Question about turning the output of Reduce into a Range

Questions
1

I’d like to turn the output of this:

f[8, n_] = n (3 n - 2);

Reduce[100000 > f[8, n] > 10000 && n > 0, n, Integers]
out[1]= n \[Element] Integers && 59 <= n <= 182

into the actual list Range[59,182].

How do I do that programatically?

Thanks in advance!

2

May be you can parse it out? Something like

Clear[f, n, x, y];
f[8, n_] = n (3 n - 2);
r = Reduce[100000 > f[8, n] > 10000 && n > 0, n, Integers];
Cases[r, Inequality[x_, __, y_] :> Range[x, y]];

gives

{{59, 60, 61, 62, 63, 64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75,
  76, 77, 78, 79, 80, 81, 82, 83, 84, 85, 86, 87, 88, 89, 90, 91, 92,
  93, 94, 95, 96, 97, 98, 99, 100, 101, 102, 103, 104, 105, 106, 107,
  108, 109, 110, 111, 112, 113, 114, 115, 116, 117, 118, 119, 120,
  121, 122, 123, 124, 125, 126, 127, 128, 129, 130, 131, 132, 133,
  134, 135, 136, 137, 138, 139, 140, 141, 142, 143, 144, 145, 146,
  147, 148, 149, 150, 151, 152, 153, 154, 155, 156, 157, 158, 159,
  160, 161, 162, 163, 164, 165, 166, 167, 168, 169, 170, 171, 172,
  173, 174, 175, 176, 177, 178, 179, 180, 181, 182}}

But have to check to see if this covers other cases. Only tried it on your input. Can’t guarantee it will work for all cases now.

3

Thanks Nasser!

Caveat noted. What if instead I wanted the 5-digit numbers that f[8,n] produces? Is there an easier way than this:

Clear[f, n, x, y];
f[8, n_] = n (3 n - 2);
r = Reduce[100000 > f[8, n] > 10000 && n > 0, n, Integers];
Map[f[8,#]&, Cases[r, Inequality[x_, __, y_] :> Range[x, y]]]

Thanks again!
-Joe

4

Reduce has an internal control for this sort of thing. Could increase the limit to how many get explicitly listed as follows.

SetSystemOptions[
  "ReduceOptions" -> {"DiscreteSolutionBound" -> 10000}];