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3
Generalizing my comment on Gareth's solution, we can
arrange Pascal's triangle as a right triangular array and ignore the right half ($n < 2k$) to obtain something like this:
1
1
1 2
1 3
1 4 6
...
We then, for any $N$,
0
Gareth has found the optimal solutions, but here is an R script if anyone wants to mess around with the upper bounds for n, just change the value of the variable"UpperBound".
require(pracma)
UpperBound<-500
n<-rep(1:UpperBound,each=UpperBound)
k<-rep(1:UpperBound,times=UpperBound)
data<-as.data.frame(cbind(n,k))
colnames(data)<-c("n","k")
...
5
Choosing
gets to within about
of the desired answer. I think this is best possible with <= 100 cards.
Found with the help of a computer, but purely as an aid to calculation. My approach was to
[EDITED to add:]
Out of curiosity, I also ran a more automated search for the larger bound of n=500 mentioned in the OP. For this,
The automated search also ...
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