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If I multiply two integers $x, y $ in $[0,2^L)$, I get an integer in $[0,2^{2L})$. Clearly, this map from $[0,2^L) \times [0,2^L) \to [0,2^{2L})$ is not bijective.

I am interested in the size of the image. It is less $2^{2L}$. By a classical result from Erdös, the size of the image divided by $2^{2L}$ goes to zero.

A simple argument shows that it cannot be much larger than $2^{2L}/2$. Indeed, pick $z$ in the image of this map, if $z$ is not a square (and there are only $2^L$ squares), then there are at least two corresponding distinct pairs of $x,y$ mapping to $z$ since multiplication is commutative.

But, clearly, this bound is naive for $L$ large. We have that the size of the image is much less than $2^{2L}/2$ (it goes to zero). I am looking for a reasonably tight bound. For example, given $L=64$ or $L=128$, I would like to bound (above and below) the size of the image with actual numbers.

The exact answer is known for up to $L=25$: see the OEIS sequence A027417.

Turbo
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lemire
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    This corresponds to a well known problem regarding the size of the general multiplication table. A factor of 1/L should appear in your answer. A search on MathOverflow for erdos reveals among others http://mathoverflow.net/questions/31663/distinct-numbers-in-multiplication-table . I think Dimitris Koukoulopoulos knows more, including the n dimensional version of this problem. Hopefully he will chime in. Gerhard "Ask Somebody Else About This" Paseman, 2013.12.02 – Gerhard Paseman Dec 03 '13 at 03:23
  • @GerhardPaseman It is indeed related to Erdös's multiplication table problem, but I am not looking for a growth characterization, I am looking for an actual bound... which I believe Erdös did not provide. – lemire Dec 03 '13 at 03:54
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    No, but Ford, Koukoulopoulos, and others have tightened it up. I don't know how good an estimate you need, but I suspect L^\delta (log L)^3/2 will be the denominator, or close to it. I will not hazard a guess as to the value of \delta. Gerhard "Will Wait For Better Responses" Paseman, 2013.12.02 – Gerhard Paseman Dec 03 '13 at 03:59
  • @GerhardPaseman One has a bound or not. Saying that the asymptotic behaviour is like c f(L) for some unknown c is not a bound... unless, of course, you can bound c. – lemire Dec 03 '13 at 04:17
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    Gerhard has given you some places to look. Why not try them? – Gerry Myerson Dec 03 '13 at 05:44
  • @GerryMyerson There are related questions on mathoverflow but though they are related, their answer does not help. The references I have found through these links point me to results of the kind "there exist a constant c... we don't know what it is... such that the size of the image is bounded by c f(L)". Let me be more precise... what is the size of the image for L=64? I do not know. I have no bound for it. – lemire Dec 03 '13 at 22:49
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    I'm willing to believe that the various people who have improved Erdös's bound did not work out explicit constants in their bounds. However, there's no reason in principle that it couldn't be done. You could look at the simplest of these proofs and see if you can work through them with explicit constants. – Greg Martin Mar 02 '14 at 08:45

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