[seqfan] Re: Intermediate result for ordering problem: 2-D \phi()

[hv at crypt.org]

hv at crypt.org wrote:
:That's very nice. Though it makes the computation less obvious, I'd rather
:express it as:
:
:  mn = sum_{k=1}^\inf { a([m/k], [n/k]) }
:
:.. in which each component of the sum counts the pairs with gcd equal to k
:(and analogously for higher dimension).

I notice also that the above expression uniquely defines a(m, n), which
leads me to wonder what other expressions in m, n on the LHS would define
interesting functions.

Putting m+n on the LHS is marginally interesting because it yields the
surprisingly UNinterestingly regular a(m, n) = c(m > 0) + c(n > 0).

Setting instead the asymmetrical m^n yields something that is at least
harder, though I'm not sure yet if it is more interesting: in this case
a(m, 0) goes (by my hand calculations) like:
  0, 1, 0, -1, -1, -2, -1, -2, -2, -2, -1, -2, -2, -3, -2, -1, -1, -2, -2, -3
.. which appears to be Mertens's function (A002321(n)).

Hugo