arithmoi-0.11.0.0: Efficient basic number-theoretic functions.

Math.NumberTheory.Recurrences.Bilinear

Contents

Description

Bilinear recurrent sequences and Bernoulli numbers, roughly covering Ch. 5-6 of Concrete Mathematics by R. L. Graham, D. E. Knuth and O. Patashnik.

Note on memory leaks and memoization. Top-level definitions in this module are polymorphic, so the results of computations are not retained in memory. Make them monomorphic to take advantages of memoization. Compare

>>> :set +s
>>> binomial !! 1000 !! 1000 :: Integer
1
(0.01 secs, 1,385,512 bytes)
>>> binomial !! 1000 !! 1000 :: Integer
1
(0.01 secs, 1,381,616 bytes)


against

>>> let binomial' = binomial :: [[Integer]]
>>> binomial' !! 1000 !! 1000 :: Integer
1
(0.01 secs, 1,381,696 bytes)
>>> binomial' !! 1000 !! 1000 :: Integer
1
(0.01 secs, 391,152 bytes)

Synopsis

# Pascal triangle

binomial :: Semiring a => [[a]] Source #

Infinite zero-based table of binomial coefficients (also known as Pascal triangle).

binomial !! n !! k == n! / k! / (n - k)!

Note that binomial !! n !! k is asymptotically slower than binomialLine n !! k, but imposes only Semiring constraint.

>>> take 6 binomial :: [[Int]]
[[1],[1,1],[1,2,1],[1,3,3,1],[1,4,6,4,1],[1,5,10,10,5,1]]


binomialRotated :: Semiring a => [[a]] Source #

Pascal triangle, rotated by 45 degrees.

binomialRotated !! n !! k == (n + k)! / n! / k! == binomial !! (n + k) !! k

Note that binomialRotated !! n !! k is asymptotically slower than binomialDiagonal n !! k, but imposes only Semiring constraint.

>>> take 6 (map (take 6) binomialRotated) :: [[Int]]
[[1,1,1,1,1,1],[1,2,3,4,5,6],[1,3,6,10,15,21],[1,4,10,20,35,56],[1,5,15,35,70,126],[1,6,21,56,126,252]]


binomialLine :: (Enum a, GcdDomain a) => a -> [a] Source #

The n-th (zero-based) line of binomial (and the n-th diagonal of binomialRotated).

>>> binomialLine 5
[1,5,10,10,5,1]


binomialDiagonal :: (Enum a, GcdDomain a) => a -> [a] Source #

The n-th (zero-based) diagonal of binomial (and the n-th line of binomialRotated).

>>> take 6 (binomialDiagonal 5)
[1,6,21,56,126,252]


binomialFactors :: Word -> Word -> [(Prime Word, Word)] Source #

Prime factors of a binomial coefficient.

binomialFactors n k == factorise (binomial !! n !! k)
>>> binomialFactors 10 4
[(Prime 2,1),(Prime 3,1),(Prime 5,1),(Prime 7,1)]


# Other recurrences

stirling1 :: (Num a, Enum a) => [[a]] Source #

Infinite zero-based table of Stirling numbers of the first kind.

>>> take 5 (map (take 5) stirling1)
[[1],[0,1],[0,1,1],[0,2,3,1],[0,6,11,6,1]]


Complexity: stirling1 !! n !! k is O(n ln n) bits long, its computation takes O(k n^2 ln n) time and forces thunks stirling1 !! i !! j for 0 <= i <= n and max(0, k - n + i) <= j <= k.

One could also consider unsignedStirling1st to compute stand-alone values.

stirling2 :: (Num a, Enum a) => [[a]] Source #

Infinite zero-based table of Stirling numbers of the second kind.

>>> take 5 (map (take 5) stirling2)
[[1],[0,1],[0,1,1],[0,1,3,1],[0,1,7,6,1]]


Complexity: stirling2 !! n !! k is O(n ln n) bits long, its computation takes O(k n^2 ln n) time and forces thunks stirling2 !! i !! j for 0 <= i <= n and max(0, k - n + i) <= j <= k.

One could also consider stirling2nd to compute stand-alone values.

lah :: Integral a => [[a]] Source #

Infinite one-based table of Lah numbers. lah !! n !! k equals to lah(n + 1, k + 1).

>>> take 5 (map (take 5) lah)
[[1],[2,1],[6,6,1],[24,36,12,1],[120,240,120,20,1]]


Complexity: lah !! n !! k is O(n ln n) bits long, its computation takes O(k n ln n) time and forces thunks lah !! n !! i for 0 <= i <= k.

eulerian1 :: (Num a, Enum a) => [[a]] Source #

Infinite zero-based table of Eulerian numbers of the first kind.

>>> take 5 (map (take 5) eulerian1)
[[],[1],[1,1],[1,4,1],[1,11,11,1]]


Complexity: eulerian1 !! n !! k is O(n ln n) bits long, its computation takes O(k n^2 ln n) time and forces thunks eulerian1 !! i !! j for 0 <= i <= n and max(0, k - n + i) <= j <= k.

eulerian2 :: (Num a, Enum a) => [[a]] Source #

Infinite zero-based table of Eulerian numbers of the second kind.

>>> take 5 (map (take 5) eulerian2)
[[],[1],[1,2],[1,8,6],[1,22,58,24]]


Complexity: eulerian2 !! n !! k is O(n ln n) bits long, its computation takes O(k n^2 ln n) time and forces thunks eulerian2 !! i !! j for 0 <= i <= n and max(0, k - n + i) <= j <= k.

bernoulli :: Integral a => [Ratio a] Source #

Infinite zero-based sequence of Bernoulli numbers, computed via connection with stirling2.

>>> take 5 bernoulli
[1 % 1,(-1) % 2,1 % 6,0 % 1,(-1) % 30]


Complexity: bernoulli !! n is O(n ln n) bits long, its computation takes O(n^3 ln n) time and forces thunks stirling2 !! i !! j for 0 <= i <= n and 0 <= j <= i.

One could also consider bernoulli to compute stand-alone values.

euler :: forall a. Integral a => [a] Source #

The same sequence as euler', but with type [a] instead of [Ratio a] as the denominators in euler' are always 1.

>>> take 10 euler :: [Integer]
[1,0,-1,0,5,0,-61,0,1385,0]


eulerPolyAt1 :: forall a. Integral a => [Ratio a] Source #

Infinite zero-based list of the n-th order Euler polynomials evaluated at 1. The algorithm used was derived from Algorithms for Bernoulli numbers and Euler numbers by Kwang-Wu Chen, third formula of the Corollary in page 7. Element-by-element division of sequences A1986631 and A006519 in OEIS.

>>> take 10 eulerPolyAt1 :: [Rational]
[1 % 1,1 % 2,0 % 1,(-1) % 4,0 % 1,1 % 2,0 % 1,(-17) % 8,0 % 1,31 % 2]


faulhaberPoly :: (GcdDomain a, Integral a) => Int -> [Ratio a] Source #

>>> sum (map (^ 10) [0..100])
959924142434241924250
>>> sum \$ zipWith (*) (faulhaberPoly 10) (iterate (* 100) 1)
959924142434241924250 % 1