# Function that displays set of integers fulfilling constrictions

I would like to make a function of three parameters $i$, $j$ and $k$ that displays a list of integers fulfilling some criteria. Specifically the set of integers is:

$\zeta_{ijk} = \Big\{p\in\{2j+2,2j+3,\ldots,2k+2\}\ |\$ $p=2i+1,\ \text{or}\ p\leq i+j,\ \text{or}\ p\geq i+k+3 \Big\}$

for $k\in\{0,1,\ldots\}$, $\ j \in \{0,1,\ldots,k\}$ and $i\in\{j,j+1,\ldots,k+1\}$.

where $|$ denotes that constrictions on $p$ follows. To clarify, some examples are:

$\zeta_{000}=\zeta_{100}=\emptyset$

$\zeta_{001}=\{4\}$, $\ \zeta_{101}=\{3\}$, $\ \zeta_{201}=\{2\}$

$\zeta_{111}=\zeta_{211}=\emptyset$

$\zeta_{002}=\{5,6\}$, $\ \zeta_{102}=\{3,6\}$, $\ \zeta_{202}=\{2,5\}$, $\ \zeta_{302}=\{2,3\}$

And so on. I have problems implementing the logical constrictions in e.g. the Table-function, inside my user-defined function. Any ideas?

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## 1 Answer

 ζ[i_, j_, k_] := Select[ Table[p, {p, 2 j + 2, 2 k + 2}] , (# == 2 i + 1 || # <= i + j || # >= i + k + 3) & ] 

will produce your function. Also for more control, conditions must be added to be complete concerning the domain.

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Thanks a lot! I'm not sure what your last comment means but I think what you wrote is perfect. – jorgen Jun 11 '13 at 16:47
Your domain is : k∈{0,1,…}, j∈{0,1,…,k} and i∈{j,j+1,…,k+1} so you could add in the conditions : (# == 2 i + 1 || # <= i + j || # >= i + k + 3) && (k>=0 && j>=0 && j<=k && i>=j && i<=k+1)& . But if you control correctly i,j,k it is not a problem. – tchronis Jun 11 '13 at 16:52
Ok! I see what you mean. – jorgen Jun 11 '13 at 16:58