Find "n" consecutive free numbers from table

This is a gaps-and-islands problem. Assuming there are no gaps or duplicates in the same id_set set:

WITH partitioned AS (
  SELECT
    *,
    number - ROW_NUMBER() OVER (PARTITION BY id_set) AS grp
  FROM atable
  WHERE status = 'FREE'
),
counted AS (
  SELECT
    *,
    COUNT(*) OVER (PARTITION BY id_set, grp) AS cnt
  FROM partitioned
)
SELECT
  id_set,
  number
FROM counted
WHERE cnt >= 3
;

Here's a SQL Fiddle demo^* link for this query: http://sqlfiddle.com/#!1/a2633/1.

UPDATE

To return only one set, you could add in one more round of ranking:

WITH partitioned AS (
  SELECT
    *,
    number - ROW_NUMBER() OVER (PARTITION BY id_set) AS grp
  FROM atable
  WHERE status = 'FREE'
),
counted AS (
  SELECT
    *,
    COUNT(*) OVER (PARTITION BY id_set, grp) AS cnt
  FROM partitioned
),
ranked AS (
  SELECT
    *,
    RANK() OVER (ORDER BY id_set, grp) AS rnk
  FROM counted
  WHERE cnt >= 3
)
SELECT
  id_set,
  number
FROM ranked
WHERE rnk = 1
;

Here's a demo for this one too: http://sqlfiddle.com/#!1/a2633/2.

If you ever need to make it one set per id_set, change the RANK() call like this:

RANK() OVER (PARTITION BY id_set ORDER BY grp) AS rnk

Additionally, you could make the query return the smallest matching set (i.e. first try to return the first set of exactly three consecutive numbers if it exists, otherwise four, five etc.), like this:

RANK() OVER (ORDER BY cnt, id_set, grp) AS rnk

or like this (one per id_set):

RANK() OVER (PARTITION BY id_set ORDER BY cnt, grp) AS rnk

---

_{* The SQL Fiddle demos linked in this answer use the 9.1.8 instance as the 9.2.1 one doesn't appear to be working at the moment.}

A simple and fast variant:

SELECT min(number) AS first_number, count(*) AS ct_free
FROM (
    SELECT *, number - row_number() OVER (PARTITION BY id_set ORDER BY number) AS grp
    FROM   tbl
    WHERE  status = 'FREE'
    ) x
GROUP  BY grp
HAVING count(*) >= 3  -- minimum length of sequence only goes here
ORDER  BY grp
LIMIT  1;

• Requires a gapless sequence of numbers in number (as provided in the question).

• Works for any number of possible values in status besides 'FREE', even with NULL.

• The major feature is to subtract row_number() from number after eliminating non-qualifying rows. Consecutive numbers end up in the same grp - and grp is also guaranteed to be in ascending order.

• Then you can GROUP BY grp and count the members. Since you seem to want the first occurrence, ORDER BY grp LIMIT 1 and you get starting position and length of the sequence (can be >= n).

Set of rows

To get an actual set of numbers, don't look up the table another time. Much cheaper with generate_series():

SELECT generate_series(first_number, first_number + ct_free - 1)
    -- generate_series(first_number, first_number + 3 - 1) -- only 3
FROM  (
   SELECT min(number) AS first_number, count(*) AS ct_free
   FROM  (
      SELECT *, number - row_number() OVER (PARTITION BY id_set ORDER BY number) AS grp
      FROM   tbl
      WHERE  status = 'FREE'
      ) x
   GROUP  BY grp
   HAVING count(*) >= 3
   ORDER  BY grp
   LIMIT  1
   ) y;

If you actually want a string with leading zeros like you display in your example values, use to_char() with the FM (fill mode) modifier:

SELECT to_char(generate_series(8, 11), 'FM000000')

SQL Fiddle with extended test case and both queries.

Closely related answer:

• Select longest continuous sequence

This is a fairly generic way to do this.

Bear in mind it depends on your number column being consecutive. If it's not a Window function and/or CTE type-solution will probably be needed:

SELECT 
    number
FROM
    mytable m
CROSS JOIN
   (SELECT 3 AS consec) x
WHERE 
    EXISTS
       (SELECT 1 
        FROM mytable
        WHERE number = m.number - x.consec + 1
        AND status = 'FREE')
    AND NOT EXISTS
       (SELECT 1 
        FROM mytable
        WHERE number BETWEEN m.number - x.consec + 1 AND m.number
        AND status = 'ASSIGNED')