Help needed with formula to consolidate data for mailmerge email list

Hi guys, I could really use some help!

Owing to the need to send customers codes via email, and email software's restrictions to one email per recipient in a mailmerge, I am trying to create a formula which will, for any customer who needs to receive more than one code, put all instances of the code into the same cell. For any customer who only needs one code, it just outputs that code.

I will then have a separate formula which identifies which rows are redundant, so I can filter them out and end up with a list with just one row per customer, with all their codes together in the same cell.

I have run into some problems with the initial formula, and am frankly stuck. It seems to be almost working, but is producing some odd errors. I've attached a sample file with some mock email addresses and codes in columns A and B respectively, and the formula and its output in column C.

Oddities:

• The first customer in the sample should be getting 20 codes (from CODE A-1 to CODE A-20). Instead they are also getting a code belonging to the next customer (CODE B-1). This same error occurs for the 2nd customer, who gets CODE C-1 from the list meant to go to the 3rd customer, and so on. I've no doubt this is caused by something stupidly simple, but I just can't see where in the formula this is happening! Maybe I've been staring at this formula too long!
• Why ON EARTH has the formula in cell C3 returned "FALSE"? Why doesn't this happen anywhere else?

I would be really appreciative if someone with more experienced eyes could take a look and see if they can figure out where I've gone wrong.

And here is the monstrous formula meant to accomplish the task:

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Thank you so much.

Hi mrmmickle1!

Thanks so much for your help.

I did have some trouble with this formula as written but after some playing around I managed to come up with something based on it which accomplishes what I need.

You've saved my bacon! I've given you some well deserved reputation.

Thanks again.