, is arbitrarily large too. In particular, it must be divisible by a large prime power. On the other hand, we can use Lemma 4, 5, or 6 to make sure that is not divisible by a large power of that prime, and we are done.

A proof of the first statement is somewhat more involved, but the idea is to take the following assumption:

There exists a finite set of primes, such that, for all , can be written as .

And derive a contradiction by using nested intervals such that for all , there exists a , such that for all . In particular, for all and all , which is, for large enough , a contradiction, since , by Lemma 3.

To do this, realize that the last sentence implies that, for all large enough, there exists an index , such that . But if is an arbitrary constant, is large enough and is the smallest integer larger than such that divides , then . Now, if we take to be suitably large, then we have a large interval where for all inside this interval. Then, inside this interval, we repeat this trick with a different index , and a slightly smaller constant , such that inside this interval, we find a slightly smaller interval, such that for all inside this smaller interval, we have . Repeat this process times, and we have that for all and all in the last, smallest, interval (which, to be clear, is thus contained in all the others).

Hopefully the above made some sense. If not, just have a little patience and look forward to more elaboration, which will come in due time. In the mean time, try to embrace the glorious notation I laid out in this post.

EDIT (9 december 2012): Have a look here.

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A set is n-summable if, and only if, is n-large.

First we will prove that if is n-large, then is n-summable. To this end, let be an infinite sequence such that for all , we have: . Define . Then we have infinitely many , but at most distinct . And thus, there exists an infinite subsequence with . Now, define and . Then, by construction, we have .

To prove the converse direction, assume is any infinite set of integers and assume , with . Then we will prove that if is not n-large, then . So assume that for all with . Now, let be any integer larger (in absolute value) than . Then is a set of integers, all larger (in absolute value) than , with the difference between largest and smallest element being: . In particular, . And since , we also have .

Now, if we say is -summable if exist with and , then, unfortunately, it is not true that is -summable if is n-summable for all . A counterexample:

Define sets as follows:

1) , where is the -th entry of https://oeis.org/A002260

2) If , then

Define . Then is n-summable for all , because is n-large for all , with . Although it is also easy to construct explicit sets and with , , and . We claim, however, that is not -summable. If were -summable, then, by some appropriate addition or subtraction, we have for some infinite sets and , with and . In particular, . Now, if with , then . And thus, either lies strictly between two elements of , or lies strictly between the largest element of and the smallest element of . Either way, . So is contained in the sets that have no more than elements. Now, let be any element of larger than , and let be any element of larger than . Then, since , we have that . And thus, is larger than the largest element of . On the other hand, , which is thus smaller than the smallest element of . And we must conclude that is not an element of ; contradiction.

I realize that the above proof that our counterexample works may be somewhat inelegant, so I might rewrite it soon. In which case I’ll probably explicitly prove something like the following: if is as above, is an infinite set, with , and , then .

At last, I have the following hunch, on which I’ll probably be working for the next few days/years:

If there exists an absolute constant , such that for all there exist infinitely many for which , then is -summable.

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Edit: under heavy influence of MathOverflow (those guys are awesome), I wrote an introduction for my paper. The above version is updated.

Edit2: Although I didn’t get any feedback from my thesis advisor, I resubmitted! The waiting can begin.

Edit3 (05-12-2012): My paper got accepted!! The Referee Report was quite positive, actually. It asked me to change, out of 23 mathematically dense pages, exactly one letter. The above version is, probably, the final version.

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,

with perhaps the most beautiful proof I have ever found.

Assume that we play a tournament with players, in which every game is a knock-out game between players. So in the first round there are games to be played, one for every people. Then in the second round only half of the players remains (i.e. ), so there are another games to be played, and so forth. Right up to the semi final and final, which are just and games, respectively. So the total amount of games played in the tournament is: . Another way to see how many games total there are played is the following: in every game, exactly player gets eliminated. There is only one winner, so the other players must get elimated somewhere. So the total amount of games played must equal . Kablam.

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Now, let be small and be the sequence of all such that . Then I’d like to bluntly conjecture that there exist infinitely many such that , for some absolute . In particular, goes to as goes to , that is:

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