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I found an answer for how to determine the coefficients of a polynomial $P(x)$ with all nonnegative integer coefficients, knowing its value on one or two points.

How could that method be extended to a multivariate polynomial on $n$ variables with all nonnegative integer coefficients? Or what is the most efficient method to determine the coefficients if it is possible to get the evaluation of the polynomial in any needed point?

In a special case I am currently interested in polynomials of the form:

$$P(x_1,\ldots,x_n)=\prod_{1 \le i \lt j \le n}{\left(1+x_{i}x_{j}\right)}$$

Fabius Wiesner
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  • Does my answer to MSE question 4049277 "Avoiding brute force:..." help you? – Somos Oct 08 '21 at 18:52
  • @Somos very nice thank you. I will think about it, although I need to get the coefficients of the "binomial" polynomials addends and adapt the method to multivariate polynomials. – Fabius Wiesner Oct 08 '21 at 19:11
  • Also, do you know a bound on the degree of the polynomial? If not, then knowing its values at any finite number of points will not uniquely determine it. – Somos Oct 08 '21 at 19:14
  • For me it would be also interesting to count the number of monomials, up to a permutation of variables like here or just the plain number of them (in this case it should be OEIS A005155. – Fabius Wiesner Oct 08 '21 at 19:16
  • Yes, you can see in the example I am mostly interested now the maximum degree is $n-1$. – Fabius Wiesner Oct 08 '21 at 19:19
  • Your question first sentence is misleading. For the last equation of a product of factors, just expand it out to get the sum of monomials. There is no need to know the value of the polynomial $P$ at particular points. – Somos Oct 08 '21 at 19:21
  • I am mostly interested in that polynomial, but it would be nice to have a general approach also. Regarding the expansion, it has time complexity $2^n$ so I was looking for something more efficient or a closed formula (in the example it would be a closed formula for OEIS A005155; and yes, this might be impossible). – Fabius Wiesner Oct 08 '21 at 19:30
  • Sorry I meant $2^{n \choose 2}$, I think that is the complexity. – Fabius Wiesner Oct 08 '21 at 19:41

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