Sorry for following up to myself again...
Ben Bacarisse <ben.use...@bsb.me.uk> writes:
>> ... But I think that is an easier way (no code yet though!) unless
>> you are set on one particular enumeration: consider the triple as a pair
>> one element of which runs over the enumeration of pairs you already
This algorithm has the disadvantage that the sequence is not any of the
usual "zig-zag" patterns. Using numbers, for example the sums of the
triples are not monotonic:
*Main> take 30 $ map sum (pn 2 [1..])
That probably makes it a non-starter for you.