Does anyone know whether any arithmetical or asymptotic results have been obtained about the AdamsWatters triangle?

$\begingroup$ I upvoted your question to balance out whoever voted it down. Thanks for adding the comment in which you added context and detail to the question. (I had no idea what you were talking about in your initial post.) $\endgroup$ – userN Oct 25 '09 at 0:14
There seems to be no response to this, but perhaps somebody knows something about it in another terminology.
Franklin T. AdamsWatters defined a triangle similar to Pascal's, but where the latter has c = a + b between and under a and b, AdamsWatters takes c = (a+b)/gcd(a,b). The first few rows look like this:
1
1 1
1 2 1
1 3 3 1
1 4 2 4 1
1 5 3 3 5 1
1 6 8 2 8 6 1
1 7 7 5 5 7 7 1
1 8 2 12 2 12 2 8 1
AdamsWatters summed the rows and obtained a sequence a(n) which is numbered A125606 in the OEIS database. He conjectures that log(a(n))/n tends to log(2*zeta(3)/zeta(2)). That was what I meant when I asked about asymptotic information. (The analogous limit is log(2) for Pascal's triangle, of course)
I am actually more interested in whether the AdamsWatters triangle has any nontrivial arithmetic structure. When you add a and b, the sum is trivially divisible by gcd(a,b), so the operation (a+b)/gcd(a,b) removes this trivial information. Pascal's triangle has a lot of arithmetic structure, but I did not see any when I factored the entries in the AW triangle out to the fiftieth row. Perhaps dividing by the gcd leaves nothing, or maybe I didn't think about it the right way.
Two caveats: The little piece above is not enough to see what happens. For example, the twelfth row is the first that has both odd and even entries (apart from the 1s). Also there seems no reason to think that the AW triangle has any combinatorial significance.

$\begingroup$ This is really obscure. There are no Google scholar hits and this post is the first Google hit. $\endgroup$ – Qiaochu Yuan Oct 25 '09 at 0:12

$\begingroup$ I'm assuming you mean log(a(n))/n tends to log(2*zeta(3)/zeta(2)), or a(n)^(1/n) tends to 2*zeta(3)/zeta(2). In any case, this doesn't seem to be all that reasonable; looking at the sums of the first 1000 rows, a(n)^(1/n) seems to fluctuate around 1.55 or so, while 2*zeta(3)/zeta(2) = 1.46. Barring some theoretical reason to expect an expression having this general form, I have trouble believing this conjecture. $\endgroup$ – Michael Lugo Oct 25 '09 at 0:37

$\begingroup$ Thanks for spotting the missing log. I never tried to check the asymptotics myself; I just factorized entries, looking for patterns. I agree that this may seem a bit obscure, but I suspect that the operation (a+b)/gcd(a,b) has arithmetic significance, because of the dividing away the trivial information. In fact I began from this quotient and found what AdamsWatters had done by searching OEIS with likely sequences. $\endgroup$ – engelbrekt Oct 25 '09 at 1:05

$\begingroup$ If we call the table entries $A(n,k)$, with the same numbering conventions as for Pascal's triangle, then the sequence $A(n,2)$ (which goes $1,3,2,3,8,7,2,\dots$) is A131134 in the Online Encyclopedia, research.att.com/~njas/sequences/A131134, but there's no significant information there other than the first 69 terms. Still, that list of terms raises some questions. $A(28,2)=88$, but $A(30,2)=2$; $A(36,2)=108$, but $A(37,2)=4$. $A(n,2)$ is unbounded above since if $p$ is prime then $A(p,2)\ge p$ or $A(p+1,2)>p$. But does $A(n,2)$ go to infinity? $\endgroup$ – Gerry Myerson Mar 19 '10 at 3:44