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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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