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Feb 3, 2018 at 0:50 comment added Alexey Ustinov @JoseBrox Too many formulae within. Looks like Rødseth's formula hidden behind complicated notations.
Feb 2, 2018 at 17:29 comment added Jose Brox @AlexeyUstinov Btw, are you aware of the recent paper by Tripathi with exact formulas for the Frobenius number of three numbers? It is "Formulae for the Frobenius number in three variables", sciencedirect.com/science/article/pii/S0022314X16301743
Feb 2, 2018 at 8:16 comment added Alexey Ustinov @JoseBrox I applied this technique in Frobenius problem, see 1 and 2. Reduction of Frobenius problem to a lattice proble is described here.
Jan 29, 2018 at 18:30 comment added Jose Brox @AlexeyUstinov Thanks! Could you please give some nice reference to learn about these techniques for lattices?
Jan 29, 2018 at 11:25 comment added Alexey Ustinov @Jose Brox The "history" goes back to the theory of numerical integration and uniform distribution theory. When you replace given function $f$ by its Fourier series you'll get error term depending on Fourier corfficients of $f$ and trigonometric sums of a given net. This argument was used by Weyl in his pioneer work Über die Gleichverteilung von Zahlen mod. Eins (1916). Basic latice problem is a distribution of vectors in reduced basis. Conditions from Lemma 5 can be replaced for another ones (see the proof).
Jan 29, 2018 at 11:09 history edited Alexey Ustinov CC BY-SA 3.0
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Jan 29, 2018 at 10:34 comment added Jose Brox Also, does this example actually satisfy the finite difference conditions on your Lemma 5, or perhaps they can be exchanged for others? If I'm not mistaken, I get $\Delta_{1,0}=y\geq 0$, $\Delta_{0,1}=x\geq 0$.
Jan 29, 2018 at 10:26 comment added Jose Brox I find your main formula very interesting. What is its "history"? Can it be traced back in the literature to more specific instances? And could you point to any simple example of its use in a lattice-connected problem? (The simpler, the better).
Dec 10, 2015 at 2:25 history edited Alexey Ustinov CC BY-SA 3.0
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Dec 9, 2015 at 3:08 comment added Alexey Ustinov And this $\log$ is impotant because usually (numerically, it is not proved) normalized error term (in our case $R_p\cdot p^{-5/2}$) is gaussian i.e. unbounded.
Dec 9, 2015 at 2:12 comment added Alexey Ustinov Yes. But it must be $O(p_n^{5/2}\log^2p_n)$ instead of $O(p_n^{5/2})$.
Dec 8, 2015 at 18:51 comment added martin thanks for the great answer! So in this case $$\left(p_n/2\right)^3-\sum\limits_{k \in A_{p_n}}{k}=\mathcal{O}\left(p_n\ ^{5/2}\right)$$?
Dec 7, 2015 at 12:14 history answered Alexey Ustinov CC BY-SA 3.0