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edited Apr 13, 2017 at 12:19

Haskell (too many bytes)

Using only operations on lists and Raymond Manzoni's result result:

c n = last $ map numerator $ zipWith (*) (scanl (*) (1) [2,3..]) (intersperse 0 $ foldr (.) id (replicate n (\xs->(xs ++ [(1%(1+2*length xs)) * (sum (zipWith (*) xs (reverse xs)))]))) [1])

Unfortunately, this overflows for modest values of n, as it uses Int values. I will try to fix the problem using Integervalues. Until then, suggestions are welcome.

Haskell (too many bytes)

Using only operations on lists and Raymond Manzoni's result:

c n = last $ map numerator $ zipWith (*) (scanl (*) (1) [2,3..]) (intersperse 0 $ foldr (.) id (replicate n (\xs->(xs ++ [(1%(1+2*length xs)) * (sum (zipWith (*) xs (reverse xs)))]))) [1])

Unfortunately, this overflows for modest values of n, as it uses Int values. I will try to fix the problem using Integervalues. Until then, suggestions are welcome.

Haskell (too many bytes)

Using only operations on lists and Raymond Manzoni's result:

c n = last $ map numerator $ zipWith (*) (scanl (*) (1) [2,3..]) (intersperse 0 $ foldr (.) id (replicate n (\xs->(xs ++ [(1%(1+2*length xs)) * (sum (zipWith (*) xs (reverse xs)))]))) [1])

Unfortunately, this overflows for modest values of n, as it uses Int values. I will try to fix the problem using Integervalues. Until then, suggestions are welcome.

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created Jun 16, 2016 at 1:58

Rodrigo de Azevedo

Haskell (too many bytes)

Using only operations on lists and Raymond Manzoni's result:

c n = last $ map numerator $ zipWith (*) (scanl (*) (1) [2,3..]) (intersperse 0 $ foldr (.) id (replicate n (\xs->(xs ++ [(1%(1+2*length xs)) * (sum (zipWith (*) xs (reverse xs)))]))) [1])

Unfortunately, this overflows for modest values of n, as it uses Int values. I will try to fix the problem using Integervalues. Until then, suggestions are welcome.