Números de Dudeney

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1. Haskell

   import Test.QuickCheck numerosDudeney :: Integer -> [Integer] numerosDudeney k = filter (esDudeney k) xs where xs = [x^k | x <- [1..]] esDudeney :: Integer -> Integer -> Bool esDudeney k x = x == (sum (digitos x))^k digitos :: Integer -> [Integer] digitos n = [read [c] | c <- show n] prop_dudeney :: (Positive Integer) -> Bool prop_dudeney (Positive n) = (not.esDudeney 3) (n+19683)

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   import Test.QuickCheck   numerosDudeney :: Integer -> [Integer] numerosDudeney k = filter (esDudeney k) xs   where xs = [x^k | x <- [1..]]   esDudeney :: Integer -> Integer -> Bool esDudeney k x = x == (sum (digitos x))^k   digitos :: Integer -> [Integer] digitos n = [read [c] | c <- show n]   prop_dudeney :: (Positive Integer) -> Bool prop_dudeney (Positive n) = (not.esDudeney 3) (n+19683)

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2. Haskell

   import Test.QuickCheck numerosDudeney :: Integer -> [Integer] numerosDudeney k = [m | n <- [1..], let m = n^k, let x = sumaDigitos m, x^k == m] prop_maximo_Dudeney3 :: (Positive Integer) -> Bool prop_maximo_Dudeney3 (Positive n) = x^3 /= m where m = 19683 + n x = sumaDigitos m sumaDigitos :: Integer -> Integer sumaDigitos = foldr (\ x y -> read [x] + y) 0 . show

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   import Test.QuickCheck   numerosDudeney :: Integer -> [Integer] numerosDudeney k = [m | n <- [1..], let m = n^k,                         let x = sumaDigitos m, x^k == m]   prop_maximo_Dudeney3 :: (Positive Integer) -> Bool prop_maximo_Dudeney3 (Positive n) = x^3 /= m   where m = 19683 + n         x = sumaDigitos m   sumaDigitos :: Integer -> Integer sumaDigitos = foldr (\ x y -> read [x] + y) 0 . show

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3. Haskell

   {- Existe un i tal que, para todo s >= i es Sum dígitos s^k < s el nº máximo de dígitos de s^k es s^k ~ 10^d d = k log10 s cuyo valor máximo es cuando todos son 9, por lo que se alcanza la cota máxima cuando 9 k log10 s = s representada en la función `sOverflow` -} sOverflow k s = let k' = fromIntegral k s' = fromIntegral s in 9 * k' * logBase 10 s' < s' -- ahora podemos calcular todos los números sin -- recurrir a QuickCheck dun k = 1: [n | s <-takeWhile (not . sOverflow) [2..] , let n = s^k , s == (sum . map (toInteger . digitToInt) . show) n] {- Ejemplos: > dun 3 [1,512,4913,5832,17576,19683] > dun 4 [1,2401,234256,390625,614656,1679616] > mapM_ print $ dun 20 1 1215766545905692880100000000000000000000 1424201691977055041360709423546231879609039601 20864448472975628947226005981267194447042584001 > mapM_ print $ dun 120 1 208305631093147769593077124832920398396424242736587139105600423547681633381448345957997431002341051384681406440179824973860848822669929122863069080109517208593661481936595852227827401123071804455068693490439387062950631497740127012356534717939472405042404759723585669336105702330636847342292588431606104304105400537826987381228458150816889796114773161175996672040171148993202310379336001 5189753014425057489701348386485062139584865957477360434844367399843084988969366791646004249212100600480717775212590831764015670415686563804660475312078146716216806311907369813125444042567504621328266159703394297053600592102596692073626252142818051725113774238337691235318550041720433923148395754801224168556458842399838213009699562748685654937790642215723109778399221560208307212649496576 -- el último número de dudeney para k=500 tiene 1967 dígitos > length $ show $ last $ dun 500 1967 (7.54 secs, 10,758,999,688 bytes) -}

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   {-     Existe un i tal que, para todo s >= i es       Sum dígitos s^k < s     el nº máximo de dígitos de s^k es       s^k ~ 10^d       d = k log10 s     cuyo valor máximo es cuando todos son 9, por   lo que se alcanza la cota máxima cuando       9 k log10 s = s     representada en la función `sOverflow`   -}   sOverflow k s = let k' = fromIntegral k                     s' = fromIntegral s                 in  9 * k' * logBase 10 s' < s'   -- ahora podemos calcular todos los números sin -- recurrir a QuickCheck dun k = 1: [n | s <-takeWhile (not . sOverflow) [2..]               , let n = s^k               , s == (sum . map (toInteger . digitToInt) . show) n]   {- Ejemplos:   > dun 3 [1,512,4913,5832,17576,19683]   > dun 4 [1,2401,234256,390625,614656,1679616]   > mapM_ print $ dun 20 1 1215766545905692880100000000000000000000 1424201691977055041360709423546231879609039601 20864448472975628947226005981267194447042584001   > mapM_ print $ dun 120 1 208305631093147769593077124832920398396424242736587139105600423547681633381448345957997431002341051384681406440179824973860848822669929122863069080109517208593661481936595852227827401123071804455068693490439387062950631497740127012356534717939472405042404759723585669336105702330636847342292588431606104304105400537826987381228458150816889796114773161175996672040171148993202310379336001 5189753014425057489701348386485062139584865957477360434844367399843084988969366791646004249212100600480717775212590831764015670415686563804660475312078146716216806311907369813125444042567504621328266159703394297053600592102596692073626252142818051725113774238337691235318550041720433923148395754801224168556458842399838213009699562748685654937790642215723109778399221560208307212649496576   -- el último número de dudeney para k=500 tiene 1967 dígitos > length $ show $ last $ dun 500 1967 (7.54 secs, 10,758,999,688 bytes)     -}

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4. Haskell

   import Test.QuickCheck numerosDudeney :: Integer -> [Integer] numerosDudeney k = [a | a <- [1..] , esDudeney a k ] esDudeney :: Integer -> Integer -> Bool esDudeney n k = sum (digitos n) ^ k == n digitos :: Integer -> [Integer] digitos n | n < 10 = [n] | otherwise = n `mod` 10 : digitos (n `div` 10) prop_Dudeney3 :: Integer -> Property prop_Dudeney3 n = esDudeney n 3 ==> n <= 19683 -- *Main> quickCheck prop_Dudeney3 -- *** Gave up! Passed only 60 tests.

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   import Test.QuickCheck   numerosDudeney :: Integer -> [Integer] numerosDudeney k = [a | a <- [1..]                       , esDudeney a k ]   esDudeney :: Integer -> Integer -> Bool esDudeney n k = sum (digitos n) ^ k == n   digitos :: Integer -> [Integer] digitos n | n < 10    = [n]           | otherwise = n `mod` 10 : digitos (n `div` 10)   prop_Dudeney3 :: Integer -> Property prop_Dudeney3 n = esDudeney n 3 ==> n <= 19683   -- *Main> quickCheck prop_Dudeney3 -- *** Gave up! Passed only 60 tests.

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   1. Haskell

      prop_Dudeney :: Integer-> Property prop_Dudeney n = n > 1 ==> not (esDudeney (19683*n) 3) -- *Main> quickCheck prop_Dudeney -- +++ OK, passed 100 tests.

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      prop_Dudeney :: Integer-> Property prop_Dudeney n = n > 1 ==> not (esDudeney (19683*n) 3)   -- *Main> quickCheck prop_Dudeney -- +++ OK, passed 100 tests.

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