Timeline for Deg $n$ integral polynomial $P(x)$ with $n+1$ integer solutions to $0\leq P\leq d$
Current License: CC BY-SA 4.0
5 events
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Jan 7, 2019 at 12:06 | comment | added | Wolfgang | Right, I neglected those. Empirically, if for $n=2^k$ the last factor is denoted $(a_{2^{k-1}} x^{2^k-2}+a_{2^{k-1}-1} x^{2^k-4}\cdots+a_2x^2+a_1)$, then the 2-valuations of the coefficients seem to be $\nu_2(a_i)=k-\nu_2(i)$ for all except the leading and the next one. Thus an even stronger conjecture (checked till $k=8$), but I don't know if that helps. | |
Jan 7, 2019 at 9:45 | comment | added | Ilya Bogdanov | @Wolfgang: I've learned from Maple that the coefficients are integer, but I'm still thinking on how to prove that... My trouble is with powers of 2 only. | |
Jan 7, 2019 at 8:59 | comment | added | Wolfgang | You may want to add the construction you found, as it generalizes indeed for $n=2^k$, just by Lagrange interpolation for the odd values $P(-n+1)=P(-n+3)=\cdots=P(n-1)=d$ between the even ones $P(-n)=P(-n+2)=\cdots=P(n)=0$. It is easy to see from the factors of $d$ that the coefficients must be integers. E.g. for $n=32$, it is $ -x^2(x^2-4)\cdots (x^2-16^2 )(x^{14} - 672x^{12} + 174200x^{10} - 22119856x^8 + 1443224772x^6 - 46275602672x^4 + 625295296152x^2 - 2334623374800)$. | |
Jan 6, 2019 at 8:06 | comment | added | Wolfgang | Yes asymptotically they differ by $\frac18\sqrt{2\pi n}\approx .31\sqrt{ n} $. | |
Jan 5, 2019 at 21:49 | history | answered | Ilya Bogdanov | CC BY-SA 4.0 |