SQL Server 2019 (Linux)

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select version();
Msg 195 Level 15 State 10 Line 1
'version' is not a recognized built-in function name.

--
--
--  Method 3 - Pure SQL
--
--
--
-- Setup of parameters...
--

CREATE TABLE param(n INT, m INT);
INSERT INTO param VALUES (3, 5);
SELECT * FROM param;


n
m


3
5


WITH cte (i, cnt) AS
(
  SELECT 1 AS i, 1 AS cnt
  UNION ALL
  SELECT 
    CASE
      WHEN cnt < (SELECT n FROM param) THEN (i + 1)
      WHEN cnt = (SELECT n FROM param) THEN cnt
      ELSE i - 1
    END AS i,
    cnt + 1
  FROM cte
  WHERE cnt < 2 * (SELECT n FROM param)
)
SELECT * FROM cte;


i
cnt


1
1

2
2

3
3

3
4

2
5

1
6


WITH cte (i, cnt) AS
(
  SELECT 1 AS i, 1 AS cnt
  UNION ALL
  SELECT 
    CASE
      WHEN cnt < (SELECT n FROM param) THEN (i + 1)
      WHEN cnt = (SELECT n FROM param) THEN cnt
      ELSE i - 1
    END AS i,
    cnt + 1
  FROM cte
  WHERE cnt < 2 * (SELECT n FROM param)
),
seq (n) AS
(
  SELECT 1
  UNION ALL
  SELECT n + 1 FROM seq WHERE n < (SELECT m FROM param)
)
SELECT i
FROM
(
  SELECT 
    s.n, c.cnt, c.i 
  FROM cte c
  CROSS JOIN
  (SELECT * FROM seq) AS s
) AS tab
ORDER BY n, cnt;


i


1

2

3

3

2

1

1

2

3

3

2

1

1

2

3

3

2

1

1

2

3

3

2

1

1

2

3

3

2

1