Hi Bruce,    I would like the following formula.     After a down trend of at least 4 days  the next day a long black candle appears. The next days open is below the open of the previous day and closes at the previous days cloe.  ( there is an equivalant close both days )  Thank You.

Hi Bruce,  I may not have been quite clear.    We have 4 down days   day 5 is a long black candle and day 6 closes at day 5   close Thanks.

Hi Bruce,    On Thur.afternoon I asked for help regarding a matching low formula. and can not find a reply. I know you are busy @ apologize if I am being impatient. Also, you may have answered  @ I do have difficulties navagating this system. It is good but so much to learn. Thank You.

By down days, do you mean the close is below the open, the close is below the previous close, both, or something else?

What makes the long black candle "long"?

Are you interested in the entire length of the candle, just the body, both, or something else?

Does it need to be longer than any of the previous X candles? Does it need to be at least a certain multiple of the average of the previous X candles? Something else entirely?

The following Condition Formula requires the close to be below both the open and previous close to be considered down and requires the 5th candle to have both a larger body and total range than any of the previous 4 candles.

C = C1 AND C1 < O1 AND C1 < C2 AND C2 < O2 AND C2 < C3  AND C3 < O3 AND C3 < C4  AND C4 < O4 AND C4 < C5  AND C5 < O5 AND C5 < C6 AND O2 - C2 < O1 - C1 AND O3 - C3 < O1 - C1 AND O4 - C4 < O1 - C1 AND O5 - C5 < O1 - C1 AND H2 - L2 < H1 - L1 AND H3 - L3 < H1 - L1 AND H4 - L4 < H1 - L1 AND H5 - L5 < H1 - L1

Hi Bruce, Thanks for the response  I  will try @ be clearer. That is my short fall. The close is below the open @ this occurs for at least 4 days in a row. The day5 opens below the open of day 4 and day 5's close is the same as day 4' s close.   Thank You

The following checks for the close being below the open for at least four days in a row. It then checks for the open of the fifth day to be below the open of the fourth day and the close of the fifth day to be exactly the same as the close of the fourth day.

C = C1 AND O < O1 AND C1 < O1 AND C2 < O2 AND C3 < O3 AND C4 < O4

It does not check for the fourth day being long in any way. Nor does it take into account the idea that the pattern was six bars long instead of five bars long as appeared to be the case in the first description.