Jon E. Schoenfield: I included one more term in the Data because there was room for it within the traditional 260-character guideline.  I figured I needed to note that in the Extensions, although it'd be fine with me to omit that.

Jon E. Schoenfield: @Editors -- Is the Prog section a good place for this program?  I have a version that's formatted with line breaks, but I think it's probably too many lines for the Prog section.  I could upload a text file containing the formatted version (it has just a few comments) to the Links section, if that seems better.

Jon E. Schoenfield: I hope to upload the program tonight. I'd welcome any help with the wording I've just added!

Joerg Arndt: Yes, better. Can you write a program (if you care)?  I find it striking how Collatz-related sequences tend to be in a terrible shape and useless on top of that.

Jon E. Schoenfield: Okay. I realized there's a much more efficient way to get these terms than the brute-force approach I was using. I hope to get a program uploaded tomorrow.

Jon E. Schoenfield: I figured out what was meant by the notation "a(n-2):(x is 1 mod 6)". I have a clumsily written Magma program that tests every vector c (for m up to 13), stores the resulting values of x that are integers, and reports how many there are, along with how many are == 1(mod 6) and how many are == 5 (mod 6).  If started at n=1, it runs out of time after reporting its results for n=26. As it turns out, just going up through m=13 is apparently sufficient to fine all the resulting integers x for all n <= 26; at least, it gets the same counts of integer x values as listed in the Data for all n <= 26.

Jon E. Schoenfield: ("fine" -> "find")

Jon E. Schoenfield: Since the first term of the sequence is a(1), and the formula for a(n) makes use of a(n-1) and a(n-2), it obviously can't apply to n=1 or n=2. It works for n=3 and all n in 5..26 (i.e., as far as I've checked). (And although my program times out before it can report any results for n=27, its outputs for n=25 and n=26, including the counts of x values congruent to 1 and 5 (mod 6), when plugged into the formula, give a(27)=403, which also agrees with the Data.)

Jon E. Schoenfield: The only value of n <= 27 at which the formula fails is n=4: at n=2, the a(2)=1 integer value of x is 1 (which is congruent to 1 (mod 6)), and at n=3, there are a(3)=0 integer values, so at n=4, "a(n-2):(x is 1 mod 6)" evaluates to 1, "a(n-1):(x is 5 mod 6)" evaluates to 0, and the formula would give a(4) = 1 + 1 + 0 = 2, but a(4) = 1 -- hence the need for the qualifier, "For n > 4" ....
a(2) = 1   #1mod6 = 1   #5mod6 = 0
[ 1 ]

a(3) = 0   #1mod6 = 0   #5mod6 = 0
[]

a(4) = 1   #1mod6 = 0   #5mod6 = 1

Jon E. Schoenfield: Arrgh!  Copy/paste error.  I'm sorry, let me try that again.  :-(

Jon E. Schoenfield: The only value of n <= 27 at which the formula fails is n=4: at n=2, the a(2)=1 integer value of x is 1 (which is congruent to 1 (mod 6)), and at n=3, there are a(3)=0 integer values, so at n=4, "a(n-2):(x is 1 mod 6)" evaluates to 1, "a(n-1):(x is 5 mod 6)" evaluates to 0, and the formula would give a(4) = 1 + 1 + 0 = 2, but a(4) = 1 -- hence the need for the qualifier, "For n > 4" ....

Jon E. Schoenfield: It may be that the anomaly at n=4 is because the one integer x value at n=2, i.e., x=1, is somehow a special case....

Jon E. Schoenfield: ... And I suppose that maybe this is somehow related to Wolfdieter's comment.  I.e., this sequence is the specified sequence of first differences, except that a(2) is too large by 1, and the given formula for a(n) appears to apply for all n > 3 except n=4, where "a(n) = a(n-2) + a(n-2):(x is 1 mod 6) + a(n-1):(x is 5 mod 6)" gives a(4) = a(2) + a(2):(x is 1 mod 6) + a(1):(x is 5 mod 6), which makes use of a(2) and yields a value that's too large by 1.

Jon E. Schoenfield: Does what I've added to the Comments entry look like a good way to clarify the formula? (Improvements welcome, of course!)

Jon E. Schoenfield: If asked, I'd be glad to add a short table to the Example section, e.g., with columns n, a(n), number of values in a(n) that are == 1 (mod 6), and number that are == 5 (mod 6).  Could also include a list of those x values. (For each n <= 8, there's at most 1.) But if no table seems necessary, that's fine, too.