You can make it pretty short too, if you allow yourself fix:
rs=1:fix(\f p n->n++f(p++n)p)[1][0]
I came up with this on the train home, but then I realised it was the
same as your solution :)
On 08/11/2007, at 12:57 PM, Alfonso Acosta wrote:
How about this,
infiniteRS :: [Int]
infiniteRS = let acum a1 a2 = a2 ++ acum (a1++a2) a1 in 1 : acum
[1] [0]
it certainly fits in one line but it's not really elegant
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