Count the number of integer solutions

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SUMMARY

The discussion focuses on counting the number of integer lattice points satisfying the inequality n + ∑_{k=1}^{n} |x_{k}| ≤ N. The solution involves generating functions, specifically finding the coefficient of a term in a polynomial expansion. The problem simplifies to determining the function f(N', n), which represents the number of solutions, defined recursively as f(N', n) = f(N', n-1) + 2∑_{k=1}^{N'} f(N'-k, n-1). It is established that if n > N, the number of solutions is zero.

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benorin
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Count the number of integer solutions of (rather, # of integer lattice points such that)

[tex]n+\sum_{k=1}^{n} \left| x_{k}\right| \leq N[/tex]

Not homework, so no rush. I have worked it through before with a prof., but he's so brilliant I didn't understand much of anything he said :redface: . His solution involed a generating function, something like the coefficient of the blah term in the expansion of blahblah.

Note that the LHS is the height of a polynomial with integer coefficients, although it has nothing to do with the solution.
 
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If n > N, the answer is 0. Otherwise, the problem reduces to finding the lattice point solutions of:

[tex]\sum _{k=1} ^n |x_k| \leq N'[/tex]

where N' > 0. Let f(N',n) denote the number of solutions. Then we want to find f(N',n) and this is just:

[tex]f(N',n) = f(N',n-1) + 2\sum _{k=1} ^{N'} f(N'-k,n-1)[/tex]

We get this because [itex]\sum _{k=1} ^n |x_k| = |x_1| + \sum_{k=2} ^n |x_k|[/itex] and when |x1| > 0, then x1 has two possible values whereas when |x1| = 0 then x1 = 0 only. It's a start.
 

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