What I find easiest to remember is the generating function for binary trees. A tree is either a node with a pair of trees as children, or a leaf. So T(x) = T(x)^2 + x or T(x)^2 - T(x) + x = 0 T(x) = [1 +/- sqrt(1 - 4x)]/2 If we plug in 1/1000, we get [1+/-sqrt(1-4/1000)]/2 = 0.001001002005014 On Thu, Feb 2, 2012 at 11:46 PM, Robert Munafo <mrob27@gmail.com> wrote:
Yesterday I found the far easier-to-remember:
500 - sqrt(499*501)
which also has secondary advantages of giving you 6 digits for each Catalan number, and being far easier to expand to an 8-digit or longer version if you want.
- Robert
mrob.com/pub/math/seq-digits.html#irrational
On Sun, Jan 29, 2012 at 10:39, Warren Smith <warren.wds@gmail.com> wrote:
Jorg Arndt's example generating Pascal's triangle and a Fibonacci-like sequence was considerably cooler... another I invented quite a while back was
1000 / (500 + sqrt(249990)) = 1.00001000020000500014000420013200429014300...
where the sequence 1,2,5,14,42,... is the Catalan numbers (which easily give you the central binomial coefficients).
How did I do that? Well, again, one way is using generating functions.
But the continued fraction of quadratic irrationals is ultimately periodic and once you know its period you can easily solve for the surd, so again, this is easy to construct from the desired sequence without any intelligence.
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